Cyclically generated finite groups - Centre for Interdisciplinary ...

Cyclically generated finite groupsElizabeth KimberPh.D. ThesisUniversity of St Andrews2006

ContentsDeclarationAcknowledgementsAbstractviiixxi1 Introduction 11.1 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Definitions and some preliminary ideas . . . . . . . . . . . . . 41.2.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . 41.2.2 Constructing automorphisms . . . . . . . . . . . . . . . 51.3 Summary of research . . . . . . . . . . . . . . . . . . . . . . . 61.4 Summary of Thesis . . . . . . . . . . . . . . . . . . . . . . . . 92 Some cyclically generated groups 112.1 Some cyclically generated groups . . . . . . . . . . . . . . . . 132.1.1 Finite abelian groups . . . . . . . . . . . . . . . . . . . 132.1.2 Nonabelian finite simple groups . . . . . . . . . . . . . 152.1.3 Some more examples of nonabelian cyclically generatedgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.2 Computational techniques . . . . . . . . . . . . . . . . . . . . 232.2.1 Conjugacy classes . . . . . . . . . . . . . . . . . . . . . 242.3 Groups that are not cyclically generated . . . . . . . . . . . . 272.3.1 ECS groups . . . . . . . . . . . . . . . . . . . . . . . . 282.3.2 Using characteristic subgroups . . . . . . . . . . . . . . 292.4 Direct products . . . . . . . . . . . . . . . . . . . . . . . . . . 32iii

iv3 Direct products 373.0.1 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . 383.1 Survey results . . . . . . . . . . . . . . . . . . . . . . . . . . . 383.1.1 Proofs . . . . . . . . . . . . . . . . . . . . . . . . . . . 403.2 Further direct product results . . . . . . . . . . . . . . . . . . 624 Minimal cyclic generating sets for abelian groups 794.0.1 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . 804.0.2 Abért’s Theorem . . . . . . . . . . . . . . . . . . . . . 814.1 Preliminary results . . . . . . . . . . . . . . . . . . . . . . . . 814.1.1 Group automorphisms as matrices . . . . . . . . . . . . 895 Rank 3 and rank 4 abelian groups with minimal cyclic generatingsets 955.1 Rank 3 abelian groups with minimal cyclic generating sets . . 955.1.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . 1005.2 Rank 4 abelian groups with minimal cyclic generating sets . . 1015.2.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . 1136 Rank 5 abelian groups 1156.0.2 Why is the rank 5 case different from ranks 3 and 4? . 1166.0.3 A note on computational results . . . . . . . . . . . . . 1176.1 Preliminary ideas . . . . . . . . . . . . . . . . . . . . . . . . . 1186.2 Rank 5 result . . . . . . . . . . . . . . . . . . . . . . . . . . . 1296.2.1 A note on rotations . . . . . . . . . . . . . . . . . . . . 1546.3 Abelian 5-groups with cyclic generating sets of size 5 . . . . . 1577 Abelian groups of higher rank 1657.1 Abelian p-groups with cyclic generating sets of prime size q . . 1667.2 Cyclic generating sets of square free size . . . . . . . . . . . . 1727.3 Abelian p-groups cyclically generated by p generators . . . . . 179A GAP and ACE codes 189A.1 GAP codes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189A.1.1 Naïve code for searching for cyclically generated groups 189

vA.1.2 Code for finding GSO generating sets . . . . . . . . . . 190A.1.3 Code for ECS groups . . . . . . . . . . . . . . . . . . . 190A.1.4 Less naïve code for searching for cyclically generatedgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . 191A.2 Example of ACE code for testing generating sets . . . . . . . . 197Notation 199Bibliography 201

DeclarationI, Elizabeth Kimber, hereby certify that this thesis, which is approximately57,000 words in length, has been written by me, that it is the record ofwork carried out by me and that it has not been submitted in any previousapplication for a higher degree.Date: ................. Signature of Candidate: .............................................I was admitted as a research student in September 2002 and as a candidatefor the degree of Doctor of Philosophy in September 2003; the higher studyfor which this is a record was carried out in the University of St Andrewsbetween 2002 and 2005.Date: ................. Signature of Candidate: .............................................In submitting this thesis to the University of St Andrews I understand thatI am giving permission for it to be made available for use in accordancewith the regulations of the University Library for the time being in force,subject to any copyright vested in the work not being affected thereby. I alsounderstand that the title and abstract will be published, and that a copyof the work may be made and supplied to any bona fide library or researchworker.Date: ................. Signature of Candidate: .............................................I hereby certify that the candidate has fulfilled the conditions of the Resoluvii

AcknowledgementsI should like to thank Professor Edmund Robertson for suggesting this topicand for his kind supervision of my research. I am also grateful to Dr JohnO’Connor and Professor James Wiegold for interesting ideas about cyclicgenerating sets and their suggestions of possible approaches to certain problems.I appreciate the assistance of Dr Alexander Konovalov in sending me a copyof reference [32], which might not otherwise have been available to me.Both Anton Evseev and Pablo Spiga very helpfully provided me with proofsof part of Theorem 5.2. I am grateful to them both, and also to Matt Towerswith whom I have had many interesting discussions about cyclic generatingsets.Finally, I should like to thank Bob for being an excellent office mate, and myfamily and friends for their support.ix

AbstractA finite group is cyclically generated if it has an automorphism that cyclesthrough a generating set for the group. We present results that help us todetermine whether a given group is cyclically generated without first identifyingits automorphism group. Using these techniques we establish conditionsunder which certain familiar groups are cyclically generated and, in particular,we prove that every finite abelian group is cyclically generated. Withthe emphasis on groups that are the direct product of a cyclic group anda nonabelian group, we investigate necessary and sufficient conditions for adirect product to be cyclically generated.The question of when a finite abelian group has a minimal cyclic generatingset is addressed in the second part of this thesis. As every symmetric generatingset is a cyclic generating set, we can use a recent result about symmetricgenerating sets for abelian groups to find examples of abelian groups thathave cyclic generating sets of minimum size. We explore the connection betweenautomorphisms, matrices, and roots of certain polynomials over ringsof prime power order to construct suitable automorphisms or prove that nosuch automorphism exists. We prove necessary and sufficient conditions fora finite abelian group of rank 3 or 4 to have a minimal cyclic generatingset, and also for an abelian group of rank 5 if the order of the group is notdivisible by 5. Finally, we present partial results for abelian groups of higherprime and square free rank.xi

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Chapter 1IntroductionThe starting point for this investigation was the question of which finitegroups have a presentation 〈X R 〉 in which the set of defining relators, R,is invariant under the action of the automorphism that cycles through theelements of X. A presentation of this form is called a generalized cyclic presentationand, for example, 〈x 1 , x 2 x 2 1, x 2 2, [x 1 , x 2 ], [x 2 , x 1 ]〉 is a generalizedcyclic presentation for the Klein 4-group. As the name indicates, a generalizedcyclic presentation is a generalization of a cyclic presentation, see [20],but it is also a generalization of a symmetric presentation, see [1], and thereforeprevious work on symmetrically presented groups as well as cyclicallypresented groups can be used to provide examples of groups with generalizedcyclic presentations. Such examples include the finite abelian groups of rank2 and certain finite abelian groups of higher rank, the alternating groups,the dihedral groups, the nonabelian groups of order pq for odd primes p andq, and the Fibonacci groups, see [1], [9], [10], [32], and [20] respectively. Finitesimple groups can be added to this list because [3] contains a note thatproves that all finite simple groups have a symmetric presentation on twogenerators.A group that has a symmetric presentation 〈X R 〉 is said to be symmetricallygenerated by X and, similarly, we will say that a group is cyclicallygenerated if it has a generalized cyclic presentation. We will see later thathaving a generalized cyclic presentation is equivalent to having an automor-1

2phism that cycles through a generating set for the group. If a group has suchan automorphism then the set it cycles through is called a cyclic generatingset.Before giving a summary of the research methods and findings, we introducemore of the background to this research and look more closely at the connectionsbetween the theory of cyclically generated groups and that of cyclicallyand symmetrically presented groups.1.1 BackgroundWe begin with formal definitions of cyclic and symmetric presentations.Definition 1.1. Let F be the free group on the generators x 1 , x 2 , . . . , x n andlet w be a reduced word in some of the x i s. Let θ be the automorphism ofF induced by x 1 ↦→ x 2 ↦→ · · · ↦→ x n ↦→ x 1 and let wθ j be the word in the x i sthat results from applying the jth power of θ to each of the x i s occurring inw. Then the group presentation〈x 1 , x 2 , . . . , x n w, wθ, wθ 2 , . . . , wθ n−1 〉is called a cyclic presentation, see [20].We will return to cyclically presented groups later.Definition 1.2. Let F be the free group on the generators x 1 , x 2 , . . . , x n andfor j = 1, 2, . . . , k, let w j be a reduced word in some of the x i s. For each σin S n , let θ σ be the automorphism of F defined by x i ↦→ x iσ and let w j θ σbe the result of applying θ σ to each of the x i s occurring in w j . The grouppresentation〈x 1 , x 2 , . . . , x n w j θ σ , (1 ≤ j ≤ k), (σ ∈ S n )〉is called a symmetric presentation, see [27].As well as the groups listed in the previous section, examples of symmetricallygenerated groups can be found in [27], [4], [12], [10], and [32].

3It has already been noted that the idea of generalized cyclic presentabilityis a generalization of cyclic and symmetric presentability. To see how thesegeneralizations arise we consider the conditions that a group must satisfy inorder to be symmetrically generated and then show that a natural weakeningof these conditions results in a condition that is equivalent to the group beingcyclically generated. We then show that the cyclically presented groups area proper subset of the cyclically generated groups.If a group G is symmetrically generated by X = {x 1 , x 2 , . . . , x n }, with symmetricpresentation 〈X R 〉, then it is implied by the definition of a symmetricpresentation that R is invariant under the action of S n on X, and therefore everypermutation of X, and hence every element of S n , must correspond to anautomorphism of G. For example, if n = 6 and σ = (1 3)(4 5 6) then σ correspondsto the map θ defined by x 1 ↦→ x 3 ↦→ x 1 , x 2 ↦→ x 2 , x 4 ↦→ x 5 ↦→ x 6 ↦→ x 4 .Replacing every x i by x i θ in any relator gives another relator because of theinvariance of R, so θ induces an automorphism. Now suppose that we replacethe requirement that every permutation corresponds to an automorphism ofthe group with the weaker requirement that only the n-cycle (1 2 · · · n) does.That is, we only require that the map x 1 ↦→ x 2 ↦→ · · · ↦→ x n ↦→ x 1 induces anautomorphism. This is the condition that a group must satisfy in order tobe cyclically generated. Only if a group is cyclically generated by two generatorscan we say with certainty that the group is symmetrically generated,and therefore, while some cyclically generated groups are also symmetricallygenerated, there are other cyclically generated groups that do not have symmetricpresentations. For example, it follows from Theorem 1 of [1] thatC 2 × C 4 × C 8 does not have a symmetric presentation, but we will see laterthat every finite abelian group is cyclically generated.An important subset of the cyclically generated groups consists of the groupsthat have an n generator n relator generalized cyclic presentation. This isthe set of cyclically presented groups, which have been studied extensively,see [20], [21], [4], [11], [5], and [18]. A cyclically presented group must necessarilyhave deficiency zero and hence only the groups that have trivial SchurMultiplier can be cyclically presented, see [6] p.140. As an example of a

4group that is cyclically generated but not cyclically presented, consider D 8 ,the dihedral group of order 8. This is generated by 2 elements of order 2 andthere is an automorphism that swaps these two generators, so D 8 is cyclicallygenerated, but the Schur Multiplier of D 8 is cyclic and hence D 8 cannot becyclically presented. It follows that cyclic generation can be thought of as ageneralization of cyclic presentability, and it differs from symmetric generationin that the set of cyclically presented groups is a proper subset of thecyclically generated groups.1.2 Definitions and some preliminary ideasFor ease of reference, we now give some formal definitions of ideas associatedwith generalized cyclic presentations and cyclic generating sets.1.2.1 DefinitionsThe following definition of a generalized cyclic presentation is based on thedefinition of a cyclic presentation in [20].Let F be the free group on n generators X = {x 1 , x 2 , . . . , x n } and letw 1 , w 2 , . . . , w k be reduced words in F. Let θ be the automorphism of F inducedby x 1 ↦→ x 2 ↦→ · · · ↦→ x n ↦→ x 1 and let R = {w i θ j : 1 ≤ i ≤ k, 1 ≤j ≤ n}.Definition 1.3. If X and R are as above then the presentation 〈X R〉 iscalled a generalized cyclic presentation.Definition 1.4. A group that is defined by a generalized cyclic presentationis said to be cyclically generated.Definition 1.5. Any set that is cyclically permuted by an automorphism iscalled a cyclic set.Definition 1.6. The generating set in a generalized cyclic presentation iscalled a cyclic generating set.

5Definition 1.7. If θ is an automorphism of a group G and there exists somex in G such that {x, xθ, xθ 2 , . . .} is a cyclic generating set for G, then θ iscalled a cyclic automorphism.Definition 1.8. If a group of rank n has a cyclic generating set of size nthen this set is called a minimal cyclic generating set.We also need the following definitions for two classes of groups.Definition 1.9. A group is called a GSO group if it is generated by elementsof the same order, i.e. if it is generated by its elements of order m for somem. See [13] and [26]. A group that is not generated by elements of the sameorder is called a non-GSO group.Definition 1.10. A generating set for a group that consists of elements ofthe same order is called a GSO generating set.Definition 1.11. If every element of a group G lies in a proper characteristicsubgroup of G then G is called an ECS group.1.2.2 Constructing automorphismsWe will often prove that a group is cyclically generated by finding a suitableautomorphism and showing that repeated applications of this automorphismto a single group element results in a generating set for the group. If weneed to construct a suitable automorphism explicitly, we usually do this bydefining a map in terms of the generators of a presentation for the group,so if the group G is defined by the n generator presentation 〈X R 〉 then wedefine a map θ by x i ↦→ y i = y i (x 1 , x 2 , . . . , x n ). We will say that θ is an ontomap if the set {y i (x 1 , x 2 , . . . , x n ) : i = 1, 2, . . . , n} generates G, becauseif we can extend θ to a homomorphism then it will be onto. Because weare only considering finite groups, this map will be an automorphism andtherefore it is enough to show that θ can be extended to a homomorphism,i.e. it induces a homomorphism. In order to prove that this is the case, it issufficient to show that for any relator r = r(x 1 , x 2 , . . . , x n ) in R, we haver(x 1 θ, x 2 θ, . . . , x n θ) = 1,

6see Substitution Test, [20], p.29. With some abuse of notation, we sometimesuse the same symbol to denote both an onto map defined on the generatorsof a group, and the automorphism that this map induces, but the meaningwill be clear from the context. It is worth noting here that in practice, if therelator r only involves those x i that are fixed by θ, there is no need to checkthat r(x 1 θ, x 2 θ, . . . , x n θ) = 1. Similarly, if a group is the direct product oftwo groups, G and H, and the map in question maps the generators of Ginto G and the generators of H into H then there is no need to check thecommutator relators of the form [g, h].1.3 Summary of researchIn the early part of the research we showed that having a generalized cyclicpresentation is equivalent to having an automorphism that cycles through agenerating set. From this point we decided to focus on determining whetherthe automorphism group of a given group contains a cyclic automorphism,rather than trying to find generalized cyclic presentations for groups. Weproved that every finite abelian group has a cyclic automorphism, and thenlooked for more examples of familiar groups that are cyclically generated.We found cyclic generating sets for the symmetric groups and the generalizedquaternion groups, among others. The fact that the problem could beconsidered as a question about the action of automorphisms on generatingsets meant that we could use GAP, [14], to approach the problem more systematicallyand identify less obvious examples of cyclically generated groups.More importantly, the use of GAP enabled us to find examples of groupsthat are not cyclically generated. The smallest group that is not cyclicallygenerated is the non-GSO group of order 16, see [26]. Clearly a necessarycondition for a group to have a cyclic generating set is that it is generatedby its elements of order m for some m, so the non-GSO groups are a classof groups that we know cannot be cyclically generated. Every non-GSOgroup of order less than 100 has order divisible by 16 and the question of theexistence of a non-GSO p-group for an odd prime p is the subject of a famousproblem in the Kourovka Notebook, [24], which remained unanswered until

72004, see [26].Searches using GAP found examples of GSO groups that are not cyclicallygenerated and thereby proved that the GSO condition is not a sufficientcondition for a group to be cyclically generated. A stronger condition thanGSO, but necessary for a group to be cyclically generated, is that the groupmust have a GSO generating set that does not intersect nontrivially with anyproper characteristic subgroup. Again, this condition was identified by usingGAP to analyse small groups that are not cyclically generated. The smallestexample of a group that is GSO but is not cyclically generated is the groupS 3 × C 4 , and by using GAP we found that this group is not cyclically generatedbecause every GSO generating set has a nontrivial intersection with thecentre of the group. We used the idea of characteristic subgroups to define asecond class of groups that we know are not cyclically generated. These arethe groups in which every element lies in a proper characteristic subgroup,the so called ECS groups. Certain subgroups are known to be characteristic,and if we can show that every element of a given group lies in one of thesesubgroups, then we will have proved that the group is not cyclically generatedwithout determining its automorphism group or identifying the GSOgenerating sets. In fact, the group S 3 × C 4 is an ECS group, as are D 8 × C 4and D 10 ×C 4 . These ECS groups prove that it is not necessarily true that thedirect product of two cyclically generated groups is cyclically generated. Itis also not true that the direct product of two groups that are not cyclicallygenerated is also not cyclically generated. We used GAP to carry out a surveyof the direct products of some small non abelian groups and cyclic groups.If a group is known to be cyclically generated then it is natural to ask whetherit has a cyclic generating set of minimum size. In 2003, shortly after the publicationof a related result concerning symmetric generating sets for abeliangroups, see [1], we started to address the question of whether finite abeliangroups have minimal cyclic generating sets. The 2003 result by Miklós Abért,[1], which gives necessary and sufficient conditions for a finite abelian groupto be symmetrically generated, and the size of a symmetric generating setwhen it exists, is a generalization of a 1969 result by Emerson, [12], in which

8it is shown that the abelianization of certain symmetrically generated groupsmust be of a certain form. Since every symmetric generating set is a cyclicgenerating set, we can use Abért’s result to show that there are finite abeliangroups that have minimal cyclic generating sets, but by searching with GAPwe found examples of abelian groups that do not satisfy the conditions inAbért’s theorem, but have minimal cyclic generating sets. We proved that,in order to establish whether a rank n ≥ 2 finite abelian group G has a cyclicgenerating set of size n, it is sufficient to establish whether each primarypart of G has a cyclic generating set of size dividing n. This is because theautomorphism group of an abelian group is the direct product of the automorphismgroups of its p-primary parts. Therefore we were able to reduce theproblem to a question about abelian p-groups. It was not, however, simplya matter of establishing whether an abelian p-group has a minimum cyclicgenerating set. For example, if the 7-primary part of a rank 5 abelian grouphas rank 3, then we had to determine whether this 7-group has a cyclic automorphismof order 5, not of order 3. In order to show that certain p-groupshave a cyclic generating set of a suitable size, we used roots of polynomialsover integer rings of the form Z p α to construct cyclic automorphisms of theappropriate order. To show that other p-groups do not have a cyclic generatingset of any suitable size, we could sometimes argue using the orderof the automorphism group, but where this was not possible, we used argumentsinvolving factor groups of the p-groups in question and conjugacyclasses of certain matrix groups. A note in Abért’s paper proves that everyfinite abelian group of rank 2 has a symmetric presentation on 2 generatorsand we used the methods outlined above to prove necessary and sufficientconditions for finite abelian groups of ranks 3 and 4 to have minimal cyclicgenerating sets, and also for finite abelian groups of rank 5 whose order isnot divisible by 5. In the case of an arbitrary finite abelian group, there is somuch choice over the size of cyclic generating sets that each of the primaryparts can have, that proving the non-existence of cyclic generating sets of appropriatesizes, and hence a classification similar to the one given by Abért,was not attempted. We did, however, find some partial results concerningfinite abelian groups of arbitrary prime or square free rank.

91.4 Summary of ThesisChapters 2 and 3 are concerned with finding cyclic generating sets for certainfinite groups and identifying techniques that can be used to determinewhether such sets exist. Some details of the use of GAP can also be foundin Chapter 2 and at the end of this chapter we present some results aboutdirect products. The focus on direct products continues in Chapter 3 wherewe present the results of a GAP survey of some groups that are the directproduct of a small nonabelian group and a cyclic group, and we use someof the results from Chapter 2 to prove the conjectures that we were able tomake using these survey results. Some of these results can be extended to amore general case and these theorems complete the first part of this thesis.The remaining chapters consist of work on minimal cyclic generating setsfor finite abelian groups. Some of the results from Chapter 2 are adaptedto establish the existence of cyclic generating sets of certain sizes and thesemodified results can be found in Chapter 4. In this chapter we also presentthe result (Theorem 1) from Abért’s 2003 paper, [1], and give a direct proof,based on the techniques used by Abért, that if a finite abelian group satisfiesthe conditions of Theorem 1, then it is cyclically generated by n generators.We use these results in Chapters 5–7. Chapter 5 contains classification resultsfor rank 3 and rank 4 finite abelian groups with minimal cyclic generatingsets, while in Chapter 6 we present a classification of finite abelian groupsof rank 5 and order not divisible by 5. Here we also highlight a connectionbetween one of the cyclic automorphisms constructed and the golden ratio,and present some partial results concerning the existence of cyclic generatingsets of size 5 for abelian 5-groups. Chapter 7 contains some partial resultsfor finite abelian groups of higher rank, including an existence result forcyclic generating sets of prime and square free size, and a proof that usesa connection with Pascal’s Triangle to prove that Z p 2 × Z p 2 is cyclicallygenerated by p generators if and only if p = 2 or p = 3. This result iscombined with previous results to give a classification of rank 2 abelian p-groups with cyclic generating sets of size p.

Chapter 2Some cyclically generatedgroupsWe begin with a result which shows that for a finite group, having a generalizedcyclic presentation is equivalent to being cyclically generated.Theorem 2.1. A finite group is cyclically generated if and only it has ageneralized cyclic presentation.Proof. Suppose a finite group G is cyclically generated by X={x 1 , x 2 ,. . .,x n }.Then there is an automorphism θ of G such that x 1 ↦→ x 2 ↦→ · · · ↦→ x n ↦→ x 1 .Let P = 〈X R〉 be a finite presentation for G on the generating set X, soany relator r is a word in the x i s, i.e. r = r(x 1 , x 2 , . . . , x n−1 , x n ). As θ is anautomorphism,1 = rθ = r(x 1 θ, x 2 θ, . . . , x n−1 θ, x n θ) = r(x 2 , x 3 , . . . , x n , x 1 ),and therefore the fact that rθ = 1 follows from the relators R. This meansthat adding the relator rθ to the presentation P does not alter the groupdefined by P and it follows that, if R = {r 1 , r 2 , . . . , r k }, then for each r i andθ j , with 1 ≤ j ≤ n − 1, we can add r i θ j to the set of relators in P withoutchanging the group defined by the resulting presentation. Therefore G is11

12defined by the generalized cyclic presentation〈X r 1 , r 1 θ, r 1 θ 2 , . . . , r 1 θ n−1 ,r 2 , r 2 θ, r 2 θ 2 , . . . , r 2 θ n−1 ,.r k , r k θ, r k θ 2 , . . . , r k θ n−1 〉.Conversely, suppose that G is defined by the generalized cyclic presentation〈x 1 , x 2 , . . . , x n r 1 1 , r 1 2 , . . . , r 1 n−1 , r 1 n ,r 2 1 , r 2 2 , . . . , r 2 n−1 , r 2 n ,.r n−1 1 , r n−1 2 , . . . , r n−1 n−1 , r n−1 nr n 1 , r n 2 , . . . , r n n−1 , r n n 〉.where r i 1 = r i (x 1 , x 2 , . . . , x n−1 , x n ) and r i j is the result of replacing x 1 withx j , x 2 with x j+1 , and so on, in r i 1 , i.e.r i j = r i (x j , x j+1 , . . . x n−1 , x n , x 1 , x 2 , . . . , x j−1 ).Let θ be the map defined byx 1 ↦→ x 2x 2 ↦→ x 3.x n−1 ↦→ x nx n ↦→ x 1 .Then clearly θ is onto, and in order to show that it is a homomorphism weneed to show that replacing x i by x i θ in any of the relators in the abovepresentation gives a word in the x j s that is also equal to 1 in G. But for anyr i j ,r i (x j θ, x j+1 θ, . . . , x n−1 θ, x n θ, x 1 θ, . . . , x j−1 θ)= r i (x j+1 , x j+2 , . . . , x n , x 1 , x 2 , . . . , x j )= r i j+1= 1

13and therefore θ is an automorphism of G, which, by definition, cycles throughthe generating set {x 1 , x 2 , . . . , x n }. Hence G is cyclically generated, as required.It is not difficult to find examples of groups that have presentations whicheither are already generalized cyclic presentations, or can be made into generalizedcyclic presentations by the addition of (redundant) relators. If thefamiliar presentations for a group do not include a generalized cyclic presentation,one of them may still give an indication of how a cyclic automorphismmight be constructed. We can then use the same presentation to prove thatthe map defined is indeed an automorphism and it then only remains to provethat the automorphism cycles through a generating set. Some of the resultsin this chapter use these techniques.2.1 Some cyclically generated groups2.1.1 Finite abelian groupsEvery cyclic group is trivially cyclically generated by one generator, and ifG = C n × C n then〈a, b a n , b n , [a, b], [b, a]〉is a generalized cyclic presentation for G. Similarly, if G is a direct product ofm copies of the cyclic group of order n, then G is defined by the presentation〈x 1 , x 2 , . . . , x m x n i , (i = 1, 2, . . . , m), [x i , x j ], (1 ≤ i < j ≤ m)〉and from this presentation it is clear that the map x 1 ↦→ x 2 ↦→ · · · ↦→ x m ↦→ x 1induces an automorphism.If G is the rank 2 abelian group C mn ×C n then G is defined by the presentationG = 〈a, b a mn , b n , [a, b]〉.In order to prove that the onto map θ defined by a ↦→ ab, b ↦→ b inducesan automorphism of G, we need to show that replacing a and b by aθ andbθ in the relators of the given presentation gives a word that is equal to

14the identity of the group. As G is abelian, we do not need to check thecommutator relators. From the remaining relators, we have(aθ) mn = (ab) mn = a mn b mn = 1,and(bθ) n = b n = 1,so θ is an automorphism and therefore G is cyclically generated by a and aθ.By generalizing this automorphism we obtain the following result.Theorem 2.2. Every finite abelian group is cyclically generated.Proof. By the Fundamental Theorem of Finite Abelian Groups, we can writeany finite abelian group G in the formG = C m1 × C m2 × · · · × C mnwhere m n m n−1 · · · m 2 m 1 . Let x i be a generator of the ith cyclic factor C mi ,soG = C m1 × C m2 × · · · × C mn= 〈x 1 , x 2 , . . . , x n x m ii , (i = 1, 2, . . . , n), [x i , x j ], (1 ≤ i < j ≤ n)〉.Define an onto map byx 1 ↦→ x 1 x 2x 2 ↦→ x 2 x 3.x n−1 ↦→ x n−1 x nx n ↦→ x n .Note that o(x i+1 ) o(x i ) and therefore(x i x i+1 ) m i= x m ii x m ii+1 = 1,

15so this map can be extended to a homomorphism θ of G. As θ maps onto G,it is an automorphism. Note also that since θ has finite order, x 1 θ k = x 1 forsome k. Applying successive powers of θ to x 1 gives the setx 1 ↦→ x 1 x 2 ↦→ x 1 x 2 2x 3 ↦→ · · · ↦→ x 1 Xx n ↦→ · · · ↦→ x 1where X is a product of powers of x 2 , x 3 , . . . , x n−1 . The set {x 1 , x 1 θ, x 1 θ 2 , . . .}clearly generates G, and hence G is cyclically generated.For example, letG = C 12 × C 6 × C 6 × C 2= 〈x 1 , x 2 , x 3 , x 4 x 121 , x 6 2, x 6 3, x 2 4, [x i , x j ], (1 ≤ i < j ≤ 4)〉.Let θ be as defined in the proof of Theorem 2.2, i.e. it is the automorphisminduced by the mapThenx 1 ↦→ x 1 x 2 , x 2 ↦→ x 2 x 3 , x 3 ↦→ x 3 x 4 , x 4 ↦→ x 4 .(x 1 x 2 ) 12 = x 121 x 122 = 1, (x 2 x 3 ) 6 = x 6 2x 6 3 = 1, and (x 3 x 4 ) 6 = x 6 3x 6 4 = 1,so θ is an automorphism of G. Now applying θ to x 1 givesx 1 ↦→ x 1 x 2 ↦→ x 1 x 2 2x 3 ↦→ x 1 x 3 2x 3 3x 4 ↦→ x 1 x 4 2x 6 3x 4 4 = x 1 x 4 2↦→ x 1 x 5 2x 4 3 ↦→ x 1 x 6 2x 9 3x 4 4 = x 1 x 3 3 ↦→ x 1 x 2 x 3 3x 3 4 ↦→ x 1 x 2 2x 4 3x 6 4 = x 1 x 2 2x 4 3↦→ x 1 x 3 2x 6 3x 4 4 = x 1 x 3 2 ↦→ x 1 x 4 2x 3 3 ↦→ x 1 x 5 2x 7 3x 3 4 = x 1 x 5 2x 3 x 3 4 ↦→ x 1 x 6 2x 6 3x 6 4 = x 1 .Let S be the set {x 1 θ j : j = 0, 1, . . . , 11} and let H be the subgroupgenerated by S. Clearly x 1 is in H, and as x 1 θ = x 1 x 2 , x 2 must also be inH. Similarly, as x 1 θ 2 = x 1 x 2 2x 3 , H must also contain x 3 . Finally, as x 1 θ 3 =x 1 x 3 2x 3 3x 4 , and x 1 , x 2 , and x 3 are in H, it follows that x 4 is in H and thereforeH = G as required.2.1.2 Nonabelian finite simple groupsIt is noted in [3] that every finite nonabelian simple group is symmetricallygenerated by two generators. The proof is as follows. Let G be a finite

16nonabelian simple group. Let a be an element of order 2 and let b be anotherelement of G such that a and b generate G. (Such a and b exist because everyfinite nonabelian simple group is generated by an element of order 2 and oneother element, see [23] and [33].) Let θ be conjugation by a and consider thesubgroup H generated by b and bθ. Suppose that H ≠ G. Then a cannot bein H, so aH ≠ H, but H has index 2 in G because aH ∪ H = G. ThereforeH is normal in G, which is a contradiction, so G is cyclically generated bytwo generators.We have shown that finite abelian groups and finite nonabelian simple groupsare cyclically generated and these groups are, in a sense, at the two extremesof the range of finite groups. In particular, this means that we have shownthat the composition factors of any group are cyclically generated and thereforethe way in which a group is built up from its composition factors affectswhether or not it is cyclically generated. For example, we shall see later thatS 3 × C 2 × C 2 is cyclically generated, but S 3 × C 4 is not.We will now provide some more examples of groups that are cyclically generated.2.1.3 Some more examples of nonabelian cyclically generatedgroupsDihedral groupsThe dihedral group of order 2n is defined by the presentation〈a, b a 2 , b 2 , (ab) n 〉, (2.1)see [10] p.6, and we can add the relator (ba) n without changing the groupdefined by the presentation, so there is an automorphism such that a ↦→ b ↦→a.Symmetric groupsIn the symmetric group S n , repeated conjugation of (1 2) by the n-cycle(1 2 . . . n) yields the set {(1 2), (2 3), . . . , (n−2 n−1), (n−1 n), (n 1)}, which

17is a generating set because for any i, j with 1 ≤ i < j ≤ n, the transposition(i j) can be written as(i i+1)(i+1 i+2) · · · (j−2 j−1)(j−1 j)(j−2 j−1)(j−3 j−2) · · · (i+2 i+3)(i+1 i+2)(i i+1)and the transpositions generate S n .Alternating groupsWe already know that for n ≥ 5, A n is cyclically generated by two generators,and the following 2 generator presentations for A n can be used to show thatin each case the map a ↦→ b ↦→ a is an automorphism.For m ≥ 3, A 2m and A 2m+1 are defined in [9], p.150, byA 2m = 〈a, b a 2m−1 , b 2m−1 , (ab) m , (a −1 b) 3 , (a −l b −1 ab l ) 2 , (1 ≤ l ≤ m − 2)〉A 2m+1 = 〈a, b a 2m+1 , b 2m+1 , (ab) m , (a −1 b) 3 , (a −l b l ) 2 , (2 ≤ l ≤ m)〉,while the presentationsA 5 = 〈a, b a 5 , b 5 , (ab) 2 , (a −1 b) 3 〉, and A 4 = 〈a, b a 3 , b 3 , (ab) 2 〉,are taken from [10] p.137 and p.7 respectively.The presentation for A 4 given above is a special case of a presentation forA n given by Carmichael in 1923 and later by Coxeter, see [8]. Using thispresentation for A n , we now verify that, for all n, the group A n has a cyclicgenerating set consisting of n − 2 elements of order 3. In 1923 Carmichaelproved that A n+2 is defined by the presentation〈x 1 , x 2 , . . . , x n x 3 i , (1 ≤ i ≤ n), (x i x j ) 2 , (1 ≤ i < j ≤ n)〉, (2.2)see [8] and [10]. Let θ be the map defined by x 1 ↦→ x 2 ↦→ . . . ↦→ x n ↦→ x 1 .This map induces an automorphism because we can see immediately fromthe presentation that (x i θ) 3 = x 3 i+1 = 1, for 1 ≤ i ≤ n − 1, (x n θ) 3 = x 3 1 = 1,and((x i θ)(x j θ)) 2 = (x i+1 x j+1 ) 2 = 1

18for 2 ≤ i + 1 < j + 1 ≤ n, i.e. for 1 ≤ i < j ≤ n − 1, while if j = n then((x i θ)(x j θ)) 2 = (x i+1 x 1 ) 2 = x i+1 (x 1 x i+1 ) 2 x −1i+1 = 1for 1 ≤ i ≤ n − 1. Therefore A n+2 is cyclically generated by x 1 , x 2 , . . . , x n .Generalized quaternion groupsWe prove that the generalized quaternion groups (as defined on p.45 of [20])are cyclically generated for all n. The quaternion group of order 4n, denotedhere by Q 4n , is defined by the presentation〈x, y x n y −2 , y −1 xyx〉.(Note that this group is denoted by Q 2n in [20].) From the second relator wehavey −1 xy = x −1 ,soand hencex n = (y −1 xy) −n = y −1 x −n y = y −1 y −2 y = y −2 ,x 2n = x n x n = y −2 y 2 = 1.Define an onto map θ byx ↦→ x −1 , y ↦→ xy.By replacing x and y with xθ and yθ in the first relator, we have(xθ) n (yθ) −2 = x −n (xy) −2= x −n y −1 (x −1 y −1 x −1 )= x −n y −2= x −2n= 1and by putting xθ and yθ in place of x and y in the second relator we get(yθ) −1 xθyθxθ = y −1 x −1 x −1 xyx −1 = y −1 x −1 yx −1 = (y −1 x −1 y)x −1 = xx −1 = 1,

19so θ induces an automorphism. Moreover, if we apply successive powers ofthis automorphism to y we gety ↦→ xy ↦→ x −1 xy = yand {y, xy} is a generating set, so the group is cyclically generated, as required.Groups of order pq where p and q are odd primesIf p and q are odd primes such that there exists a nonabelian group G oforder pq, then G is defined by the following generalized cyclic presentation〈x, y x p , y p , (x −1 y) k x(x −1 y) kn x −1 , (y −1 x) k y(y −1 x) kn y −1 〉,where p = 2k + 1 and n p ≡ 1 (mod q), see Example 6 in [32].Groups of order p 3 where p is an odd primeIf p is an odd prime and G is a nonabelian group of order p 3 then either Ghas exponent p or it has exponent p 2 .If G has exponent p then it is defined by the 3 generator presentation〈a, b, c a p , b p , c p , [b, a]c −1 , [a, c], [b, c] 〉,see [15] p.52. We can use the relator [a, b] = c to re-write this presentationas the 2 generator presentation〈a, b a p , b p , [a, b] p , [a, [a, b]], [b, [b, a]]〉and by adding relators we can make this into a generalized cyclic presentation〈a, b a p , b p , [a, b] p , [b, a] p , [a, [a, b]], [b, [b, a]]〉.If G has exponent p 2 , then it is defined by the presentation〈a, b a p2 , b p , b −1 aba −(1+p) 〉,see [15] p.52. We define an automorphism of order 2 that cycles through agenerating set. Let θ be the map defined by a ↦→ ba −1 b, b ↦→ b. Then θ is

20onto, and it remains to check that replacing a with aθ and b with bθ in eachof the relators in the above presentation still gives the identity.Clearly (bθ) p = 1 and as G has exponent p 2 , we know that (aθ) p2 = 1,so we only need to prove that (bθ) −1 (aθ)(bθ)(aθ) −(1+p) = 1, or equivalently(bθ) −1 aθbθ = (aθ) 1+p , so we need to show thatb −1 (ba −1 b)b = (ba −1 b) 1+p . (2.3)The left hand side of (2.3) is a −1 b 2 and from the third relator we haveb −1 ab = a 1+p , so we can rewrite the right hand side of (2.3) as (b 2 a −(1+p) ) 1+p .Therefore we want to prove thata −1 b 2 = (b 2 a −(1+p) ) 1+p . (2.4)We begin by expressing both sides of (2.4) in the form b k a l , proceeding byinduction.Claim 1 a −i b = ba −i(1+p) .ProofBase step a −1 b = ba −(1+p) .Inductive stepa −(i+1) b = a −1 (a −i b) =a −1 (ba −i(1+p) )= (a −1 b)a −i(1+p) = ba −(1+p) a −i(1+p) = ba −(i+1)(1+p) .Claim 2 a −i b 2 = b 2 a −i(1+p)2 .ProofBase stepa −1 b 2 = (a −1 b)b = (ba −(1+p) )b = b(a −(1+p) b) = b 2 a −(1+p)2 .

21Inductive stepa −(i+1) b 2 = a −1 (a −i b 2 ) = a −1 (b 2 a −i(1+p)2 )= (a −1 b 2 )a −i(1+p)2 = (b 2 a −(1+p)2 )a −i(1+p)2 = b 2 a −(i+1)(1+p)2 .Therefore the left hand side of (2.4) is b 2 a −(1+p)2 , but as a p2 = 1, this isb 2 a −1−2p .We need the following result before we can find an expression for the righthand side of (2.4) in the form b k a l .Claim 3(b 2 a −(1+p) ) i = b 2i a −(1+p)[1+(1+p)2 +(1+p) 4 +···+(1+p) 2(i−1)] .ProofBase step There is nothing to prove if i = 1.Inductive step(b 2 a −(1+p) ) i+1 = (b 2 a −(1+p) ) i b 2 a −(1+p)= b 2i a −(1+p)[1+(1+p)2 +(1+p) 4 +···+(1+p) 2(i−1)] b 2 a −(1+p)= b 2i+2 a −(1+p)3 [1+(1+p) 2 +(1+p) 4 +···+(1+p) 2(i−1)] a −(1+p) by Claim 2= b 2(i+1) a −(1+p)3 [1+(1+p) 2 +(1+p) 4 +···+(1+p) 2(i−1)] a −(1+p)= b 2(i+1) a −(1+p)[1+(1+p)2 +(1+p) 4 +···+(1+p) 2i] .Therefore(b 2 a −1 ) 1+p = b 2(1+p) a −(1+p)[1+(1+p)2 +(1+p) 4 +···+(1+p) 2p ]= b 2 a −(1+p)[1+(1+p)2 +(1+p) 4 +···+(1+p) 2p ]because b p = 1.As a p2 = 1, we need to find−(1 + p)[1 + (1 + p) 2 + (1 + p) 4 + · · · + (1 + p) 2p ] (mod p 2 ).

22Using (1 + p) i ≡ 1 + ip (mod p 2 ), we have−(1 + p)[1 + (1 + p) 2 + (1 + p) 4 + · · · + (1 + p) 2p ]p∑= −(1 + p) (1 + p) 2i≡ −(1 + p)i=0p∑(1 + 2ip) (mod p 2 )i=0( p∑= −(1 + p) 1 + 2pi=0(= −(1 + p) 1 + p + 2p)p∑ii=0p(p + 1)2≡ −(1 + p)(1 + p) (mod p 2 )≡ −1 − 2p (mod p 2 ))so(b 2 a −(1+p) ) 1+p = b 2 a −1−2pas required. Therefore θ induces an automorphism, which will also be denotedby θ.The set {a, aθ} generates G becausea −(1+p) (aθ) −1 = a −(1+p) (ba −1 b) −1 = a −(1+p) (b −1 a)b −1 = a −(1+p) a 1+p b −2 = b −2and p is odd, so b must be in the subgroup generated by a and aθ. ThereforeG is cyclically generated.By combining the results in this section we have proved the following theorem.Theorem 2.3. For any primes p and q, all groups of order p, p 2 , pq, and p 3are cyclically generated.All the cyclic automorphisms defined in this section, except for the one inthe proof for groups of odd order p 3 and exponent p 2 , were found withoutusing GAP, see [14]. The automorphism constructed to prove this result was

23found by setting GAP to search for cyclic automorphisms for a specific primep using the code in Appendix A. By searching through the results of thistest by hand, we were able to find a number of examples that might easily begeneralized to prove the result for a general prime p. These candidates weregeneralized by hand and then tested on the groups for p ≤ 13 using ACE, see[17]. On finding an automorphism that was suitable for all of these cases, itonly remained to prove the general result. By using GAP and ACE in thisway we were able to find a suitable automorphism much more easily.2.2 Computational techniquesHaving established that certain well known groups are cyclically generated,a more systematic approach was made possible by using GAP, see [14], tosearch through the groups in the small groups library. This provided furtherexamples of cyclically generated groups as well as some examples of smallgroups that are not cyclically generated and these examples provided muchof the information used to develop conjectures. Initial searches for cyclicgenerating sets for specific groups were carried out simply by setting GAPto run through all possible choices for a group element x and automorphismθ. The code for this can be found in Appendix A. Later, particularly whenwe were searching for cyclic generating sets of a particular size, we wantedto make the search quicker. The method above involves many redundantcomputations. For example, if the elements of order n do not generate thegroup, there is no need to test the elements of order n. Therefore, as a firststep to improving the searching method, we set GAP to compute the set ofn such that the group is generated by its elements of order n.If a group is cyclically generated, then in general there appear to be manypairings of a group element with an automorphism that give rise to cyclicgenerating sets. In order to prove that a given group is cyclically generated,it is therefore often enough to use GAP to apply randomly chosen automorphismsto elements of appropriate orders until it finds a pair that producea cyclic generating set. Unfortunately this method can, at best, only give

24us a partial answer to the the question of whether or not a given group iscyclically generated. We can, however, obtain a complete answer by first settingGAP to compute the conjugacy classes of the automorphism group, andthen testing the action of a representative of each class on each of the groupelements of the appropriate orders. This reduces the number of possibilitiesthat GAP must test. Where GAP is unable to compute the conjugacy classesof the automorphism group, we can, as an alternative, set it to computethe conjugacy class representatives for the group and then test the action ofeach automorphism on the representatives that have the appropriate orders.The results in the next subsection show that these methods are sufficient togive a complete answer to the question of whether a given group is cyclicallygenerated.2.2.1 Conjugacy classesTheorem 2.4. Let G be a finite group and let θ be an automorphism ofG. Then θ cycles through a generating set for G if and only if, for each ϕconjugate to θ in Aut G, the map ϕ cycles through a generating set.Proof. Suppose first that {g, gθ, gθ 2 , . . . gθ n−1 } generates G, where n = o(θ),and let ϕ be conjugate to θ in Aut G. Then there exists some automorphismβ such that ϕ = β −1 θβ. Let x = gβ, so g = xβ −1 and xβ −1 θ i = gθ i fori = 0, 1, . . . , n − 1.Therefore G is generated by the set S = {xβ −1 , xβ −1 θ, xβ −1 θ 2 , . . . , xβ −1 θ n−1 }.As β is an automorphism of G, it must map a generating set onto a generatingset, so the image of S under β must also generate G, butSβ = {xβ −1 β, xβ −1 θβ, xβ −1 θ 2 β, . . . , xβ −1 θ n−1 β}= {x, xβ −1 θβ, x(β −1 θβ) 2 , . . . , x(β −1 θ n−1 β)}= {x, xϕ, xϕ 2 , . . . , xϕ n−1 },so ϕ cycles through a generating set as required.In the case of a group generated by 2 generators, Theorem 2.4 is Lemma1 of [32]. After the lemma, Tomaszewski comments that there might be

25two automorphisms of order 2, both of which swap generators, but are notconjugate. He illustrates this fact by providing two automorphisms of A 4that each cycle through generating sets, but are not conjugate.We now show that, in order to establish the existence or non-existence of acyclic generating set for a group, it is sufficient to apply each automorphismonly to a representative of each conjugacy class in the group, rather than toevery group element.Theorem 2.5. If x lies in a cyclic generating set for a finite group G and yis conjugate to x, then y also lies in a cyclic generating set for G.Proof. Suppose that θ, an automorphism of G of order n, is such that theset {x, xθ, xθ 2 , . . . , xθ n−1 } generates G. Let S = {x, xθ, xθ 2 , . . . , xθ n−1 } andlet y be conjugate to x. Then there exists some z such that y = z −1 xz, solet β be the automorphism defined by conjugation by z. Now let ϕ be theautomorphism of G defined byϕ = β −1 θβ,so o(ϕ) = o(θ) = n. We show that ϕ cycles through a generating set containingy.Note thatyϕ j = y(β −1 θβ) j = yβ −1 θ j β = (zyz −1 )θ j βand therefore, if T = {y, yϕ, yϕ 2 , . . . , yϕ n−1 }, then= z −1 ((zyz −1 )θ j )z = z −1 (xθ j )z = (xθ j )βT = {xβ, (xθ)β, (xθ 2 )β, . . . , (xθ n−1 )β} = Sβ.As S generates G and β is an automorphism, T also generates G, as required.For practical purposes, and because of the high incidence of cyclic generatingsets in cyclically generated groups, it is often enough to restrict the searchesto one automorphism from each conjugacy class and test its action on only

26one element from each conjugacy class in the group. However, if this doesnot produce a cyclic generating set then it does not prove that no such setexists. For example, if G = D 10 = 〈a, b a 2 , b 5 , (ab) 2 〉, thenb 2 ab −2 = b 2 ab 3 = b 2 abb 2 = b(bab)b 2 = bab 2 = (bab)b = ab,so a and ab are conjugate in G. If, however, θ is the automorphism induced bya ↦→ a, b ↦→ b −2 , then clearly applying θ to a does not produce a generatingset, but applying it to ab givesab ↦→ ab −2 ↦→ ab 4 ↦→ ab −3 ↦→ ab,andabab 4 = b −1 b 4 = b −3 ,so b is in the subgroup generated by {b −1 a, b 2 a, b −4 a, b 3 a}, and hence this setgenerates G. Therefore, if GAP were to compute θ as a representative of aconjugacy class of Aut D 10 and a as a representative of a conjugacy class ofD 10 then it would not conclude that D 10 is cyclically generated, but if insteadit selected θ and ab as representatives then it would. The reason for this isthat the map ϕ defined in the proof of Theorem 2.5 depends on the map β,that is, it depends on the relationship between y and x.These methods allow us to search for further examples of groups that are notcyclically generated. For example, we have the following result.Theorem 2.6. Of the groups of order less than 31, only 3 are not cyclicallygenerated. These are SmallGroup([16,8]), SmallGroup([24,5]), and Small-Group([24,8]) in the GAP small groups library.Proof. The result was proved using the GAP code in Appendix A.Further investigations using GAP gave an indication of why these groups arenot cyclically generated. We now look more closely at some conditions thatensure that there is no automorphism that cycles through a generating set.

272.3 Groups that are not cyclically generatedA group cannot be cyclically generated if it is not generated by elements ofthe same order, i.e. if it is a non-GSO group, see [26] and Definition 1.9.The smallest non-GSO group is the semidihedral group of order 16, whichis SmallGroup([16,8]) in the GAP small groups library and is defined by thepresentation 〈a, b a 8 , b 2 , baba −5 〉. Searches using GAP have provided furtherexamples of non-GSO groups of order up to 1000, all of which have orderdivisible by 16. The question of whether there is a non-GSO p-group for anodd prime p is the subject of Problem 11.6 in the Kourovka Notebook, see[24], which readsLet p be an odd prime. Is it true that every finite p-group possesses a set ofgenerators of equal orders?This was answered in 2004 by E. A. O’Brien, Carlo M. Scoppola, and M.R. Vaughan-Lee, see [26]. They proved that for any odd prime p, there is agroup of exponent p p+1 that is not generated by elements of the same orderand it follows from this that, for every prime p, there is a p-group that is notcyclically generated. However, before this result was proved, searches thatwe had carried out using GAP had already identified examples of groups thatare not cyclically generated, some of which are p-groups of odd order. Forexample, SmallGroup([243,5]), which is defined by the presentation〈a, b, c, d a 9 , b 9 , a 3 b −3 , c 3 , d 3 , [c, d], [a, b]c, [a, c]a 3 , [b, c]d 〉,is one of 4 groups of order 243 that are not cyclically generated. Thereare also examples of groups of odd order that are not p-groups and are notcyclically generated. For example, SmallGroup([567,9]), which is defined bythe presentation〈a, b, c a 9 , b 9 , c 7 , [a, b]a 3 , [a, c]c, [b, c], [a 3 , c]〉,is one of 7 groups of size 567 that are not cyclically generated. As all ofthese examples are GSO, they prove that it is not true that a group thatis GSO is necessarily cyclically generated. The smallest GSO groups that

28are not cyclically generated are SmallGroup([24,5]), which is S 3 × C 4 , andSmallGroup([24,8]), which is defined by the presentation〈a, b, c a 2 , b 3 , (ab) 2 , c 2 , [b, c], [a, c] 2 〉.Further investigation of these groups found that they cannot have a cyclicgenerating set because every element lies in a proper characteristic subgroup.GAP was then used to prove that SmallGroup([243,5]) and Small-Group([567,9]) also have this property.2.3.1 ECS groupsRecall that a characteristic subgroup H of a finite group G is a subgroup Hthat is preserved by automorphisms of G, i.e. Hθ ⊆ H, for all θ in Aut G, soif x ∈ H and y ∉ H then there is no automorphism of G such that x ↦→ y.It follows immediately that a group cannot be cyclically generated if everyelement lies in a proper characteristic subgroup. We will call a group withthis property an ECS group, see Definition 1.11. The smallest examples ofECS groups are the semidihedral group of order 16, SmallGroup([24,8]), andS 3 × C 4 . No other group of order less than 32 is ECS.Certain subgroups of a group G are known to be characteristic. For example,the centre, Z(G), and the Frattini subgroup, Φ(G), as are the subgroupsgenerated by all the elements of order m, for any relevant m. The subgroupsmG and G(m) defined bymG = 〈g m : g ∈ G〉andG(m) = 〈g ∈ G : o(g)|m〉are characteristic, see [30] p.42 and p.85. Sometimes we can use these subgroupsto prove that a given group is ECS and if this is the case then wehave proved that the group in question is not cyclically generated withoutexplicitly considering any of its automorphisms. We can also prove that certainsubgroups are characteristic by proving that they are a characteristic

29subgroup of a known characteristic subgroup. For example, for any groupG, and any relevant integer m, the subgroup mZ(G) is characteristic. It iseasy to search for (small) examples of ECS groups using GAP, and suitablecode can be found in Appendix A. Every non-GSO group is an ECS groupbecause every element x lies in the subgroup generated by elements of ordero(x). As the group is non-GSO, these characteristic subgroups are all propersubgroups. It is not true, however, that a group that is not ECS is necessarilycyclically generated. SmallGroup([32,24]) in the GAP small groups library isthe smallest example of a group that is neither ECS nor cyclically generated.It is defined by the presentation〈a, b, c a 4 , b 2 , c 4 , [a, b]c 2 , [a, c], [b, c]〉.2.3.2 Using characteristic subgroupsSuppose that we want to prove that a GSO group G is not cyclically generated.If S is a GSO generating set for G then we already know that anyelement of S that lies in a proper characteristic subgroup cannot be in acyclic generating set, so in particular we can consider those subgroups thatare known to be characteristic, such as the centre of the group, and removefrom S any element that lies in one of these subgroups. Let T be the subsetof S consisting of the elements that do not lie in any of these familiar characteristicsubgroups. If we can show that T is empty for each possible GSOgenerating set S then we will have proved that G cannot be cyclically generated.If, however, there is some S for which T is non-empty then we mightstill be able to prove that G is not cyclically generated without determiningits automorphism group. This is because if the elements of a GSO generatingset satisfy the following condition then there cannot be an automorphismthat cycles through the set.Lemma 2.7. Let G be a finite group and let X be a GSO generating set for Gconsisting of elements of order n. Suppose there exists a proper characteristicsubgroup C of G, and elements g, h of X, such that for some m dividing n,g m ∈ C but h m ∉ C.

30Then g and h cannot belong to the same cyclic set.Proof. Suppose X, C, g, h, m, n satisfy the conditions in the lemma. If g andh were in the same cyclic set then there would exist an automorphism θ suchthat gθ i = h for some i. But thenh m = (gθ i ) m = (g m )θ i ∈ C,which is a contradiction.Therefore we might be able to use Lemma 2.7 to split each non-empty T intosmaller subsets. If we can then show that, for each T, none of these subsetsgenerate G, we will have proved that G is not cyclically generated. Again wecan use subgroups that are known to be characteristic and consider whethercertain powers of elements of T are in these subgroups. This method is usedin Chapter 3 to prove that D 2m × C 4 is not cyclically generated for any m.Another way in which we can use characteristic subgroups is by noting that ifa cyclically generated group is factored by a characteristic subgroup then theautomorphism that cycles through the generating set for the original groupinduces an automorphism on the factor group, and this induced automorphismalso cycles through a generating set.Lemma 2.8. Let G be a cyclically generated group. If K is a characteristicsubgroup of G then the factor group G/K is also cyclically generated.Proof. Assume x ∈ G and θ ∈ Aut G cyclically generate G, and define amap ˆθ on G/K by(gK)ˆθ = gθK.We show that ˆθ is an automorphism and then show that it cycles through agenerating set.Suppose y, z in G are such that yK = zK. ThenyK = zK ⇒z −1 y ∈ K ⇒ (z −1 y)θ ∈ Kθ ⊆ K ⇒ z −1 θyθ ∈ K⇒ (zθ) −1 yθ ∈ K ⇒ yθK = zθK ⇒ (yK)ˆθ = (zK)ˆθ,

31so θ is well defined. It is a homomorphism because(yK)ˆθ(zK)ˆθ = yθKzθK = yθzθK = (yz)θK = (yzK)ˆθ = [(yK)(zK)]ˆθ.Finally, let ˆθ −1 be defined by(gK)ˆθ −1 = gθ −1 K.Then ˆθ −1 is also a well defined homomorphism by the above arguments, and(gK)ˆθ −1ˆθ = (gθ −1 K)ˆθ = gθ −1 θK = gK, and similarly (gK)ˆθˆθ −1 = gK,so ˆθ is an automorphism of G/K.Let gK be an arbitrary element of G/K. As x and θ cyclically generate G,g can be written as a word w = w 1 w 2 · · · w l where each w j is xθ i for some i.ThengK = wK = (w 1 w 2 · · · w l )K = (w 1 K)(w 2 K) · · · (w l K)where each w j K is xθ i K = (xK)ˆθ i , so gK is a word in the elements (xK)ˆθ i ,and hence ˆθ and xK cyclically generate G/K.It is important to note here that we are factoring by a characteristic ratherthan just a normal subgroup, as every finite group is the homomorphic imageof a finitely generated free group and these groups are cyclically generated.The converse of Lemma 2.8 is not true because, for example, for any group G,the derived subgroup G ′ is characteristic and the factor group G/G ′ is abelianand hence cyclically generated. This means that we cannot use Lemma 2.8to build infinite families of cyclically generated groups from small examples.However, to see how Lemma 2.8 might be used, consider the example of thenonabelian group of order 21. This was proved to be cyclically generatedby Tomaszewski, see [32], but if we didn’t already know that this group iscyclically generated, we could prove that it is by showing that it is isomorphicto the factor group F (4, 3)/Z(F (4, 3)) where F (4, 3) is the Fibonacci groupdefined by the cyclic presentation〈x 1 , x 2 , x 3 , x 4 x 1 x 2 x 3 x −14 , x 2 x 3 x 4 x −11 , x 3 x 4 x 1 x −12 , x 4 x 1 x 2 x −13 〉.

32As the centre of any group is characteristic, the result follows by Lemma 2.8.In general, the harder part of the problem is proving that a given groupis not cyclically generated. The following result, which is an immediateconsequence of Lemma 2.8, enables us to construct infinite families of groupsthat are not cyclically generated by showing that all the groups in the familyare extensions of a characteristic subgroup by a group that we know is notcyclically generated.Theorem 2.9. If a group G has a characteristic subgroup K such that G/Kis not cyclically generated then G is not cyclically generated.For example, if H is the non-GSO group of order 16 and K is a group ofodd order, any group G that is an extension of K by H cannot be cyclicallygenerated because K is a characteristic subgroup of G and G/K = H, whichis not cyclically generated. In particular, the direct product of H and a groupof odd order is not cyclically generated.Another example of a direct product that is not cyclically generated is S 3 ×C 4 ,which is SmallGroup([24,5]), but since D 12 = S 3 × C 2 and Q 8 × C 2 is not isomorphicto SmallGroup([16,8]), we know that some nonabelian direct productsare cyclically generated. It is therefore natural to ask whether otherdirect products of an abelian group and a nonabelian group are cyclicallygenerated. A survey of the direct products of cyclic groups with all nonabeliangroups of order at most 16 is the subject of Chapter 3. The proofsof the results of this survey use some of the results from this chapter includingone from the next section, in which we look more closely at some directproducts.2.4 Direct productsThe example of S 3 × C 4 shows that it is not true that the direct product oftwo cyclically generated groups is necessarily cyclically generated. This is incontrast to the equivalent result for GSO groups, where the direct productdoes inherit the GSO property from its direct factors. The following proof

33of this result was given by Edmund Robertson and James Wiegold as partof their investigation into non-GSO groups, see [13]. Suppose that G and Hare generated by the GSO sets X and Y, consisting of elements of orders mand n respectively. Let k be the least common multiple of m and n. Withoutloss of generality, we can assume that the highest power of 2 dividing m alsodivides n and therefore, for each x in X and each y in Y, the element (x 2 , y)of G × H has order k. Since (x −1 , y −1 ) also has order k, the elements (x, 1)and hence (1, y) can be written as words in the elements of order k in G × H.There are also examples of cyclically generated groups that are the directproduct of groups that are not cyclically generated. For example, S 3 × C 4 isnot cyclically generated, but (S 3 × C 4 ) × (S 3 × C 4 ) is cyclically generated, soin particular, it is not an ECS group, even though S 3 ×C 4 is ECS. However, ifG is SmallGroup([24,8]) then both G and G × G are ECS groups, so whetherthe direct product of two copies of an ECS group is itself ECS depends onthe group in question.As these examples indicate that a classification result for general direct productswill be difficult to find, we focus instead on situations in which we canprove partial results. We begin by considering the question of whether thedirect product of a group with itself is cyclically generated. By Theorem 2.2,the abelian case has already been answered, but in the nonabelian case wehave the following partial result.Theorem 2.10. If a finite group G is cyclically generated by n generatorsthen for any m coprime to n, the groupis cyclically generated.K = G × G × · · · × G} {{ }mProof. We write any elements of K as the m-tuple (g 1 , g 2 , . . . , g m ) whereeach g i is in G. Let θ ∈ AutG and x ∈ G be such that x and θ cyclicallygenerate G and let θ ′ be the automorphism of K defined by(g 1 , g 2 , . . . , g m )θ ′ = (g 1 θ, g 2 θ, . . . , g m θ).

34Let ϕ be the automorphism of K that cycles through the m copies of G, so(g 1 , g 2 , . . . , g m )ϕ = (g m , g 1 , g 2 , . . . , g m−1 ).Let x i denote the element (1, 1, . . . , 1, x, 1, . . . , 1) where x is in the ith position.By successively applying the mn powers of ϕ ◦ θ ′ to x 1 we getx 1 ↦→ x 2 θ ↦→ · · · ↦→ x m θ n−1 ↦→ x 1 .Let S = {x 1 (φ ◦ θ ′ ) j : j = 0, 1, 2, . . . , mn − 1}. Then because m and n arecoprime, x i θ j is in S for each j between 0 and n − 1, and each i between 1and m. Therefore, the subgroup of K generated by S is G.From this result we know that, in particular, if G is cyclically generatedby an odd number of generators then G × G is also cyclically generated.For example, as S 3 is cyclically generated by 3 generators, we can use theproof of Theorem 2.10 to construct an automorphism that cycles through agenerating set for S 3 × S 3 .Let K = S 3 × S 3 , let x, y, z be a cyclic generating set for S 3 , and let θ denotethe automorphism x ↦→ y ↦→ z ↦→ x. A general element of K will be writtenas a pair (g 1 , g 2 ) and we define θ ′ by (g 1 , g 2 )θ ′ = (g 1 θ, g 2 θ). Let ϕ be theautomorphism of K that swaps the two copies of S 3 . Then applying ϕ ◦ θ ′ to(x, 1) we get(x, 1) ↦→ (1, y) ↦→ (z, 1) ↦→ (1, x) ↦→ (y, 1) ↦→ (1, z) ↦→ (x, 1),which is clearly a generating set for K.In the following situation we consider the direct product of two cyclicallygenerated groups.Lemma 2.11. Let G be cyclically generated by elements of order m and letH be cyclically generated by elements of order n. If m and n are coprime,then the direct product G × H is cyclically generated.Proof. Let x ∈ G and θ ∈ Aut G cyclically generate G, and y ∈ H andϕ ∈ Aut H cyclically generate H. Consider the elements of G × H as pairs

35(g, h) and define a map ψ : G × H → G × H by(g, h)ψ = (gθ, hϕ).From the assumption that θ and ϕ are automorphisms it follows that ψ isan automorphism and therefore we only have to show that ψ cycles througha generating set.Applying successive powers of ψ to the element (x, y) gives(x, y) ↦→ (xθ, yϕ) ↦→ (xθ 2 , yϕ 2 ) ↦→ · · · ↦→ (x, y).As m and n are coprime there exist integers s and t such that 1 = sm + tn,so for any j we have((x, y)ψ j ) tn = (xθ j , yϕ j ) tn = ((xθ j ) tn , (yϕ j ) tn ) = ((xθ j ) 1−sm , 1) = (xθ j , 1)and similarly ((x, y)ψ j ) sm = (1, yϕ j ). As the xθ j s generate G and the yϕ j sgenerate H, the (x, y)ψ j s must generate G × H.For example, if H is the extraspecial group of order 27 and exponent 9 thenH is cyclically generated by 2 elements of order 9, and if K = C 7 × C 7 thenK is cyclically generated by 2 elements of order 7. We show that the directproduct G = H × K is cyclically generated.Let x and y cyclically generate H, and a and b cyclically generate K, with theautomorphisms θ and ϕ induced by x ↦→ y ↦→ x and a ↦→ b ↦→ a respectively.Let ψ be the map defined in the proof of Lemma 2.11 and apply powers ofψ to (x, a) to get(x, a) ↦→ (y, b) ↦→ (x, a).Let N be the subgroup generated by {(x, a), (y, b)}. As 1 = −27 + 28 wehave (x, a) 28 = (x, 1) and (y, b) 28 = (y, 1). Similarly, (x, a) −27 = (1, a) and(y, b) −27 = (1, b), so N = G.The condition in Lemma 2.11 that m and n are coprime is sufficient but notnecessary. For example, S 3 × C 2 and S 3 × Q 8 are both cyclically generated.We use Lemma 2.11 to prove some of the results in Chapter 3.

Chapter 3Direct productsThe smallest example of a group that is not cyclically generated, but is generatedby elements of the same order, is the direct product S 3 × C 4 . The firstsection of this chapter contains the results of a survey to find further examplesof groups that are not cyclically generated and are the direct product ofa small nonabelian group and a cyclic group. The survey was carried out usingGAP and covered groups that are the direct products of small nonabeliangroups of orders up to 16 and a cyclic group. The results of this search allowedus to make conjectures for the case when the cyclic group had order n.These conjectures are given in a table at the beginning of the section and theproofs follow. In some cases, these proofs led to more general results. Forexample, the results for small dihedral groups led to a result for D 2m ×C n forany m and n, and the results for Q 8 ×C n and Q 8 ×C 2 ×C n led to a result forQ 8 ×A, where A is an arbitrary finite abelian group. These more general theoremsare proved in the second section. We use similar methods throughout,and in most cases we rely to some extent on computational proofs. The basicmethod can be summarized in the following way. In order to prove that agroup G is cyclically generated, we either construct a suitable automorphism,or use Lemma 2.11 to show that such an automorphism exists. To prove thatG is not cyclically generated, we factor by characteristic subgroups to reduceG to a group that we can prove is not cyclically generated using GAP. It thenfollows by Theorem 2.9 that G is not cyclically generated. The GAP codeused to prove that certain group are not cyclically generated can be found37

38in Appendix A.One result that will be used repeatedly is the following.Lemma 3.1. If k is odd then for any λ ≥ 1, the numbers k + 2 λ and 2k arecoprime.Proof. Let d = (k +2 λ , 2k) and suppose d > 1. Then as d divides k +2 λ andk is odd, d must be odd, and hence d must divide k because d divides 2k.But as d divides k and k +2 λ , it must divide 2 λ , which is a contradiction.3.0.1 NotationFor ease of notation, elements of the direct product H × K will be writtenas words of the form hk where h ∈ H and k ∈ K, rather than as pairs (h, k).Unless otherwise stated, elements of the nonabelian group H in the directproduct H ×K will be denoted by letters from the beginning of the alphabet,and elements of the abelian group K will be denoted by letters from the endof the alphabet.If θ is an automorphism of H and ϕ is an automorphism of K, then we willsometimes use θ and ϕ to construct an automorphism of H × K. By thephrase ‘extend θ to an automorphism of H × K in the natural way’ we meandefine the automorphism of H × K that acts in the same way as θ on H andas the identity on K.3.1 Survey resultsThe table below lists the conjectures we were able to make, based on resultsobtained using GAP. Note that in the table CG stands for cyclically generatedand n/a written in the second column of the table refers to the fact that thegroup G is cyclically generated for all n. The groups T and G 1 , . . . , G 5 aredefined as followsT = 〈a, b a 2 b −3 , b 6 , baba −1 〉 is SmallGroup([12,1]),

39G 1 = 〈a, b a 2 , b 8 , abab −3 〉 is SmallGroup([16,8]),G 2 = 〈a, b a 8 , b 2 , aba −5 b〉 is SmallGroup([16,6]),G 3 = 〈a, b a 4 , b 4 , abab −1 〉 is SmallGroup([16,4]),G 4 = 〈a, b, c a 4 , b 2 , c 2 , cbca 2 b, baba −1 , caca −1 〉 is SmallGroup([16,13]),G 5 = 〈a, b a 4 , b 4 , (ab) 2 , (ba −1 ) 2 〉 is SmallGroup([16,3]).G Smallest not CG example Conjectured condition on n̸̸̸̸̸̸̸̸̸̸̸̸̸S 3 × C n S 3 × C 4 S 3 × C n is CG ⇔ 4 |nD 8 × C n D 8 × C 4 D 8 × C n is CG ⇔ 4 |nQ 8 × C n n/a Q 8 × C n is CG for every nD 10 × C n D 10 × C 4 D 10 × C n is CG ⇔ 4 |nA 4 × C n n/a A 4 × C n is CG for every nT × C n T × C 8 T × C n is CG ⇔ 8 |nD 12 × C n D 12 × C 4 D 12 × C n is CG ⇔ 4 |nD 14 × C n D 14 × C 4 D 14 × C n is CG ⇔ 4 |nD 16 × C n D 16 × C 4 D 16 × C n is CG ⇔ 4 |nQ 16 × C n Q 8 × C 4 Q 16 × C n is CG ⇔ 4 |n(D 8 × C 2 ) × C n D 8 × C 2 × C 4 D 8 × C 2 × C n is CG ⇔ 4 |n(Q 8 × C 2 ) × C n n/a Q 8 × C 2 × C n is CG for every nG 1 × C n G 1 × C 2 G 1 × C n is not CG for any nG 2 × C n G 2 × C 4 G 2 × C n is CG ⇔ 4 |nG 3 × C n G 3 × C 4 G 3 × C n is CG ⇔ 4 |nG 4 × C n G 4 × C 4 G 4 × C n is CG ⇔ 4 |nG 5 × C n G 5 × C 4 G 5 × C n is CG ⇔ 4 |nConjectures made from survey results

403.1.1 ProofsTheorem 3.2. S 3 ×C n is cyclically generated if and only if 4 does not dividen.Proof. If 4 does not divide n then either n is odd or n = 2k where k is odd.We consider these cases separately.Suppose first that n is odd and note that S 3 is cyclically generated by theelements ( 1 2), ( 2 3), and ( 3 1). As n is odd, any generator of C n has oddorder, and the result follows from Lemma 2.11.Now suppose n = 2k where k is odd.Let C 2k be generated by x. Then S 3 × C 2k is defined by the presentation〈a, b, x a 2 , b 2 , (ab) 3 , x 2k , [a, x], [b, x]〉.Define an onto map θ on S 3 × C 2k bya ↦→ bx k , b ↦→ abax k , x ↦→ x.Note that(aθ) 2 = (bx k ) 2 = b 2 x 2k = 1(bθ) 2 = (abax k ) 2 = (aba) 2 x 2k = 1(aθbθ) 3 = (bx k abax k ) 3 = (baba) 3 x 6k = 1and since x is fixed by θ,(xθ) 2k = [aθ, xθ] = [bθ, xθ] = 1,so this map can be extended to an automorphism, and from now on θ willdenote this automorphism.Applying successive powers of θ to the element ax 2 givesax 2 ↦→ bx k+2 ↦→ abax 2 ↦→ ax k+2 ↦→ bx 2 ↦→ abax k+2 ↦→ ax 2and it remains to show that if S = {ax 2 , bx k+2 , abax 2 , ax k+2 , bx 2 , abax k+2 },then S generates S 3 × C 2k . Let H be the subgroup generated by S.

41Note that(ax 2 )(ax k+2 ) = ax 2 ax k+2 = a 2 x k+4 = x k+4and since k is odd, k + 4 is coprime to 2k by Lemma 3.1. Therefore x ∈ Hand hence a, b ∈ H, so H = S 3 × C 2k .Conversely, we use Theorem 2.9 to show that S 3 ×C 4k is not cyclically generatedfor any k. We begin by showing that S 3 × C 4 is not cyclically generated.Note thatS 3 × C 4 = 〈a, b, xa 2 , b 2 , (ab) 3 , x 4 , [a, x], [b, x]〉.Clearly the elements of order 2 do not generate S 3 × C 4 , and the sets ofelements of orders 3, 6, and 12 are not generating sets for S 3 × C 4 becausethe only elements of these orders in S 3 ×C 4 have a 3-cycle as their permutationpart. Therefore a GSO generating set for S 3 ×C 4 can only consist of elementsof order 4. These are the elements ax, bx, abax, ax 3 , bx 3 , abax 3 , x, x 3 , of whichx and x 3 are in the centre of the group and therefore cannot be in a cyclicgenerating set. LetH = 〈ax, bx, abax, ax 3 , bx 3 , abax 3 〉.As abax = (abax 3 ) −1 and abax 3 = axbxax, we have H = 〈ax, bx, ax 3 , bx 3 〉,butax 3 = ax −1 = (ax) −1 ,andsobx 3 = bx −1 = (bx) −1 ,H = 〈ax, bx〉.If x ∈ H then x can be written as a word in ax and bx, and this word musthave length congruent to 1 modulo 4. But the permutation part of such aword will be odd, and hence it cannot be the identity. Therefore x is notin H. It follows that any GSO generating set must contain both an elementfrom the centre and an element from outside the centre, and by definitionthere cannot be an automorphism that cycles through such a set because

42the centre of a group is characteristic. Therefore S 3 × C 4 is not cyclicallygenerated.Now suppose that 4 divides n and write n = 4k for some k, which is notnecessarily odd. Let G = S 3 × C 4k . Note that Z(G) = 〈x〉 and let 4Z(G)denote the subgroup of Z(G) generated by the set {g 4 : g ∈ Z(G)}. As4Z(G) is characteristic in Z(G), which is characteristic in G, we know that4Z(G) is characteristic in G. We prove that for any k, G/4Z(G) = S 3 × C 4 .Noting that 4Z(G) = 〈x 4 〉, we see immediately thatG/4Z(G) = 〈a, b, x a 2 , b 2 , (ab) 3 , x 4k , [a, x], [b, x], x 4 〉= 〈a, b, x a 2 , b 2 , (ab) 3 , x 4 , [a, x], [b, x]〉and as S 3 × C 4 is not cyclically generated, it follows by Theorem 2.9 that Gcannot be cyclically generated.The same condition on n holds for D 8 × C n .Theorem 3.3. D 8 ×C n is cyclically generated if and only if 4 does not dividen.Proof. If n is odd then the result follows by Lemma 2.11 because D 8 iscyclically generated by elements of order 2. If n = 2k where k is odd thenwe construct an automorphism and prove that it cycles through a generatingset.As D 8 × C 2k is defined by the presentation〈a, b, x a 2 , b 2 , (ab) 4 , x 2k , [a, x], [b, x]〉,we define an onto map θ bya ↦→ b, b ↦→ ax k , x ↦→ x.If we apply θ to each of the generators in the relators in the presentationabove we get(aθ) 2 = b 2 = 1, (bθ) 2 = a 2 x 2k = 1, (aθbθ) 4 = (bax k ) 4 = (ba) 4 x 4k = 1,

43and θ fixes x so (xθ) 2k = [aθ, xθ] = [bθ, xθ] = 1. Therefore θ can be extendedto a homomorphism, and as it maps onto a generating set for the group, it isan automorphism. Now let θ denote this automorphism. We apply successivepowers of θ to ax :ax ↦→ bx ↦→ ax k+1 ↦→ bx k+1 ↦→ ax 2k+1 = ax.Suppose that H is the subgroup generated by the set {ax, bx, ax k+1 , bx k+1 }.Then H contains the element (ax k+1 )(ax) 3 = a 4 x k+4 = x k+4 and as k is odd,k + 4 is coprime to 2k by Lemma 3.1. Therefore x is in H and hence a andb are in H, so H = D 8 × C 2k as required.Now suppose 4 divides n, so n = 4k for some k. Let C 4k be generated by xand let G = D 8 × C 4k . Then G is defined by the presentation〈a, b, x a 2 , b 2 , (ab) 4 , x 4k , [a, x], [b, x]〉.The subgroup generated by the 4th powers of the elements of D 8 × C 4k is4G = 〈(σx i ) 4 : σ ∈ D 8 , i = 0, 1, 2, 3 〉= 〈σ 4 x 4i : σ ∈ D 8 , i = 0, 1, 2, 3 〉= 〈x 4 〉and henceG/4G = 〈a, b, x a 2 , b 2 , (ab) 4 , x 4k , [a, x], [b, x], x 4 〉= 〈a, b, x a 2 , b 2 , (ab) 4 , x 4 , [a, x], [b, x]〉,which is a presentation for D 8 × C 4 . Either GAP or a similar argument to theone used in the proof of Theorem 3.2 can be used to show that D 8 × C 4 isnot cyclically generated, and therefore D 8 × C 4k is not cyclically generated,by Theorem 2.9.The details of the proof that D 8 ×C 4 is not cyclically generated can be foundin the proof of Theorem 3.18 in the next section.Theorem 3.4. For any integer n, Q 8 × C n is cyclically generated.

44Proof. Recall that Q 8 , the quaternion group of order 8, is the Fibonaccigroup F (2, 3) and is therefore defined by the cyclic presentation〈a, b, c abc −1 , bca −1 , cab −1 〉,see [20] p.75. It follows that Q 8 is cyclically generated by a, b, and c, whichhave order 4. If x generates C n , then Q 8 × C n is presented by〈a, b, c, x abc −1 , bca −1 , cab −1 , x n , [a, x], [b, x], [c, x]〉.We know there is an automorphism a ↦→ b ↦→ c ↦→ a, so we can defineθ ∈ Aut(Q 8 × C n ) to be the automorphism induced by the mapa ↦→ b, b ↦→ c, c ↦→ a, x ↦→ x.Applying successive powers of θ to ax givesax ↦→ bx ↦→ cx ↦→ axand(ax)(bx)(cx) −1 = abc −1 x 2 x −1 = x,so θ cycles through a generating set for Q 8 × C n .Theorem 3.5. The group D 10 × C n is cyclically generated if and only if 4does not divide n.Proof. As D 10 is cyclically generated by elements of order 2, we can useLemma 2.11 to show that D 10 × C n is cyclically generated if n is odd.If n = 2k where k is odd then we use the usual presentation for D 10 to saythat G = D 10 × C 2k is presented by〈a, b, x a 2 , b 5 , (ab) 2 , x 2k , [a, x], [b, x]〉.Define an onto map θ : G → G bya ↦→ b −1 abx k , b ↦→ b, x ↦→ x.

45As θ fixes b and x we just need to check that replacing each generator by itsimage under θ preserves the first and third relators:(aθ) 2 = (b −1 abx k ) 2 = (b −1 ab) 2 x 2k = b −1 a 2 bx 2k = b −1 b = 1,and(aθbθ) 2 = (b −1 abx k b) 2 = (b −1 abbb −1 abb)x 2k = b −1 (ab) 2 b = 1.Therefore θ induces an automorphism of G, which we will now denote byθ. It remains to show that θ cycles through a generating set. By applyingsuccessive powers of θ to ax 2 we getax 2 ↦→ b −1 abx k+2↦→ b −1 (b −1 abx k )bx k+2 = b −2 ab 2 x 2↦→ b −2 (b −1 abx k )b 2 x 2 = b −3 ab 3 x k+2↦→ b −3 (b −1 abx k )b 3 x k+2 = b −4 ab 4 x 2↦→ b −4 (b −1 abx k )b 4 x 2 = b −5 ab 5 x k+2 = ax k+2↦→ b −1 abx k x k+2 = b −1 abx 2↦→ b −2 ab 2 x k+2 ↦→ b −3 ab 3 x 2 ↦→ b −4 ab 4 x k+2 ↦→ ax 2 .Let H be the subgroup generated by these ten elements. As(ax 2 )(ax k+2 ) = a 2 x k+4 = x k+4 ,and k is odd, x must be in H because 2k and k + 4 are coprime by Lemma3.1. Therefore a and b −1 ab are also in H and it follows that H contains bbecause(b −1 ab)a = b −1 (aba) = b −2and b has order 5. Therefore D 10 × C 2k is cyclically generated if k is odd.If n = 2 λ+2 r, where r is odd and λ ≥ 0, then we can show that D 10 × C n isnot cyclically generated by using Theorem 2.9 and the fact that GAP can beused to show that D 10 × C 4 is not cyclically generated. LetG = D 10 × C n = 〈a, b, x a 2 , b 5 , (ab) 2 , x 2λ+2r , [a, x], [b, x]〉.

46The centre of G is 〈x〉 because the centre of D 10 is trivial, and hence4Z(G) = 〈x 4 〉,soG/4Z(G) = 〈a, b, x a 2 , b 5 , (ab) 2 , x 2λ+2r , [a, x], [b, x], x 4 〉= D 10 × C 4 ,which is not cyclically generated.We generalize the proofs of Theorems 3.3 and 3.5 in the next section to provethat D 2m × C n is cyclically generated if and only if 4 does not divide n. Asthe results for D 12 × C n , D 14 × C n , and D 16 × C n follow from this result, weomit the proofs of these results.Theorem 3.6. For any n, the group A 4 × C n is cyclically generated.Proof. Recall that A 4 is cyclically generated by two generators of order 3(see (2.2)), so if 3 does not divide n then the result is immediate from Lemma2.11.Now suppose that 3 divides n and let 3 i be the highest power of 3 dividingn, so n = 3 i m for some m not divisible by 3. We decompose C n as C 3 i × C mand show that A 4 × C 3 i is cyclically generated by 2 elements of order 3 i . Itwill then follow by Lemma 2.11 that A 4 × C n is cyclically generated.From the generalized cyclic presentation for A 4 in the previous chapter, weknow that A 4 × C 3 i is defined by the presentation〈a, b, x a 3 , b 3 , (ab) 2 , x 3i , [a, x], [b, x]〉,and there is an automorphism of A 4 such that a ↦→ b ↦→ a. We extend thisto an automorphism of A 4 × C 3 i by x ↦→ x. Applying this automorphism toax givesax ↦→ bx ↦→ ax,and as((ax)(bx)) 2 = (ab) 2 x 4 = x 4

47and x has order a power of 3, this shows that ax and bx generate A 4 × C 3 i.As ax has order 3 i , it follows by Lemma 2.11 that A 4 × C n is cyclicallygenerated.Theorem 3.7. Let T be the group of order 12 defined by the presentation〈a, b a 2 b −3 , b 6 , baba −1 〉.Then the group T × C n is cyclically generated if and only if 8 does not dividen.Proof. We begin with a proof that T is cyclically generated by two elementsof order 4. Let θ : T → T be the map defined by a ↦→ ba and b ↦→ b −1 . Thenclearly (bθ) 6 = 1, and(aθ) 2 (bθ) −3 = (ba) 2 b 3 = babab 3 = (bab)ab 3 = a 2 b 3 = b 3 b 3 = 1,and(bθ)(aθ)(bθ)(aθ) −1 = b −1 bab −1 a −1 b −1 = ab −1 a −1 b −1 = (baba −1 ) −1 = 1,so θ extends to an automorphism of T. Under this automorphism,a ↦→ ba ↦→ b −1 ba = a,so T has a cyclic generating set.Therefore, if n is odd, T × C n is cyclically generated by Lemma 2.11. If n iseven then we consider separately the two cases 4 ̸ |n and 4|n.Suppose first that 4 does not divide n, so n = 2k where k is odd. Let xgenerate C n , so T × C n has presentation〈a, b, x a 2 b −3 , b 6 , baba −1 , x 2k , [a, x], [b, x]〉.Define an onto map θ bya ↦→ ab −1 x k , b ↦→ b −1 , x ↦→ x −1 .Clearly replacing a, b, and x by their images under θ does not affect therelators involving only b or x or the commutator relators, but we need to

48verify that this is also the case for the other relators. Note that from thethird relator, ba = ab −1 , so from the first relator we get(aθ) 2 (bθ) −3 = (ab −1 x k )(ab −1 x k )(b 3 )= ab −1 ab −1 x 2k b 3= (ab −1 ) 2 b 3= (ba) 2 b 3= (bab)ab 3= a 2 b 3= a 2 b −3= 1,and from the third relator we get(bθ)(aθ)(bθ)(aθ) −1 = b −1 ab −1 x k b −1 x −k ba −1= b −1 ab −1 b −1 ba −1= (b −1 a)b −1 a −1= (ab)b −1 a −1= aa −1= 1,so θ induces an automorphism, which will now be denoted by θ.We apply successive powers of θ to ax to getax ↦→ ab −1 x k−1 ↦→ ab −1 x k bx −k+1 = ax.Let H = 〈ax, ab −1 x k−1 〉. Then(ax)(ab −1 x k−1 ) = a 2 b −1 x k = b 3 b −1 x k = b 2 x kand(b 2 x k ) 4 = b 8 x 4k = b 8 = b 2so b 2 is in H. Also,(b 2 x k )(ax) k = b 2 a k x 2k = b 2 a k ,

49so b 2 a k in in H, and as b 2 is in H, this means that a k and hence a are in H.It follows that x and b are in H, so T × C 2k is cyclically generated.Now suppose n = 4k where k is odd:T × C 4k = 〈a, b, x a 2 b −3 , b 6 , baba −1 , x 4k , [a, x], [b, x]〉.Define an onto map θ bya ↦→ b −1 a −1 x −k , b ↦→ b −1 x −2k , x ↦→ x −1 .Then (bθ) 6 = b −6 x −12k = 1, and [aθ, xθ] = [bθ, xθ] = 1, and(aθ) 2 (bθ) −3 = (b −1 a −1 x −k ) 2 (b −1 x −2k ) −3= (b −1 a −1 ) 2 b 3 x 4k= (b −1 a −1 ) 2 b 3= (b −1 a −1 b −1 )a −1 b 3= a −1 a −1 b 3= 1,and(bθ)(aθ)(bθ)(aθ) −1 = (b −1 x −2k )(b −1 a −1 x −k )(b −1 x −2k )(x k ab)= b −2 a −1 b −1 abx −2k−k−2k+k= b −2 a −1 b −1 ab= b −1 (b −1 a −1 b −1 )ab= b −1 a −1 ab= 1,so θ induces an automorphism. Let θ now denote this automorphism. Applyingθ to ax givesax ↦→ b −1 a −1 x −k−1 ↦→ x 2k bx k abx k+1 = babx 4k+1 = ax.Let H be the subgroup generated by ax and b −1 a −1 x −k−1 . Then H containsthe element((ax)(b −1 a −1 x −k−1 ) −1 ) 3 = (a 2 bx 2+k ) 3 = (a 2 b) 3 x 6+3k = b 12 x 6+3k = x 6+3k ,

50and therefore H contains(ax) −(6+3k) x 6+3k = a 3k+6 .As k is odd, so is 3k + 6, and hence (4, 3k + 6) = 1, so a is in H andH = T × C n as required.Finally we prove that if 8 divides n then T ×C n cannot be cyclically generated.It can be proved using GAP, see Appendix A, that T × C 8 is not cyclicallygenerated and we prove that if 8 divides n, then T × C n has a characteristicsubgroup H such that (T × C n )/H is isomorphic to T × C 8 .Let n = 8k, where k is not necessarily odd, and let G = T × C n , soG = 〈a, b, x a 2 b −3 , b 6 , baba −1 , x 8k , [a, x], [b, x]〉.We want to identify the generators of the centre of G. Clearly x is a generatorof Z(G), so we only need to identify the generators of the centre of T. Asb 3 = a 2 , we know that b 3 commutes with a, andT/〈b 3 〉 = 〈a, b a 2 , b 3 , baba −1 〉 = S 3 ,which has trivial centre, and therefore Z(T ) = 〈b 3 〉. It follows that Z(G) isgenerated by b 3 and x and hence8Z(G) = 〈(b 3 ) 8 , x 8 〉 = 〈(b 6 ) 4 , x 8 〉 = 〈x 8 〉.The subgroup 8Z(G) is characteristic in G, so if G/8Z(G) is not cyclicallygenerated then neither is G, by Theorem 2.9. AsG/8Z(G) = 〈a, b, x a 2 b −3 , b 6 , baba −1 , x 8k , [a, x], [b, x], x 8 〉 = T × C 8 ,T × C 8k is not cyclically generated.Theorem 3.8. Q 16 × C n is cyclically generated if and only if 4 does notdivide n.Proof. As Q 16 has order 16, the result follows from Lemma 2.11 if n is odd.If n = 2k where k is odd we construct an automorphism and prove that itcycles through a generating set. LetG = Q 16 × C 2k = 〈a, b, x a 4 b −2 , b −1 aba, x 2k , [a, x], [b, x]〉.

51Note thata 8 = a 4 b 2 = a 3 ba −1 b = a 2 ba −2 b = aba −3 b = ba −4 b = bb −2 b = 1.We define an onto map θ bya ↦→ a −1 x −k , b ↦→ ab, x ↦→ x.As θ fixes x, we know that (xθ) 2k = [aθ, xθ] = [bθ, xθ] = 1, and as(aθ) 4 (bθ) −2 = (a −1 x −k ) 4 (ab) −2= a −4 x −4k b −1 a −1 b −1 a −1= a −4 b −1 a −1 b −1 a −1= a −4 b −1 (a −1 b −1 a −1 )= a −4 b −1 b −1= a −8= 1and(bθ) −1 (aθ)(bθ)(aθ) = (ab) −1 a −1 x −k aba −1 x −k= (ab) −1 a −1 aba −1= b −1 a −1 ba −1= (b −1 ab) −1 a −1= aa −1= 1,we know that θ induces an automorphism, which we now denote by θ.Applying θ repeatedly to bx givesbx ↦→ abx ↦→ a −1 x −k abx = bx 1−k ↦→ abx 1−k ↦→ a −1 x −k abx 1−k = bx.Let S = {bx, abx, bx 1−k , abx 1−k }. As (abx)(bx) −1 = a, the subgroup generatedby the set S must contain a. By Lemma 3.1, k + 4 is coprime to 2k, and(bx) 5 (bx 1−k ) −1 = b 5 b −1 x 5+k−1 = b 4 x k+4 = x k+4

52so x is in the subgroup generated by S. Hence b is also in this subgroup andtherefore Q 16 × C 2k is cyclically generated when k is odd.Now suppose that 4 divides n. We need to show that Q 16 ×C n is not cyclicallygenerated. As usual, we do this by showing that there is a characteristicsubgroup K such that Q 16 × C n factored by K is isomorphic to Q 16 × C 4 ,and as GAP can be used to prove that Q 16 × C 4 is not cyclically generated,the result follows by Theorem 2.9.LetG = Q 16 × C n = 〈a, b, x a 4 b −2 , b −1 aba, x n , [a, x], [b, x]〉where 4 divides n. We begin by identifying the centre of G. Clearly x lies inZ(G), so we just need to identify the centre of Q 16 . By the first relator, a 4 iscentral in Q 16 , so if H = 〈a 4 〉 thenQ 16 /H = 〈a, b a 4 b −2 , b −1 aba, a 4 〉= 〈a, b a 4 , b 2 , b −1 aba〉= D 8 .Therefore, if Z(Q 16 ) is not H, then the only other elements that could bein Z(Q 16 ) are a 2 , and a −2 . But in Q 16 , a 2 b = aba −1 = ba −2 and ba −2 ≠ ba 2 ,so a 2 is not in Z(Q 16 ). Therefore Z(G) is generated by a 4 and x, and hence4Z(G) = 〈g 4 : g ∈ Z(G)〉 = 〈x 4 〉, andG/4Z(G) = 〈a, b, x a 4 b −2 , b −1 aba, x n , [a, x], [b, x], x 4 〉= 〈a, b, x a 4 b −2 , b −1 aba, x 4 , [a, x], [b, x]〉= Q 16 × C 4 .Theorem 3.9. The group D 8 × C 2 × C n is cyclically generated if and onlyif 4 does not divide n.Proof. LetG = D 8 × C 2 × C n= 〈a, b, x, y a 2 , b 2 , (ab) 4 , x 2 , y n , [a, x], [b, x], [a, y], [b, y], [x, y]〉.

53If n is odd then the result follows by Lemma 2.11 and Theorem 3.3. If n iseven then we write n as 2 λ m, where m is odd and λ ≥ 1, soD 8 × C 2 × C n = D 8 × C 2 × C 2 λ × C m .The subgroup G(m) is C m , so by Theorem 2.9 and Lemma 2.11, G is cyclicallygenerated if and only if D 8 × C 2 × C 2 λ is. We show that D 8 × C 2 × C 2 λ iscyclically generated if and only if λ = 1.We construct an automorphism that cycles through a generating set for D 8 ×C 2 × C 2 as follows. The group D 8 × C 2 × C 2 is defined by the presentation〈a, b, x, y a 2 , b 2 , (ab) 4 , x 2 , y 2 , [a, x], [b, x], [a, y], [b, y], [x, y]〉and we define an automorphism θ by combining the usual cyclic automorphismof D 8 with a cyclic automorphism of C 2 × C 2 of order 3. This is theautomorphism induced by the mapa ↦→ b, b ↦→ a, x ↦→ y, y ↦→ xy.Repeatedly applying θ to ax givesax ↦→ by ↦→ axy ↦→ byxy = bx ↦→ ay ↦→ bxy ↦→ ayxy = ax.Let H = 〈ax, by, axy, bx, ay, bxy〉. Then H contains(axy)(ax) = a 2 x 2 y = yand hence H also contains a and b. From this it follows that x is in H, soH = D 8 × C 2 × C 2 .Conversely, we show that D 8 × C 2 × C 2 λ is not cyclically generated if λ ≥ 2.Let H = D 8 × C 2 × C 2 λ. If λ ≥ 2 then H/4H = D 8 × C 2 × C 4 and GAP canbe used to show that this is not cyclically generated.Theorem 3.10. For any n, the group Q 8 × C 2 × C n is cyclically generated.Proof. Let G = Q 8 × C 2 × C n and let n = 2 λ m where m is odd and λ ≥ 0.We prove that Q 8 × C 2 × C 2 λ is cyclically generated and then the result forG follows by Lemma 2.11.

54Note that Q 8 × C 2 × C 2 λis defined by the presentation〈a, b, c, x, y abc −1 , bca −1 , cab −1 , x 2 , y 2λ , [a, x], [b, x], [c, x], [a, y], [b, y], [c, y], [x, y]〉.Let θ be the automorphism defined by combining the automorphism a ↦→b ↦→ c ↦→ a with the cyclic automorphism of C 2 × C 2 λ defined in the proof ofTheorem 2.2, i.e. under θ,a ↦→ b, b ↦→ c, c ↦→ a, x ↦→ x, y ↦→ xy.Applying successive powers of θ to ay givesay ↦→ bxy ↦→ cx 2 y = cy ↦→ axy ↦→ bx 2 y = by ↦→ cxy ↦→ ax 2 y = ay.Let H be the subgroup generated by {ay, bxy, cy, axy, by, cxy}. From the firstrelator in the above presentation for Q 8 × C 2 × C 2 λ we have(ay)(by)(cy) −1 = abc −1 y = y,so y lies in H, and hence a, b, c, and x are also in H. Therefore H = Q 8 ×C 2 × C 2 λ and Q 8 × C 2 × C 2 λ is cyclically generated by elements of ordero(ay) = l.c.m(4, 2 λ ). AsG = Q 8 × C 2 × C n = Q 8 × C 2 × C2 λ × C m ,G must also be cyclically generated by Lemma 2.11.Theorem 3.11. If G 1 = 〈a, b a 2 , b 8 , abab −3 〉 then G 1 × C n is not cyclicallygenerated for any n.Proof. Recall that G 1 is the group of order 16 that cannot be generated byelements of the same order. It is the smallest group that is not cyclically generatedand GAP can be used to show that G 1 × C 2 is not cyclically generatedbecause it is not generated by elements of the same order, while G 1 × C 4 andG 1 × C 8 are not cyclically generated because every element lies in a propercharacteristic subgroup.Let G = G 1 × C n and write n = 2 λ m where m is odd and λ ≥ 0. As G(m) ischaracteristic in G andG/G(m) = G 1 × C 2 λ,

55it is enough to prove that for any λ, G 1 × C 2 λis not cyclically generated.There are two cases to consider. If λ ≤ 3 then, as mentioned above, G 1 ×C 2 λcan be shown to be not cyclically generated using GAP. If λ ≥ 4 then letH = G 1 × C 2 λ = 〈a, b, x a 2 , b 8 , abab −3 , x 2λ , [a, x], [b, x]〉so8H = 〈x 8 〉and thereforeH/8H = 〈a, b, x a 2 , b 8 , abab −3 , x 2λ , [a, x], [b, x], x 8 〉= 〈a, b, x a 2 , b 8 , abab −3 , x 8 , [a, x], [b, x]〉= G 1 × C 8which is not cyclically generated.Theorem 3.12. If G 2 = 〈a, b a 8 , b 2 , aba −5 b〉 then G 2 × C n is cyclically generatedif and only if 4 does not divide n.Proof. LetG = G 2 × C n = 〈a, b, x a 8 , b 2 , aba −5 b, x n , [a, x], [b, x]〉.As G 2 has order 16, the result for odd n follows from Lemma 2.11. If n = 2kwhere k is odd then we define an onto map θ bya ↦→ ab, b ↦→ b −1 x −k , x ↦→ x.As θ fixes x and clearly (bθ) 2 = 1, we only need to check that (aθ) 8 = 1,and (aθ)(bθ)(aθ) −5 (bθ) = 1. Note that from the third relator in the abovepresentation for G we haveaba −5 b = 1 ⇒ ab = b −1 a 5 ⇒ bab = a 5 ,so(aθ) 8 = (ab) 8 = (a(bab)) 4 = (aa 5 ) 4 = a 24 = 1

56and hence(aθ)(bθ)(aθ) −5 (bθ) = abb −1 x −k (ab) −5 b −1 x −k= a(ab) −5 b −1= a(ab) 3 b −1 as (ab) 8 = 1= a 2 (bab)abb −1= a 2 (bab)a= a 8= 1.Therefore θ induces an automorphism, which will be denoted by θ. We provethat θ cycles through a generating set by applying successive powers of θ toax :ax ↦→ abx ↦→ abb −1 x −k x = ax 1−k↦→ abx 1−k ↦→ abb −1 x −k x 1−k = ax.Let H be the subgroup generated by the set {ax, abx, ax 1−k , abx 1−k }. Thenas (ax)(ax 1−k ) −1 = x k and (ax) 8 = x 8 , H must contain the element x k+8 ,and as k is odd, 2k and k + 8 are coprime by Lemma 3.1, so x must lie in H.It follows that a and b are also in H.Now suppose that 4 divides n. If we write n = 2 λ m where λ ≥ 2 and m, is oddthen if H is the factor group G/G(m) then H = G 2 × C 2 λ. GAP can be usedto show that G 2 × C 4 and G 2 × C 8 are not cyclically generated, so for λ ≥ 4we show that H has a characteristic subgroup K such that H/K = G 2 × C 8 .The result then follows by Theorem 2.9.As G 2 has exponent 8, the subgroup 8H is 〈x 8 〉, soH/8H = 〈a, b, x a 8 , b 2 , aba −5 b, x 2λ , [a, x], [b, x], x 8 〉 = G 2 × C 8 ,as required.Theorem 3.13. If G 3 = 〈a, b a 4 , b 4 , abab −1 〉 then G 3 × C n is cyclicallygenerated if and only if 4 does not divide n.

57Proof. LetG = G 3 × C n = 〈a, b, x a 4 , b 4 , abab −1 , x n , [a, x], [b, x]〉.If n is odd, then the result follows from Lemma 2.11 because G 3 is cyclicallygenerated by elements of order dividing 16. If n = 2k where k is odd, definean onto map θ : G → G bya ↦→ a −1 x −k , b ↦→ ba, x ↦→ x.Then(aθ) 4 = (a −1 x −k ) 4 = a −4 x −4k = 1,and from the third relator we have ab = ba −1 , so(bθ) 4 = (ba) 4 = (baba) 2 = (b(ba −1 )a) 2 = b 4 = 1.Also, by rearranging the third relator we get a = ba −1 b −1 , so(aθ)(bθ)(aθ)(bθ) −1 = a −1 x −k baa −1 x −k a −1 b −1= a −1 (ba −1 b −1 )x −2k= a −1 ax −2k= 1,and therefore θ induces an automorphism, which will be denoted by θ. Applyingsuccessive powers of θ to bx givesbx ↦→ bax ↦→ baa −1 x −k x = bx 1−k ↦→ bax 1−k ↦→ baa −1 x −k x 1−k = bx.Let H = 〈bx, bax, bx 1−k , bax 1−k 〉. Then H contains the element(bx) 5 (bx 1−k ) −1 = b 5 x 5 b −1 x k−1 = b 4 x k+4 = x k+4and as k is odd, k + 4 and 2k are coprime by Lemma 3.1, so H contains x.It follows immediately that H also contains b and a, so H = G.Finally, suppose that n = 2 λ m for odd m and λ ≥ 2. ThenG/G(m) = 〈a, b, x a 4 , b 4 , abab −1 , x 2λ , [a, x], [b, x]〉.

58Let H = G/G(m). The exponent of G 3 is 4, so 4H = 〈x 4 〉 andH/4H = 〈a, b, x a 4 , b 4 , abab −1 , x 2λ , [a, x], [b, x], x 4 〉 = G 3 × C 4 ,which is not cyclically generated and this can be shown using GAP. By Theorem2.9 it follows that for λ ≥ 2, the group G 3 × C 2 λ m is not cyclicallygenerated.Theorem 3.14. If G 4 = 〈a, b, c a 4 , b 2 , c 2 , cbca 2 b, baba −1 , caca −1 〉 then thegroup G 4 × C n is cyclically generated if and only if 4 does not divide n.Proof. LetG = G 4 × C n= 〈a, b, c, x a 4 , b 2 , c 2 , cbca 2 b, baba −1 , caca −1 , x n , [a, x], [b, x], [c, x]〉.As G 4 is cyclically generated by elements of order dividing 16, the result forodd n follows from Lemma 2.11. If n = 2k for some odd k then we define anonto map θ bya ↦→ ax −k , b ↦→ bac, c ↦→ b, x ↦→ x.Clearlyand(aθ) 4 = (cθ) 2 = (xθ) n = (cθ)(aθ)(cθ)(aθ) −1 = 1,[aθ, xθ] = [bθ, xθ] = [cθ, xθ] = 1,and(bθ) 2 = (bac) 2= bacbac= ba(cb)ac= ba(b −1 a −2 c −1 )ac= ba(ba 2 c)ac= (bab)a 2 (cac)= aa 2 a= 1.

59As ac = ca,(cθ)(bθ)(cθ)(aθ) 2 (bθ) = bbacba 2 x −2k bac= (ac)ba 2 b(ac)= (ac)ba −2 bac= (ca)ba −2 bac= c(aba −1 )(a −1 ba)c= cb −2 c= c 2= 1,and(bθ)(aθ)(bθ)(aθ) −1 = bacax −k baca −1 x k= bacab(aca −1 )= bacabc= ba(ca)bc= ba 2 cbc= b −1 a −2 c −1 b −1 c −1= (cbca 2 b) −1= 1,so θ induces an automorphism of G, which we will denote by θ.We apply powers of θ to cx :cx ↦→ bx↦→ bacx↦→ bacax −k bx = bacabx 1−k = bca 2 bx 1−k = cx 1−k↦→ bx 1−k↦→ bacx 1−k↦→ bacax −k bx 1−k = bacabx = bca 2 bx = cx.Let H be the subgroup generated by the set{cx, bx, bacx, cx 1−k , bx 1−k , bacx 1−k }.

60As k is odd, 4 + k is coprime to 2k by Lemma 3.1, so as(cx) 5 (cx 1−k ) −1 = c 4 x 4+k = x 4+k ,x 4+k is in H and therefore x is in H. Hence c, b, and a are in H, as required.The exponent of G 4 is 4, so if n = 4k for some k, then G/4G is defined bythe presentation〈a, b, c, x a 4 , b 2 , c 2 , cbca 2 b, baba −1 , caca −1 , x n , [a, x], [b, x], [c, x], x 4 〉,which is a presentation for G 4 × C 4 . GAP can show that this is not cyclicallygenerated, and hence G is not, by Theorem 2.9.Theorem 3.15. If G 5 = 〈a, b a 4 , b 4 , abab, ba −1 ba −1 〉 then G 5 × C n is cyclicallygenerated if and only if 4 does not divide n.Proof. Clearly if n is odd then the result follows by Lemma 2.11. If n = 2kfor some odd k then we can construct a suitable automorphism.LetG = G 5 × C 2k = 〈a, b, x a 4 , b 4 , abab, ba −1 ba −1 , x 2k , [a, x], [b, x]〉and define an onto map θ bya ↦→ b −1 x −k , b ↦→ a, x ↦→ x.Now clearly (aθ) 4 = (b −1 x −k ) 4 = b −4 x −4k = 1, and (bθ) 4 = a 4 = 1, and(xθ) 2k = [aθ, xθ] = [bθ, xθ] = 1, but also(aθ)(bθ)(aθ)(bθ) = b −1 x −k ab −1 x −k a= b −1 ab −1 ax −2k= b −1 ab −1 a= a −1 (ba −1 ba −1 ) −1 a= a −1 a= 1,

61and(bθ)(aθ) −1 (bθ)(aθ) −1 = a(b −1 x −k ) −1 a(b −1 x −k ) −1= ababx 2k= abab= 1,so θ induces an automorphism, which we will also denote by θ. If we apply θto ax we getax ↦→ b −1 x −k+1 ↦→ a −1 x −k+1 ↦→ bx k x −k+1 = bx ↦→ ax.Let H = 〈ax, b −1 x −k+1 , a −1 x −k+1 , bx〉. Then H contains the element(ax) 3 (a −1 x 1−k ) −1 = a 4 x 3+k−1 = x 2+kand as k is odd, 2 + k and 2k are coprime, by Lemma 3.1. Therefore x is inH, and hence a and b are in H, so H = G and G is cyclically generated.Finally we prove that if 4 divides n then G 5 × C n is not cyclically generated.Suppose 4 divides n, so we can write n as 2 2+λ m where m is odd and λ ≥ 0.LetG = G 5 × C m = 〈a, b, x a 4 , b 4 , abab, ba −1 ba −1 , x 22+λm , [a, x], [b, x]〉.The exponent of G 5 is 4, so4G = 〈g 4 : g ∈ G〉 = 〈x 4 〉and thereforeG/4G = 〈a, b, x a 4 , b 4 , abab, ba −1 ba −1 , x 22+λm , [a, x], [b, x], x 4 〉= 〈a, b, x a 4 , b 4 , abab, ba −1 ba −1 , x 4 , [a, x], [b, x]〉= G 5 × C 4 ,which we can prove is not cyclically generated using GAP.This completes the results of the survey. Theorems 3.2, 3.3, 3.4, 3.5, 3.6, 3.9,and 3.10, are generalized in the next section.

623.2 Further direct product resultsWe begin with two results that are extensions of Theorems 3.2 and 3.3. Thesecond also uses Theorem 3.9.Theorem 3.16. Let A be a finite abelian group of rank 2. We can writeA = C 2 m kl × C 2 n l where m ≥ n, and k and l are both odd. The groupG = S 3 × A is cyclically generated if and only if m = n or m = 1 and n = 0.Proof. Let G = S 3 × C 2 m kl × C 2 n l. There are three cases to consider.First we consider the case m = n = 0, so G = S 3 × C kl × C l . Note that S 3 iscyclically generated by elements of order 2, and C kl ×C l is cyclically generatedby elements of odd order, so by Lemma 2.11, G is cyclically generated.Now suppose that n = 0, and m ≥ 1. We prove that G is cyclically generatedif and only if m = 1.As n = 0,G = S 3 × C 2 m kl × C l = (S 3 × C 2 m) × (C kl × C l ).If m = 1 then we know from Theorem 3.2 that S 3 ×C 2 m is cyclically generatedby elements of order 2, so G must be cyclically generated by Lemma 2.11.If m ≥ 2 then note that the centre of G is C 2 m kl × C l and that C kl × C l isa characteristic subgroup of Z(G) and hence a characteristic subgroup of G.Factoring G by C kl ×C l gives a factor group isomorphic to S 3 ×C 2 m. If m ≥ 2then by Theorem 3.2 we know that S 3 × C 2 m is not cyclically generated andtherefore, by Theorem 2.9, it follows that G is not cyclically generated.Now consider the final case, where m ≥ n ≥ 1. We show that G is cyclicallygenerated if and only if m = n. First suppose that m = n, soG = S 3 × C 2 m × C 2 m × C kl × C l .Let H = S 3 ×C 2 m ×C 2 m. We prove that H is cyclically generated by elementsof order 2 m and then it follows by Lemma 2.11 that G is cyclically generated.The group H is defined by the presentation〈a, b, x, y a 2 , b 2 , (ab) 3 , x 2m , y 2m , [a, x], [b, x], [a, y], [b, y] 〉.

63Define a map θ : H → H bya ↦→ b, b ↦→ aba, x ↦→ y, y ↦→ x −1 y −1 .Clearly θ extends to an automorphism of H because we know there is anautomorphism of S 3 under which a ↦→ b ↦→ aba ↦→ a and C 2 m × C 2 m hasa cyclic automorphism of order 3 such that x ↦→ y ↦→ x −1 y −1 ↦→ x. Theautomorphism induced by θ will be denoted by θ from now on and we showthat it cycles through a generating set. Applying successive powers of θ toax givesax ↦→ by ↦→ abax −1 y −1 ↦→ ay −1 yx = ax.We prove that S 3 × C 2 m × C 2 m = 〈ax, by, abax −1 y −1 〉. Note thatabax −1 y −1 (axbyax) = abaabax −1 y −1 xyx = xandabax −1 y −1 (byaxby) = abababx −1 y −1 yxy = y,so x, y, a, b are in the subgroup 〈ax, by, abax −1 y −1 〉. Note that the order of(ax) is 2 m , so S 3 × C 2 m × C 2 m is cyclically generated by 3 elements of ordercoprime to kl, and therefore, by Lemma 2.11, H is cyclically generated.Conversely, we show that if m > n ≥ 1 then G is not cyclically generated.As before, we decompose A as a direct product of a 2-group and a group ofodd order, i.e. we write G as G = S 3 × (C 2 m × C 2 n) × (C kl × C l ) and notethat C kl × C l is a characteristic subgroup of Z(G), and hence a characteristicsubgroup of G. Factoring G by this subgroup gives the group S 3 ×C 2 m ×C 2 n.We prove that this group is not cyclically generated and then the resultfollows by Theorem 2.9. LetH = S 3 × C 2 m × C 2 n= 〈a, b, x, y a 2 , b 2 , (ab) 3 , x 2m , y 2n , [a, x], [a, y], [b, x], [b, y], [x, y]〉.The Frattini subgroup of H, Φ(H), is generated by {x 2 , y 2 }, and thereforethe subgroup Φ(H)(2 n−1 ) is generated by the set {h ∈ Φ(H) : o(h)|2 n−1 }.Note that(x 2m−(n−1) ) 2n−1 = x 2m−(n−1) 2 n−1 = x 2m−n+1+n−1 = 1 and (y 2 ) 2n−1 = y 2n = 1,

64and therefore Φ(G)(2 n−1 ) = 〈x 2m−(n−1) , y 2 〉.LetK = H/Φ(H)(2 n−1 ) = S 3 × C 2 m−(n−1) × C 2 .Either m − n = 1, or m − n ≥ 2. If m − n = 1 then K = S 3 × C 4 × C 2 , which,as GAP can be used to show, is not cyclically generated. If m − n ≥ 2 thensinceK = 〈a, b, x, y a 2 , b 2 , (ab) 3 , x 2m−n+1 , y 2 , [a, x], [b, x] 〉,we have Φ(K)(2 m−n−1 ) = 〈x 4 〉 and thereforeK/Φ(K)(2 m−n−1 ) = S 3 × C 4 × C 2 ,so again, K is not cyclically generated.Theorem 3.9 suggests that the conditions on the Sylow 2-subgroup of theabelian group in Theorem 3.16 are also necessary and sufficient for the directproduct of D 8 and a rank 2 finite abelian group to be cyclically generated.Theorem 3.17. Let A be a finite abelian group of rank 2. If we write A =C 2 m kl ×C 2 n l where m ≥ n, and k and l are both odd, then D 8 ×A is cyclicallygenerated if and only if m = n or m = 1 and n = 0.Proof. Note first that since we can writeD 8 × A = D 8 × C 2 m × C 2 n × (C kl × C l ),D 8 ×A is cyclically generated if D 8 ×C 2 m ×C 2 n is, by Lemma 2.11. Moreover,since C kl × C l is a characteristic subgroup of D 8 × A and(D 8 × A)/(C kl × C l ) = D 8 × C 2 m × C 2 n,it follows from Theorem 2.9 that if D 8 ×C 2 m ×C 2 n is not cyclically generatedthen D 8 × A is not. Therefore we only need to prove that D 8 × C 2 m × C 2 n iscyclically generated if and only if m = n or m = 1 and n = 0.If n = 0 and m = 1 then we already know that the group is cyclicallygenerated by Theorem 3.3. If m = n then we consider an automorphism

65similar to the one constructed in the proof of Theorem 3.9. In other words,ifG = D 8 × C 2 m × C 2 m = 〈a, b, x, y a 2 , b 2 , (ab) 4 , x 2m , y 2m , [a, x], [b, x], [x, y]〉,then let θ be the automorphism induced by the mapa ↦→ b, b ↦→ a, x ↦→ y, y ↦→ x −1 y −1 .Repeatedly applying θ to ax givesax ↦→ by ↦→ ax −1 y −1 ↦→ bx ↦→ ay ↦→ bx −1 y −1 ↦→ axand by the same argument as the one used in the proof of Theorem 3.9, the set{ax, by, ax −1 y −1 , bx, ay, bx −1 y −1 } is a cyclic generating set for D 8 ×C 2 m×C 2 m.If n = 1 and m > n, then the fact that D 8 × C 2 m × C 2 n is not cyclicallygenerated follows from Theorem 3.9. For small m and n with n ≥ 2 we canuse GAP to obtain basic examples of groups that are not cyclically generated.As usual, for larger m > n we can use these examples and Theorem 2.9 toprove that D 8 × C 2 m × C 2 n is not cyclically generated.First assume that n ≥ 3. LetG = D 8 × C 2 m × C 2 n= 〈a, b, x, y a 2 , b 2 , (ab) 4 , x 2m , y 2n , [a, x], [b, x], [x, y]〉,so Z(G) = 〈(ab) 2 , x, y〉. We consider the subgroup of Z(G) generated byelements of order 2 n−1 , and denote this characteristic subgroup by H. Asn ≥ 3, we know that 2 n−1 ≥ 4, so H is generated by x 2m−n+1 and y 2n−n+1 = y 2 .ThereforeG/H = 〈a, b, x, y a 2 , b 2 , (ab) 4 , x 2m , y 2n , [a, x], [b, x], [x, y], x 2m−n+1 , y 2 〉= 〈a, b, x, y a 2 , b 2 , (ab) 4 , x 2m−m+1 , y 2 , [a, x], [b, x], [x, y]〉= D 8 × C 2 m−n+1 × C 2 .By Theorem 3.9, G/H is not cyclically generated because m − n + 1 ≥ 2 andtherefore D 8 × C 2 m × C 2 n is not cyclically generated if n ≥ 3.

66Finally, putting n = 2, we prove that for any m > n, D 8 × C 2 m × C 4 hasa characteristic subgroup K such that (D 8 × C 2 m × C 4 )/K is isomorphic toD 8 × C 8 × C 4 . LetG = D 8 × C 2 m × C 4= 〈a, b, x, y a 2 , b 2 , (ab) 4 , x 2m , y 4 , [a, x], [b, x], [x, y]〉and letThenK = 8Z(G) = 〈((ab) 2 ) 8 , x 8 , y 8 〉 = 〈x 8 〉.G/K = 〈a, b, x, y a 2 , b 2 , (ab) 4 , x 8 , y 4 , [a, x], [b, x], [x, y]〉and GAP can be used to prove that D 8 × C 8 × C 4 is not cyclically generated.Therefore D 8 × C 2 m × C 4 is not cyclically generated for any m greater than2.While Theorems 3.16 and 3.17 extend previous results by generalizing theabelian group in the direct product, the next result generalizes the resultsfor D 8 × C n and D 10 × C n to the case of a direct product of a cyclic groupwith an arbitrary finite dihedral group.Theorem 3.18. For any m the group D 2m ×C n is cyclically generated if andonly if 4 does not divide n.Proof. As D 2m is cyclically generated by elements of order 2, the result forodd n follows by Lemma 2.11.If n = 2k for odd k we construct an automorphism that cycles through agenerating set. We consider separately the cases m odd and m even. Supposefirst that m is odd. LetG = D 2m × C 2k = 〈a, b, x a 2 , b m , (ab) 2 , x 2k , [a, x], [b, x]〉and let θ be the map defined bya ↦→ b −1 abx k , b ↦→ b, x ↦→ x.(This is the map defined in the proof of the result for D 10 × C n .) Clearly θ isonto, so we only need to check that the first and third relators are preservedif we replace a with b −1 abx k . As(b −1 abx k ) 2 = b −1 abb −1 abx 2k = b −1 abb −1 ab = b −1 a 2 b = 1

67and(b −1 abx k b) 2 = b −1 abbb −1 abbx 2k = b −1 ababb = b −1 b = 1,we know that θ induces an automorphism of G, and we will now denote thisautomorphism by θ. We apply successive powers of θ to ax 2 :ax 2 ↦→ b −1 abx k+2 ↦→ b −2 ab 2 x 2↦→ b −3 ab 3 x k+2 ↦→ · · · ↦→ b −m+1 ab m−1 x 2↦→ b −m ab m x k+2 = ax k+2↦→ b −1 abx 2 ↦→ · · · ↦→ b −m+1 ab m−1 x k+1 ↦→ ax 2 .Let H be the subgroup generated by these elements. Asax 2 ax k+2 = x k+4 ,x k+4 is in H, but k + 2 and 2k are coprime by Lemma 3.1, so x is in H. Itfollows that a and therefore b −1 ab are in H. But b −1 = aba and therefore(b −1 ab)a = b −1 aba = b −2 ,so b −2 is in H. As m is odd, it follows that b must be in H.Now suppose m is even. LetG = D 2m × C 2k = 〈a, b, x a 2 , b 2 , (ab) m , x 2k , [a, x], [b, x]〉.We define the same map as that used in the proof of the result for D 8 × C n ,i.e. θ is defined bya ↦→ b, b ↦→ ax k , x ↦→ x.As there is an automorphism of D 2m that swaps a and b, and clearly (bθ) 2 = 1,we only need to check that (aθbθ) m = 1 to prove that θ induces an automorphism,and(aθbθ) m = (bax k ) m= (ba) m (x 2k ) m/2= (ba) m= 1,

68as required. Let θ denote this automorphism.We apply θ to ax to getax ↦→ bx ↦→ ax k+1 ↦→ bx k+1 ↦→ ax 2k+1 = ax.Let H be the subgroup generated by ax, bx, ax k+1 , and bx k+1 . Then as(ax k+1 ) 2 (ax) 2 = a 4 x 2k+4 = x 4 ,x 4 is in H, and(ax k+1 )(ax) −1 x 4 = x k+4so x k+4 is in H, but k is odd, so x must be in H, by Lemma 3.1. It followsimmediately that a and b are in H.We now show that if 4 divides n then D 2m ×C n is not cyclically generated andwe begin by showing that there is a characteristic subgroup K of D 2m × C 4ksuch that (D 2m × C 4k )/K is D 2m × C 4 . Again we need to consider separatelythe cases m odd and m even. Suppose first that m is odd. LetG = D 2m × C 4k = 〈a, b, x a 2 , b m , (ab) 2 , x 4k , [a, x], [b, x]〉.Then Z(G) = 〈x〉 and the subgroup Z(G)(k) is a characteristic subgroup ofG generated by x 4 . ThereforeG/Z(G)(k) = 〈a, b, x a 2 , b m , (ab) 2 , x 4k , [a, x], [b, x], x 4 〉= 〈a, b, x a 2 , b m , (ab) 2 , x 4 , [a, x], [b, x]〉= D 2m × C 4 .If m is even, then we write m = 2 λ r where r is odd. LetG = D 2m × C 4k = 〈a, b, x a 2 , b 2λr , (ab) 2 , x 4k , [a, x], [b, x]〉.As Z(G) is generated by b 2λ−1r and x,so4Z(G) = 〈(b 2λ−1r ) 4 , x 4 〉 = 〈b 2λ+1r , x 4 〉 = 〈x 4 〉,G/4Z(G) = 〈a, b, x a 2 , b 2λr , (ab) 2 , x 4k , [a, x], [b, x], x 4 〉= 〈a, b, x a 2 , b 2λr , (ab) 2 , x 4 , [a, x], [b, x]〉= D 2m × C 4 .

69For any m, let H = D 2m × C 4 . By proving that the elements of order otherthan 4 all lie in proper characteristic subgroups, we show that the only GSOgenerating set for H must consist of elements of order 4. Then we eithershow that any generating set consisting only of elements of order 4 musthave a nontrivial intersection with a proper characteristic subgroup of H,or we use Lemma 2.7 to show that there cannot be an automorphism thatcycles through the elements of such a set.If m is odd then the possible forms of elements of H, and their orders, aregiven in the table below. Note that in the table, 0 ≤ i ≤ m − 1.Group element Order Group element Orderb imb i a 2(m,i)b i mx l.c.m.(4, ) = 4m b i ax 4(m,i) (m,i)b i x 2 l.c.m.(2,m(m,i) ) = 2m(m,i)b i ax 2 2b i x 3 l.c.m.(4,m(m,i) ) = 4m(m,i)b i ax 3 4The subgroup generated by b contains all the elements of odd order andis the characteristic subgroup H(m). Now we consider the elements of evenorder and use the fact that every element g lies in the characteristic subgroupgenerated by the elements of order o(g).The elements of order 2 are x 2 , and b i a and b i ax 2 , for any i. Therefore thesubgroup generated by the elements of order 2 does not contain x. For anyi ≥ 1, the elements of order 4m/(m, i) are of the form b i x and b i x 3 . Thereforethe subgroup generated by elements of order 4m/(m, i) does not contain a.Similarly, for any i ≥ 1 the elements of order 2m/(m, i) are of the formb i x 2 , so a is not in any of the subgroups generated by the elements of order2m/(m, i). This proves that all the elements of H, other than those of order

704, lie in proper characteristic subgroups of H.The only remaining possibility for a GSO generating set for H is one consistingof elements of order 4. These elements are x, x 3 , and elements of theform b i ax and b i ax 3 for any i. As x and x 3 are in Z(H), they cannot lie inany cyclic generating set, so we only need to show that elements of the formsb i ax and b i ax 3 do not generate H. Suppose thatH = 〈b i ax, b i ax 3 : i = 0, 1, 2, . . . , m − 1〉.Then, by assumption, x can be written as a word w in these elements, and wmust have odd length because in any word of even length the exponent sumof x is even. However, since ab i = b −i a, the exponent sum of a is odd in anyword of odd length in b i ax and b i ax 3 , so x cannot be written as a word inthese elements. This contradiction proves that H cannot be generated by itselements that have order 4 and are of the form b i ax and b i ax 3 , so it cannotbe cyclically generated. By Theorem 2.9 it follows that, for any k, D 2m ×C 4kis not cyclically generated if m is odd.Now suppose that m is even. Once again we show that H cannot be cyclicallygenerated by showing that any GSO generating set intersects nontriviallywith a proper characteristic subgroup. The possible order for a generalelement b i a j x l of H depends on the powers of 2 dividing i and m. The ordersof the elements of H are given in the following table.Note that in the table, 0 ≤ i ≤ m − 1.Group element Order Group element Orderb im(m,i)b i a 2b i mx l.c.m.(4, (m,i) bi ax 4b i x 2 ml.c.m.(2, (m,i) bi ax 2 2b i x 3 ml.c.m.(4, (m,i) bi ax 3 4

71We consider the sets of elements of each possible order in turn.As m is even, we can write m as 2 λ r for some λ ≥ 1 and odd r. Any i in therange 0 ≤ i ≤ m − 1 can be written as 2 µ s for µ ≥ 0 and odd s. Therefore(m, i) = 2 min{λ,µ} (r, s), som(m, i) = r2λ−min{λ,µ}(r, s) .The elements of odd order are b i for i divisible by 2 λ , so the elements ofodd order lie in the characteristic subgroup H(r). Elements of even order aremore complicated.Clearly b m/2 , b m/2 x 2 and x 2 have order 2, and the other elements of order 2are b i a, and b i ax 2 for any i. Therefore the exponent sum of x is even in aword in elements of order 2, so the subgroup generated by elements of order2 is a proper subgroup.For any t ≥ 3 such that H has an element of order 2t, the elements of order2t are all of the form b i , b i x, b i x 2 , or b i x 3 , for some i. Therefore, for eachrelevant t, the subgroup generated by the elements of order 2t is a propercharacteristic subgroup.Finally, we consider the elements of order 4. For any i, b i ax and b i ax 3 haveorder 4, and if i = m/4 or i = 3m/4, then b i , b i x, b i x 2 , and b i x 3 have order4. Also, b i x and b i x 3 have order 4 if i = m/2. Therefore the set of elementsof order 4 is{x, x 3 , b i ax, b i ax 3 , (i = 0, 1, . . . , m − 1),b m/4 , b 3m/4 , b m/4 x, b m/2 x, b 3m/4 x, b m/4 x 2 , b 3m/4 x 2 , b m/4 x 3 , b m/2 x 3 , b 3m/4 x 3 }.Of these elements, x, x 3 , b m/2 x, and b m/2 x 3 , are in the centre of H, so theycannot be in a cyclic generating set. We need to show that if S is the set{b i ax, b i ax 3 , (i =1, 2, . . . , m − 1),b m/4 , b 3m/4 , b m/4 x, b 3m/4 x, b m/4 x 2 , b 3m/4 x 2 , b m/4 x 3 , b 3m/4 x 3 },then S does not contain a cyclic generating set for H. Note that 2Z(H) isgenerated by x 2 . We consider the squares of each of the elements of S.

72s s 2 s 2 ∈ 2Z(H)? s s 2 s 2 ∈ 2Z(H)?b i ax x 2 yes b i ax 3 x 2 yesb m/4 b m/2 no b 3m/4 b m/2 nob m/4 x b m/2 x 2 no b 3m/4 x b m/2 nob m/4 x 2 b m/2 no b 3m/4 x 2 b m/2 nob m/4 x 3 b m/2 x 2 no b 3m/4 x 3 b m/2 x 2 noBy Lemma 2.7, a cyclic generating set must either contain only elementswhose squares are in 2Z(H), or only elements whose squares are not in2Z(H). Let T be the set containing the elements of S whose squares arein 2Z(H). Then T does not generate H because if x were in the subgroupgenerated by T, then x could be written as a word of odd length in theelements of T, but the exponent sum of a in such a word would be odd.Therefore a cyclic generating set for H can only contain those elements ofS \ T. Clearly a is not is the subgroup generated by these elements. ThereforeH is not cyclically generated by its elements of order 4, and hence it isnot cyclically generated. It follows by Theorem 2.9 that D 2m × C 4k is notcyclically generated if m is even.In the next two theorems we extend results for Q 8 × C n and A 4 × C n to thedirect products of Q 8 and A 4 with arbitrary finite abelian groups. We usethe following lemma to prove these results.Lemma 3.19. For any prime p, every abelian p-group is cyclically generatedby a set of size p α for some α.Proof. For a notational convenience, which will become apparent later, westart indexing at 0. LetG = C p λ 0 × C p λ 1 × · · · × C pλ n−1 ,

73where λ 0 ≥ λ 1 ≥ λ 2 ≥ · · · ≥ λ n−1 and let x i generate C p λ i for each i. Let θbe the automorphism defined in the proof of Theorem 2.2, so with the newnotation,x 0 ↦→ x 0 x 1 , x 1 ↦→ x 1 x 2 , . . . , x n−2 ↦→ x n−2 x n−1 , x n−1 ↦→ x n−1under this automorphism. We show that the order of θ is a power of p.By applying the first few powers of θ we getx 0 ↦→ x 0 x 1 ↦→ x 0 x 2 1x 2 ↦→ x 0 x 3 1x 3 2x 3 ↦→ x 0 x 4 1x 6 2x 4 3x 4 ↦→ · · ·so the power of x i in x 0 θ j is ( ji), which is the j i th entry in Pascal’s Triangleif, by the usual convention, we start counting rows and entries at 0.If x 0 θ k = x 0 for some k > 0 then it follows that θ k fixes x 0 θ j , for any j,because(x 0 θ j )θ k = x 0 θ j+k = x 0 θ k+j = x 0 θ j ,and as the x 0 θ j s generate G, this means that θ k fixes all the elements of agenerating set. Therefore the order of θ must divide k. Moreover, θ k fixesx 0 if the entries labelled 1, 2, . . . , n − 1 in the kth row of Pascal’s Triangleare zero if they are reduced modulo p λ 1because the orders of x 1 , x 2 , . . . , x n−1divide p λ 1and the power of x i in x 0 θ k is the result of reducing ( ki)modulop λ i. We show that θ k fixes x 0 if k is a power of p.Let µ be the highest power of p dividing i! for any i in the range 1, 2, . . . , n−1.Then for each i in this range we can write i! = m i p µ−ν iwhere (m i , p) = 1.Let β = p λ1+µ . Then( ) ( β pλ 1)+µ= = pλ 1+µ (p λ1+µ − 1) · · · (p λ1+µ − i + 1)≡0 (mod p λ 1),i im i p µ−ν iso the power of x i in x 0 θ β is 0 for i = 1, 2, . . . , n − 1.Therefore the order of θ must divide p λ 1+µ .Theorem 3.20. If A is a finite abelian group, then the group Q 8 × A iscyclically generated.

74Proof. Once again we use the fact that we can decompose A into the directproductA = A 2 × A oddwhere A 2 is the 2-primary part of A and A odd is the direct product of all theother primary parts of A. We show first that Q 8 × A 2 is cyclically generatedby elements of order a power of 2, and then it follows by Lemma 2.11 thatQ 8 × A 2 × A odd is cyclically generated.By Lemma 3.19, A 2 has an automorphism θ that cycles through a generatingset of size a power of 2. Let us denote these generators by y 1 , y 2 , etc.Recall that Q 8 is cyclically presented by〈a, b, c abc −1 , bca −1 cab −1 〉,so there is a cyclic automorphism ϕ such that a ↦→ b ↦→ c ↦→ a. We definean automorphism ψ of Q 8 × A 2 by extending θ and ϕ to automorphisms ofQ 8 × A 2 and composing them in the natural way. Then applying ψ to ay 1we getay 1 ↦→ by 2 ↦→ cy 3 ↦→ ay 4 ↦→ · · · ↦→ ay 1 .Let H be the subgroup of Q 8 ×A 2 generated by the set S = {ay 1 , (ay 1 )ψ, . . . }.As o(ϕ) and o(θ) are coprime, ay i , by i , cy i are in S for each i, and thereforeeach y i is in H, becauseay i by i (cy i ) −1 = abc −1 y i y i y −1i = y i .Hence a, b, c are also in H, and Q 8 × A 2 is cyclically generated by elements oforder a power of 2. As A odd is cyclically generated by elements of odd order,it follows by Lemma 2.11 that Q 8 × A 2 × A odd is cyclically generated.We use a similar method to prove the next result.Theorem 3.21. If B is a finite abelian group then A 4 × B is cyclicallygenerated.

75Proof. We decompose B into the direct product of a 3-group, B 3 , and agroup of order coprime to 3, denoted by ˆB. We show first that A 4 × B 3 iscyclically generated by elements of order 3 i for some i, and then the resultfollows by Lemma 2.11.Note that A 4 is defined by the presentation 〈a, b a 3 , b 3 , (ab) 2 〉 and thereforemap a ↦→ b, b ↦→ a induces an automorphism, ϕ.Let θ be the automorphism of B 3 defined in the proof of Theorem 2.2 andsuppose that x 1 , x 2 , . . . , x k is a generating set for B 3 that θ cycles through,i.e. x i+1 = x i θ for i = 1, 2, . . . , k − 1 and x 1 = x k θ. We know by Lemma3.19 that o(θ) is a power of 3, so k is a power of 3. Extend θ and ϕ toautomorphisms of A 4 × B 3 in the natural way and let ψ be the compositionof the resulting maps. We show that ψ cycles through a generating set forA 4 × B 3 by applying successive powers of ψ to ax 1 :ax 1 ↦→ bx 2 ↦→ ax 3 ↦→ · · · bx k ↦→ ax 1 .Let H be the subgroup generated by the set S, whereS = {ax 1 , (ax 1 )ψ, (ax 1 )ψ 2 , . . .}.We show that H = A 4 × B 3 . As the order of θ is a power of 3 and the orderof ϕ is a power of 2, we know that (ax 1 )ψ k = bx 1 , and therefore, for each j,ax j and bx j are both in H. Now note thatax j bx j ax j bx j = (ab) 2 x 4 j = x 4 j,so x j ∈ H because (o(x j ), 4) = 1. Therefore each x j is in H, and it followsthat a, b are also in H, as required. Therefore A 4 × B 3 is cyclically generatedby elements of order 3 i where 3 i is the order of the x j s. Since we know thatˆB is cyclically generated by elements of order coprime to 3, it follows byLemma 2.11 that A 4 × B 3 × ˆB is cyclically generated.A similar method can be used to show that for any abelian group B and forany n coprime to 3, the group A n+2 × B is cyclically generated.

76A natural extension of Theorem 3.20 is that for any group G with a cyclicgenerating set of size coprime to 3, the group Q 8 × G is cyclically generated.This can be proved using the same method as that used to prove Theorem3.20. Also, by Lemma 2.11, we can use the fact that Q 8 is cyclically generatedby elements of order 4, to show that if G is any group that is generated byelements of odd order then Q 8 ×G is cyclically generated. Therefore if G is acyclically generated group such that Q 8 × G is not cyclically generated, then(i) every cyclic generating set for G must have size divisible by 3, and (ii)every cyclic generating set must consist of elements of even order. The smallestsuch groups are SmallGroup([32,26]), SmallGroup([32,27]), and Small-Group([32,33]). The direct product of Q 8 with SmallGroup([32,26]) is cyclicallygenerated, so conditions (i) and (ii) are not sufficient for Q 8 × G not tobe cyclically generated. It has not yet been possible to establish whether thedirect product of Q 8 with either SmallGroup([32,27]) or SmallGroup([32,33])is cyclically generated, although neither group is an ECS group.Initial tests to find examples of cyclically generated groups that are the directproduct of Q 8 with a small nonabelian group led to the following result, whichconcludes this chapter.Theorem 3.22. The direct product Q 8 × S n is cyclically generated for anyn.Proof. As mentioned previously, S n is cyclically generated by n elements oforder 2, and Q 8 is cyclically generated by 3 elements of order 4, so if 3 doesnot divide n, we can use the method used to prove Theorem 3.20 to provethat Q 8 × S n is cyclically generated.It remains to consider the case n = 3k for some k.As Q 8 is cyclically presented by〈x, y, z, xyz −1 , yzx −1 , zxy −1 〉,there is an automorphism θ such that x ↦→ y ↦→ z ↦→ x, and therefore thereis an automorphism θ ′ of Q 8 × S n that fixes elements of S n and acts in the

77same way as θ on Q 8 . Let ϕ be conjugation by (1 2 · · · n) and let ψ = θ ′ ◦ ϕ.We show that ψ cycles through a generating set for Q 8 × S n .We begin with the case k = 1. Applying ψ to the element x(1 2) gives the setx(1 2) ↦→ y(2 3) ↦→ z(3 1) ↦→ x(1 2)andx(1 2)y(2 3)(z(3 1)) −1 = xyz −1 (1 2)(2 3)(3 1) = (2 3)y(2 3)z(3 1)(x(1 2)) −1 = yzx −1 (2 3)(3 1)(1 2) = (3 1)z(3 1)x(1 2)(y(2 3)) −1 = zxy −1 (3 1)(1 2)(2 3) = (1 2)so Q 8 × S 3 is cyclically generated.Now suppose that k = 2. Applying ψ to x(1 2) givesx(1 2) ↦→ y(2 3) ↦→ z(3 4) ↦→ x(4 5) ↦→ y(5 6) ↦→ z(6 1) ↦→ x(1 2).Let S = {(x(1 2))ψ j : j = 0, 1, . . . , 5}. We show that S generates Q 8 × S 6 .Leta = x(1 2)y(2 3)(z(3 4)) −1 = xyz −1 (1 2)(2 3)(3 4) = (1 4 3 2)andb = y(2 3)z(3 4)(x(1 2)) −1 = yzx −1 (2 3)(3 4)(1 2) = (1 2 4 3),and note thataba = (1 4 3 2)(1 2 4 3)(1 4 3 2) = (1 2).Let H be the subgroup generated by S. As S and hence H are invariantunder ψ and ψ acts by conjugation by (1 2 3 4 5 6) on any permutation in H,H must contain any permutation in S 6 and hence it contains x, y, and z, soH = Q 8 × S 6 as required.Finally, we generalize this method to prove the result for k ≥ 3. Let ϕ beconjugation by (1 2 · · · 3k − 1 3k) and ψ be as defined previously. Applyingψ repeatedly to x(1 2) givesx(1 2) ↦→ y(2 3) ↦→ z(3 4) ↦→ x(4 5) ↦→ · · ·· · · ↦→ x(3(k−1)+1 3(k−1)+2) ↦→ y(3(k−1)+2 3k) ↦→ z(3k 1) ↦→ x(1 2),

78so we want to show that Q 8 × S 3k is generated by the setS = {x(3i+1 3i+2), y(3i+2 3i+3), z(3i+3 3i+4) : i = 0, 1, . . . k − 2}∪ {x(3(k − 1)+1 3(k − 1)+2), y(3(k − 1)+2 3k), z(3k 1)}.Let H = 〈S〉. Leta = x(1 2)y(2 3)(z(3 4)) −1= xyz −1 (1 2)(2 3)(3 4)= (1 4 3 2)and letb = y(2 3)z(3 4)(x(1 2)) −1= yzx −1 (2 3)(3 4)(1 2)= (1 2 4 3).As aba = (1 4 3 2)(1 2 4 3)(1 4 3 2) = (1 2), we know that (1 2) is in H, and asH is invariant under ψ and ψ acts by conjugation by (1 2 · · · 3k) on (1 2), itfollows that every element of S 3k is in H and hence H = Q 8 × S 3k .

Chapter 4Minimal cyclic generating setsfor abelian groupsEvery finite abelian group is cyclically generated, see Theorem 2.2, and thereforea natural question to ask is what are the possible orders of the automorphismsthat cycle through generating sets for abelian groups. In particular,are there abelian groups that are cyclically generated by a generating set ofsize equal to the rank of the group. Recall that a cyclic generating set iscalled a minimal cyclic generating set if its size is equal to the rank of thegroup. In 2003 a paper by Miklós Abért appeared in the Proceedings of theAmerican Mathematical Society in which the author gave necessary and sufficientconditions for a finite abelian group to be symmetrically generated byn generators, for any n > 2, see Theorem 1 in [1], and mentioned that everyfinite abelian group of rank 2 is symmetrically generated by two generators,see also Example 2 in [22]. From Abért’s result we were immediately ableto produce examples of finite abelian groups that are cyclically generated byminimal generating sets. However, searches using GAP found examples ofgroups that do not satisfy the conditions in Abért’s theorem and yet havea minimal cyclic generating set. Concentrating these searches on groups ofranks 3 and 4 produced enough examples to suggest necessary and sufficientconditions for groups of rank 3 and 4 to have minimal cyclic generating sets.These conditions include, but are looser than, the conditions in Abért’s resultin the case n = 3 or n = 4, which was to be expected as the existence of a79

80cyclic generating set is a weaker requirement than the existence of a symmetricgenerating set. There are some surprising examples of groups that do nothave minimal cyclic generating sets even though they have automorphismsof the appropriate order.This chapter contains various preliminary results that will be used to determinewhether certain finite abelian groups are cyclically generated by aminimal generating set. The first result is due to Miklós Abért. Before statingAbért’s result, we need to describe his notation and the notation that wewill be using for the remainder of this thesis.4.0.1 NotationFor consistency we use Abért’s slightly unusual notation for finite abeliangroups, so for any natural number n, Z n will denote the cyclic group of ordern with group operation addition, and a finite abelian group G will be writtenas a direct product of cyclic groups rather than as a direct sum, i.e.G = Z a1 × Z a1 a 2× · · · × Z a2 a 2···a n.The cyclic subgroups Z a1 a 2···a ithan direct summands.will therefore be called direct factors ratherIt is worth noting that although we will use Z n to denote the additive groupof order n, we will write group elements in terms of multiples of generators,so for example, if G = Z a × Z ab then we will define generators x and y ofZ a and Z ab respectively, and an element of G will be written as λx + µy forsome integers λ and µ. As ax = 0 and aby = 0, we know that λ and µ can beassumed to be reduced modulo a and ab respectively, and therefore lie in theinteger rings of size a and ab respectively. These rings will also be denotedby Z a and Z ab , but the context should make it clear whether Z a is denotingthe additive group or the ring.As we are using addition as the group operation, it follows that the characteristicsubgroups mG and G(m) will be generated by the sets {mg : g ∈ G}

81and {g ∈ G : o(g)|m} respectively, and the commutator −x − y + x + y willbe written as [x, y].For any prime p dividing the order of G, the p-primary part of G will bedenoted by G p .4.0.2 Abért’s TheoremTheorem 4.1. (Miklós Abért, 2003 ) Let G be a finite abelian group. ThenG can be symmetrically generated by n > 2 elements if, and only if, there arepositive integers a, b, c such that(i) G = Z a × Z n−2ab(ii) (b, c)|n.× Z abc , andFrom now on, we will call this result Abért’s Theorem and conditions (i) and(ii) will be called Abért’s conditions. Abért also notes in [1] that every finiteabelian group of rank 2 is symmetrically generated by 2 generators, since thegroup Z a × Z ab is defined by the presentation〈x, y x ab = y ab = (xy −1 ) a = x −1 y −1 xy = 1 〉.If a group G is symmetrically generated by the set X = {x 1 , x 2 , . . . , x n }, thenevery permutation of X induces an automorphism of G, so in particular,x 1 ↦→ x 2 ↦→ · · · ↦→ x n−1 ↦→ x n ↦→ x 1induces an automorphism of G and hence G is cyclically generated by ngenerators. We can therefore use Abért’s result to say that certain finiteabelian groups of rank n are cyclically generated by n generators.4.1 Preliminary resultsThe general results proved in this section will be used in later chapters. Webegin with a proof of the fact that a group satisfying Abért’s conditions for

82n > 2 is cyclically generated by n generators. Although we have alreadynoted that the existence of a cyclic generating set follows from the existenceof a symmetric generating set, this is a direct proof of the result and isincluded for completeness. The generating set produced is the same as inAbért’s proof, but we prove that it is a cyclic generating set without provingfirst that it is symmetric, which changes some of the details.Theorem 4.2. If G is a finite abelian group that satisfies Abért’s conditionsfor some n > 2, i.e. it is of the formG = Z a × Z ab × · · · × Z } {{ ab }n−2×Z abcwhere a, b, c ≥ 1 and (b, c)|n, then G can be cyclically generated by n generators.Proof. Assume that G is written in the above form, and that x 1 , x 2 , . . . , x ngenerate G, with x i generating the ith factor of G. (Note that this is a changefrom Abért’s proof as he numbered the factors right to left.) We define a newgenerating set {g 1 , g 2 , . . . , g n } and prove that there must be an automorphisminduced by the map g 1 ↦→ g 2 ↦→ g 3 ↦→ · · · ↦→ g n ↦→ g 1 .As (b, c)|n, the Diophantine equation bx − cy = n has solutions, see [29] p.4.Therefore we let F = bx and define g 1 , . . . , g n byg 1 = x ng 2 = x n − x n−1.g i = x n + x n+2−i − x n+1−i 3 ≤ i ≤ n − 1.g n = (F − n + 1)x n + x 2 − x 1 .We show first that the g i s generate G and then show that the map g 1 ↦→g 2 ↦→ g 3 ↦→ · · · ↦→ g n ↦→ g 1 induces an automorphism.

83Let K = 〈g 1 , g 2 , . . . , g n 〉. Clearly x n is in K, and therefore x n − g 2 is in K,but from the definition of g 2 ,x n − g 2 = x n−1 ,so x n−1 is in K. Now note that for 3 ≤ i ≤ n − 1,x n+1−i = x n + x n+2−i − (x n + x n+2−i − x n+1−i ) = x n + x n+2−i − g i ,but taking each i in turn, x n+1−i is in K because g i , x n and x n+2−i are in K.Therefore x n , x n−1 , x n−2 , . . . , x 3 , x 2 are in K so it only remains to prove thatx 1 is in K. As x 1 = (F − n + 1)x n + x 2 − g n , x 1 is in K, and hence K = G.Let θ be the map defined by g 1 ↦→ g 2 ↦→ g 3 ↦→ · · · ↦→ g n ↦→ g 1 . As G isfinite and θ is onto, θ is invertible, so it is enough to prove that θ inducesa homomorphism. Unlike in examples in the previous chapters, we do notknow what the relators are in a presentation for G on the g i s, even thoughwe know that such a presentation exists. Therefore we consider a generalrelator ∑ ni=1 r ig i and show that replacing each of the g i s by g i+1 (or g n byg 1 ) results in a word that is equal to the identity in G. In other words, wewant to show that ∑ ni=1 r i(g i θ) = 0. First of all we rewrite the g i s in termsof the x i sn∑∑n−1r i g i = r 1 x n + r 2 (x n − x n−1 ) + r i (x n + x n+2−i − x n+1−i )i=1i=3+ r n ((F − n + 1)x n + x 2 − x 1 ).By collecting the x i s we get(n∑n−1)) ∑r i g i = x n(r 1 + r 2 + r i + (F − n + 1)r n + x n−1 (−r 2 + r 3 )i=1i=3∑n−2+ x n+1−i (−r i + r i+1 ) + x 2 (−r n−1 + r n ) − x 1 r ni=3and as x 1 , x 2 , . . . , x n is the usual basis for G, we know that the r i s mustsatisfy the following conditions:

84(i) r 1 + r 2 + ∑ n−1i=3 r i + (F − n + 1)r n ≡ 0 (mod abc),(ii) −r 2 + r 3 ≡ 0 (mod ab),(iii) −r i + r i+1 ≡ 0 (mod ab) for i = 3, 4, . . . , n − 1,(iv) −r n ≡ 0 (mod a).Points (i)-(iv) can be combined to give(A) r n ≡ 0 (mod a),(B) r i − r i+1 ≡ 0 (mod ab) for i = 2, 3, . . . , n − 1, and(C) ∑ n−1i=1 r i + (F − n + 1)r n ≡ 0 (mod abc).Now if we write ∑ ni=1 r i(g 1 θ) in terms of the x i s we getn∑∑n−1r i (g i θ) = r n g 1 + r i g i+1i=1= r n g 1 +i=1n∑r i−1 g iand expressing the g i s in terms of the x i s we have(n∑n−1)∑r i (g i θ) = r n x n + r 1 (x n − x n−1 ) + r i−1 (x n + x n+2−i − x n+1−i )i=1i=3Collecting the x i s givesn∑r i (g i θ) = x n(r n + r 1 +i=1+i=2+ r n−1 ((F − n + 1)x n + x 2 − x 1 ).( ∑n−1)r i−1)+ r n−1 (F − n + 1) + x n−1 (−r 1 + r 2 )i=3( ∑n−2)x n+1−i (−r i−1 + r i ) + x 2 (−r n−2 + r n−1 ) − x 1 r n−1 .i=3To show that ∑ ni=1 r i(g i θ) = 0 we want to prove that(Â)(∑ n−2i=1 r i)+ rn−1 (F − n + 1) + r n ≡ 0 (mod abc),

85( ˆB) r i − r i+1 ≡ 0 (mod ab), for i = 1, 2, . . . , n − 2, and(Ĉ) −r n−1 ≡ 0 (mod a).We use (A),(B), and (C) to prove that the r i s satisfy (Â),( ˆB), and(A) we know that r n ≡ 0 (mod a), and by (B) we know that(Ĉ). Byr n−1 − r n ≡ 0 (mod ab) ≡ 0 (mod a),so r n−1 − r n = ka for some integer k. But then r n−1 = ka + r n ≡ 0 (mod a),so (Ĉ) must hold.Now we look at condition (Â). Note that as F = bx and n = bx − cy, wemust have cy = F − n, so F − n ≡ 0 (mod c). Therefore( ∑n−2)r i + r n−1 (F − n + 1) + r ni=1( n−1) ∑= r i − r n−1 + r n−1 (F − n + 1) + r ni=1( n−1) ∑= r i + (r n − r n−1 ) + r n−1 (F − n + 1)i=1( n−1) ∑= r i + (r n − r n−1 ) + r n−1 (F − n + 1) + r n (F − n + 1) − r n (F − n + 1)i=1( n−1) ∑= r i + r n (F − n + 1) +(r n − r n−1 ) + (r n−1 − r n )(F − n + 1)i=1} {{ }≡0 (mod abc) by (C)≡ (r n − r n−1 ) + (r n−1 − r n )(F − n + 1)= −(r n−1 − r n ) + (r n−1 − r n )(F − n + 1)= (r n−1 − r n )(−1 + F − n + 1)= (r n−1 − r n ) (F − n)} {{ } } {{ }≡0(mod ab) ≡0(mod c)≡ 0(mod abc),as required for condition (Â).(mod abc)

86Finally, we prove that condition ( ˆB) holds. It follows immediately from (B)that r i − r i+1 ≡ 0 (mod ab), for i = 2, 3, . . . , n − 2, but it remains to provethat r 1 − r 2 ≡ 0 (mod ab). First note that for i = 2, 3, . . . , n − 1,r n − r i ≡ 0 (mod ab) (4.1)because r n − r i = ∑ n−1j=i (r j+1 − r j ) and r j+1 − r j ≡ 0 (mod ab) for each j inthe required range.Also note that by (B) we haver 1 − r 2 ≡( ∑n−1)r i − r i+1i=1(mod ab)and thereforer 1 − r 2 ≡ r 1 − r n (mod ab). (4.2)But b divides F and r n ≡ 0 (mod a) by (A) sor 1 − r n ≡ r 1 + (F − 1)r n (mod ab). (4.3)By rearranging (C) we have∑n−1F r n ≡ − r i + (n − 1)r ni=1(mod ab)and therefore∑n−1(F − 1)r n ≡ − r i + (n − 2)r n (mod ab). (4.4)i=1

87Therefore we haver 1 − r 2 ≡ r 1 − r n (mod ab) by (4.2)≡ r 1 + (F − 1)r n (mod ab) by (4.3)( n−1) ∑≡ r 1 − r i + (n − 2)r n (mod ab) by (4.4)i=1( n−1) ∑≡ − r i + (n − 2)r n (mod ab)i=2∑n−1∑n−1∑n−1= − r i + r n since (n − 2)r n =i=2n−1i=2∑= (r n − r i )i=2n−1∑≡ 0 (mod ab) by (4.1)≡ 0i=2(mod ab), as required.This completes the proof that θ induces a homomorphism, and hence anautomorphism because we already know that it is onto. By construction, θcycles through a generating set, so G is cyclically generated by n generators.i=2r nThe next two results provide ways of breaking down the problem of determiningwhether a given abelian group is cyclically generated by a set of acertain size. The first is an extension of Lemma 2.8 and leads to an extensionof Theorem 2.9, while the second tells us that it is enough to consider eachof the primary parts of the group separately.Lemma 4.3. If G is cyclically generated by n generators and K is a characteristicsubgroup of G then G/K is cyclically generated by m generators forsome m dividing n.Proof. It was shown in the proof of Lemma 2.8 that if θ cycles through agenerating set of size n then θ induces an automorphism θ K on G/K givenby (g + K)θ K = gθ + K. As (g + K)θ n K = gθn + K = g + K, the order of θ Kmust be a factor of n.

88An immediate consequence of Lemma 4.3 is the following result.Theorem 4.4. If G has a characteristic subgroup K such that G/K is notcyclically generated by m generators for any m dividing n then G is notcyclically generated by n generators.Now we come to the result about the p-primary parts of G.Lemma 4.5. If G is a finite abelian group of order p α 11 p α 22 · · · p α kk, wherep 1 , p 2 , . . . , p k are distinct primes, then G is cyclically generated by n generatorsif and only if, for each i, G pi is cyclically generated by m i generatorsfor some m i dividing n and n = l.c.m.(m 1 , m 2 , . . . , m k ).Proof. As G = G p1 ×G p2 ×. . .×G pk and each G pi is a characteristic subgroupof G,Aut G ∼ = Aut G p1 × Aut G p2 × · · · × Aut G pk , see [31] p. 117.If we assume that there exists an automorphism θ of G of order n and anelement g of G such that 〈g, gθ, gθ 2 , . . . , gθ n−1 〉 = G, then by the decompositionabove, θ can be written as (θ 1 , θ 2 , . . . , θ k ) where each θ i is in Aut G piand g can be written as the k-tuple (g 1 , g 2 , . . . , g k ) where each g i is in G pi .Therefore we have gθ j = (g 1 θ1, j g 2 θ2, j . . . , g k θ j k), for j = 0, 1, . . . , n − 1.For any i, letH i = G p1 × G p2 × · · · × G pi−1 × G pi+1 × . . . × G pk .The subgroup H i is characteristic in G, and G pi∼ = G/Hi , so G pi is cyclicallygenerated by m i generators for some m i dividing n, by Lemma 4.3.Conversely, suppose that each G pi is cyclically generated by m i generators.Then for each i, there exists a g i in G pi and an automorphism θ iof G pi such that 〈g i , g i θ i , g i θi 2 , . . . , g i θ n i−1i 〉 = G pi . Let θ = (θ 1 , θ 2 , . . . , θ k ),let g = (g 1 , g 2 , . . . , g k ), and let n = l.c.m(m 1 , m 2 , . . . , m k ). We show thatG = 〈g, gθ, gθ 2 , . . . , gθ n−1 〉.For each l let λ l = ∏ i≠l o(g i), so λ l g i = 0 for i ≠ l and note that because(λ l , o(g l )) = 1 there is an integer µ l such that µ l λ l g = (0, . . . , 0, g l , 0, . . . , 0)

89and hence µ l λ l gθ j = (0, . . . , 0, g l θ j l, 0, . . . , 0) for j = 0, 1, 2, . . . , n − 1. Therefore〈g, gθ, gθ 2 , . . . , gθ n−1 〉 generates G.Lemma 4.5 shows that in order to show that a group G has a minimal cyclicgenerating set, it is enough to prove that each primary part of G has a cyclicgenerating set of the appropriate size. This does not, however, mean thatthe question is reduced to finding minimal cyclic generating sets for abelianp-groups. For example, as mentioned in the introduction, if the 7-primarypart of a rank 5 abelian group G has rank 3, then we need to determinewhether this 7-group has a cyclic automorphism of order 5, not of order 3,in order to determine whether G has a minimal cyclic generating set.4.1.1 Group automorphisms as matricesAs we are using addition as the group operation it is useful to define groupautomorphisms as integer matrices. To clarify, ifG = Z m1 × Z m2 × · · · × Z mnwhere m 1 |m 2 | · · · |m n= 〈x 1 〉 × 〈x 2 〉 × · · · 〈x n 〉then any g in G can be written as λ 1 x 1 +λ 2 x 2 +· · ·+λ n x n , or equivalently asthe n-tuple (λ 1 , λ 2 , . . . , λ n ). By saying that θ : G → G is the map defined withrespect to the basis {x 1 , x 2 , . . . , x n } by the n × n integer matrix A = [a i j ],we mean that θ is the map such thatgθ = (λ 1 , λ 2 , . . . , λ n )A,but we will usually writegA = (λ 1 x 1 + λ 2 x 2 + · · · + λ n x n )A = λ 1 x 1 A + λ 2 x 2 A + · · · + λ n x n Awherex i A = a i 1 x 1 + a i 2 x 2 + · · · + a i n x n .Note also that, with respect to the same basis, two matrices A = [a i j ] andB = [b i j ] represent the same map if, for each i and j, the entries satisfya i j ≡ b i j (mod m j ) where m j is the order of the jth basis element. Because

90of this we do not always have to consider integer matrices. For example, itis sometimes useful to think of the entries in the jth column as elements ofthe ring Z mj , especially for counting arguments, as we will see in Corollary4.7. At other times it is useful to think of all the entries of the matrix lyingin the same ring, for example, as m n is the exponent of G, we can think of amap from G to itself as represented by a matrix with all entries in Z mn .For abelian p-groups of a given form we can sometimes determine necessaryand sufficient conditions on the entries of a matrix for the map defined bythe matrix to be an automorphism. The next result gives such conditions fora rank n abelian p-group with no repeated cyclic direct factors.Lemma 4.6. Let p be prime. Let A = [a i j ] be an n × n integer matrix andlet θ be the map defined by A with respect to the basis {x 1 , x 2 , . . . , x n } for thegroupG = Z p α 1 × Z p α 1 +α 2 × · · · × Z p α 1 +α 2 +···+αn (α 1 , α 2 , . . . , α n ≥ 1)= 〈x 1 〉 × 〈x 2 〉 × · · · × 〈x n 〉.Then θ : G → G is an automorphism if and only if(1) p α i+1+α i+2 +···+α jdivides a i j when j > i, and(2) (a i i , p) = 1 for each i.Proof. We show first that θ is a homomorphism if and only if condition (1)holds. Then we show that in addition, θ is invertible if and only if condition(2) also holds.Firstly, suppose condition (1) holds, so if j > i we can writea i j = p α i+1+α i+2 +···+α jb i j for some integer b i j .To show that θ is a homomorphism of G it is enough to show thatp α 1+α 2 +···+α i(x i θ) = 0 for each i,

91but( n∑)p α 1+α 2 +···+α i(x i θ) = p α 1+α 2 +···+α ia i j x jj=1n∑= p α 1+α 2 +···+α ia i j x jj=1i∑n∑= a i j (p α 1+α 2 +···+α ix j ) + p α 1+···+α ip α i+1+···+α jb i j x jj=1j=i+1= 0 as required.Conversely, suppose that θ is a homomorphism, but that p α k+1+α k+2 +···+α ldoes not divide a k l for some k, l with l > k. Thereforea k l = p α k+1+α k+2 +···+α l −λ b k lfor some b k l and λ with (b k l , p) = 1 and λ ≥ 1. Note that as θ is a homomorphism,p α 1+α 2 +···+α kxk must be mapped to 0 by θ and therefore0 = (p α 1+α 2 +···+α kx k )θ= p α 1+α 2 +···+α k(x k θ) because θ is a homomorphism,( n∑)= p α 1+α 2 +···+α ka k j x jj=1n∑= p α 1+α 2 +···+α ka k j x jj=1n∑= p α 1+α 2 +···+α ka k j x jj=k+1( n∑)= p α 1+α 2 +···+α ka k j x j + p α 1+α 2 +···+α ka k l x l=≠ 0j=k+1, j≠l( n∑j=k+1, j≠lp α 1+α 2 +···+α ka k j x j)+ p α 1+α 2 +···+α k +α k+1 +α k+2 +···+α l −λ b k l x lbecause the last term is not 0 and the x i s are a basis.Therefore θ is not a homomorphism.

92Now, assuming θ is a homomorphism, we prove that θ is invertible if andonly if condition (2) holds.Recall now that two n × n integer matrices M and N represent the samemap of G if and only if m i j ≡ n i j (mod p α 1+α 2 +···+α j). Therefore there arematrices that are not equal as matrices, but that represent the same map ofG. In particular, there are integer matrices other than I n that represent theidentity map. We show that if condition (2) holds then we can find an integermatrix B such that the map represented by AB and BA is the identity map.Let C be the matrix of cofactors of A. Then AC = (det A)I n . As θ is ahomomorphism, condition (1) holds, so by expanding det A about the firstcolumn we can writedet A = a 1 1 a 2 2 · · · a n n + λpfor some integer λ, not necessarily coprime to p. By the assumption that(2) holds, a 1 1 a 2 2 · · · a n n is coprime to p, and hence det A is coprime to p.Therefore, for each i, there exist s i and t i such thats i det A + t i p α 1+α 2 +···+α i= 1.Let S be the diagonal matrix with diagonal entries s i i = s i . Then ACS is adiagonal matrix with ith diagonal entry s i det A, buts i det A = 1 − t i p α 1+α 2 +···+α i≡ 1 (mod p α 1+α 2 +···+α i),so the map represented by ACS is the identity map, and therefore CS representsthe map that is the inverse of θ. Note that(CS)A = (SC)A = S(CA) = S(det A)I nalso.Finally, we show that if condition (2) does not hold, then θ cannot be invertible.

93All integer matrices, M, representing the identity map have the form⎛⎞1 + m 1 1 p α 1m 1 2 p α 1+α 2· · · m 1 n p α 1+α 2 +···+α nm 2 1 p α 11 + m 2 2 p α 1+α 2· · · m 2 n p α 1+α 2 +···+α n⎜.⎝ .. .. ⎟. ⎠m n 1 p α 1m n 2 p α 1+α 2· · · 1 + m n n p α 1+α 2 +···+α nso det M ≡ 1 (mod p). Suppose that p divides a k k for some k. We showthat θ cannot be an automorphism. For if θ is an automorphism, then thereexists a matrix B such that AB represents the identity map, i.e. det AB ≡ 1(mod p). This means that det A (mod p) must be a unit in Z p , but sincep divides a k k , we have det A ≡ 0 (mod p). Hence det A (mod p) is not aunit, and therefore there does not exist an integer matrix B such that ABrepresents the identity map.Corollary 4.7. The automorphism group of G has p β (p − 1) n elements forsome β.Proof. Again we use the fact that two matrices A and B represent the sameautomorphism (with respect to the same basis) if and only if their entriessatisfy a i j ≡ b i j (mod p α 1+α 2 +···+α j). In other words, if the jth columns ofeach matrix are reduced modulo the order of the jth basis element then thematrices obtained are identical.This reduces the number of choices for the i jth entry to at most p α 1+α 2 +···+α j.The actual number will be further determined by the position of a i j in thematrix. There are 3 cases to consider: j > i, j = i, and j < i.If j > i then p α i+1+α i+2 +···+α ja i j .divides a i j , so there are p α 1+α 2 +···+α ichoices forIf j = i then (a i j , p) = 1, so the number of choices for a i j is the size of thegroup of units of Z pα 1 +α 2 +···+α j , which is p α 1+α 2 +···+α j −1 (p − 1).If j < i then there is no restriction on the choice of a i j , so there arep α 1+α 2 +···+α jchoices.Therefore the number of possibilities for the jth column isψ(j) = (p α 1+α 2 +···+α j) j−1 p α 1+α 2 +···+α j −1 (p − 1) ( Π j−1i=1 pα 1+α 2 +···+α i)

94and the order of the automorphism group is given byn∏ψ(j) = p β (p − 1) n for some β.j=1Corollary 4.8. Let G be a finite abelian p-group written using the notationin the statement of Lemma 4.6. If we remove the requirement thatα 1 , α 2 , . . . , α n ≥ 1, then a matrix satisfying conditions (1) and (2) defines anautomorphism of G.Lemmas 4.5 and 4.3 suggest a method of approaching the problem of determiningwhether a given finite abelian group is cyclically generated by aminimal set, and in certain cases combining these results with Corollary 4.7gives a simple proof that a given group does not have a minimal cyclic generatingset. For example, ifG = Z 3 × Z 18 × Z 18 × Z 36 × Z 72 × Z 144then by Lemma 4.5 , if G is cyclically generated by 6 generators, then eachprimary part of G must be cyclically generated by 2, 3, or 6 generators, soin particular, G 2 must be cyclically generated by 2, 3, or 6 generators. ButG 2 = Z 2 × Z 2 × Z 4 × Z 8 × Z 16 ,so it must have a cyclic generating set of size 6 because it has rank 5. Itfollows by Lemma 4.3 that G 2 /G 2 (2) must be cyclically generated by 2, 3,or 6 generators, butG 2 /G 2 (2) ∼ = Z 2 × Z 4 × Z 8 ,so this factor group must be cyclically generated by 3 or 6 generators. However,there are 2 β automorphisms of this group by Lemma 4.7, and thereforethere is no automorphism of order 3 or 6. Therefore G cannot be cyclicallygenerated by 6 generators, by Theorem 4.4. Similar arguments are used inthe next chapters.

Chapter 5Rank 3 and rank 4 abeliangroups with minimal cyclicgenerating setsThe results from the previous chapter are used here to prove necessary andsufficient conditions for an abelian group of rank 3 or rank 4 to have a cyclicgenerating set of minimum size. We begin with the rank 3 result.5.1 Rank 3 abelian groups with minimal cyclicgenerating setsTheorem 5.1. Let G = Z a × Z ab × Z abc where a ≥ 2 and let A be the setof natural numbers that are the product of primes of the form 6λ + 1, and 1.Then G is cyclically generated by 3 generators if and only if either (b, c) isin A, or 1 (b, c) is in A.3In other words, Theorem 5.1 says that if 9, 2, or any prime of the form 6λ−1divides (b, c) then G does not have a minimal cyclic generating set.Proof. Let G = Z a ×Z ab ×Z abc . By Lemma 4.5 a group is cyclically generatedby 3 generators if and only if each of its p-primary parts is either cyclic orcyclically generated by 3 generators.95

96The possible forms for the p-primary part G p of G are:(1) Z p λ(2) Z p λ × Z p λ(3) Z p λ × Z p λ+µ(4) Z p λ × Z p λ × Z p λ(5) Z p λ × Z p λ × Z p λ+µ(6) Z p λ × Z p λ+µ × Z p λ+µ(7) Z p λ × Z p λ+µ × Z p λ+µ+νwhere λ, µ, ν ≥ 1.If G p is of the form (1), then G p is cyclic and hence trivially cyclically generatedby one generator. If G p is of one of the forms (2) or (4)–(6) then itsatisfies Abért’s conditions for n = 3, and hence it is cyclically generated by3 generators by Abért’s Theorem, [1]. (Note that if G p has rank 2 then it canbe thought of as a rank 3 group with trivial first component. For exampleZ p λ × Z ∼ p λ = Z p 0 × Z p λ × Z p λ.) Therefore it remains to consider p-primaryparts of the form (3) or (7), i.e. G p = Z p λ × Z p λ+µ × Z p λ+µ+ν where µ, ν ≥ 1and λ ≥ 0. We show that a G p of one of these forms is cyclically generatedby 3 generators if and only if p ≡ 1 (mod 6), or p = 3 and at least one of µand ν is 1.We begin by constructing an automorphism of order 3 that cycles through agenerating set for G p if p ≡ 1 (mod 6).Let {x, y, z} be the usual basis for G p , i.e.G p = Z p λ × Z p λ+µ × Z p λ+µ+ν= 〈x〉 × 〈y〉 × 〈z〉.As p ≡ 1 (mod 6) there exists α ≠ 1 in Z p λ+µ+ν such that α 3 ≡ 1 (mod p λ+µ+ν ),and 1 + α + α 2 ≡ 0 (mod p λ+µ+ν ), see Lemma 7.4 in the proof of Theorem

977.1. Define a map θ with respect to {x, y, z} by the matrix⎛ ⎞α 0 0⎜⎝0 α 2 ⎟0⎠ .0 0 1(Note that if G p has rank 2 then we can put x = 0 and still define anautomorphism by this matrix.) The map θ is an automorphism by Lemma4.6 and clearly it has order 3. It remains to show that θ cycles through agenerating set for G p .Applying θ repeatedly to the group element x + y + z givesx+y+z ↦→ αx+α 2 y+z ↦→ α 2 x+α 4 y+z = α 2 x+αy+z ↦→ α 3 x+α 3 y+z = x+y+z.Let A = x + y + z, B = αx + α 2 y + z, and C = α 2 x + αy + z, and letH = 〈A, B, C〉. We want to show that H = G p .Adding the generators givesA + B + C = (x + y + z) + (αx + α 2 y + z) + (α 2 x + αy + z) = 3zso z is in H because (3, p) = 1. We can also take the following linear combinationof A, B, and C to show that 3y is in H :A + αB + α 2 C = (x + y + z) + α(αx + α 2 y + z) + α 2 (α 2 x + αy + z) = 3y.Therefore y also lies in H, so G p is cyclically generated by 3 generators.If p ≡ −1 (mod 6) or p = 2 then by Corollary 4.7,|Aut(G p )| = p r rank Gp(p − 1)for some r and therefore G p has no automorphism of order 3.Now consider the case p = 3. If at least one of µ and ν is 1, then G 3 iscyclically generated by 3 generators by Abért’s Theorem. In the case thatboth µ and ν are at least 2 we prove that G 3 is not cyclically generated by3 generators.

98As µ, ν ≥ 2, we can write µ = 2 + µ ′ and ν = 2 + ν ′ where µ ′ , ν ′ ≥ 0, soG 3 = Z 3 λ × Z 3 λ+µ ′ +2 × Z 3 λ+µ ′ +ν ′ +4= 〈x, y, z 3 λ x, 3 λ+µ′ +2 y, 3 λ+µ′ +ν ′ +4 z, [x, y], [y, z], [z, x]〉.We prove that G 3Lemma 4.3.cannot be cyclically generated by 3 generators usingLet H = G 3 /G 3 (3 λ+µ′ ). ThenH = 〈x, y, z 3 λ x, 3 λ+µ′ +2 y, 3 λ+µ′ +ν ′ +4 z, [x, y], [y, z], [z, x], x, 3 2 y, 3 ν′ +4 z〉= 〈y, z 3 2 y, 3 ν′ +4 z, [y, z]〉= Z 3 2 × Z 3 ν ′ +4,and if K = H/3 4 H thenK = 〈y, z 3 2 y, 3 ν′ +4 z, [y, z], 3 4 y, 3 4 z〉= 〈y, z 3 2 y, 3 4 z, [y, z]〉= Z 9 × Z 81 .We prove that Z 9 × Z 81 is not cyclically generated by 3 generators, and thenby Theorem 4.4 it follows that H and hence G 3 cannot be cyclically generatedby 3 generators. GAP, [14], can be used to prove that Z 9 ×Z 81 is not cyclicallygenerated by 3 generators, but we provide an alternative proof here.Let {v, w} be a basis for K whereK = Z 9 × Z 81= 〈v〉 × 〈w〉and suppose that θ is an automorphism of order 3 that cycles through agenerating set for K. If θ is represented (with respect to the basis {v, w}) bythe 2 × 2 matrix A = [a i j ], then by Lemma 4.6, we know that 9 divides a 1 2and (a 1 1 , 3) = (a 2 2 , 3) = 1. We consider the automorphisms induced by θ ontwo factor groups of K to obtain further information about the entries of A.

99Let ϕ be the natural homomorphism K → K/Φ(K) = Z 3 × Z 3 and let θ Fdenote the automorphism induced by θ on K/Φ(K) as defined in the proofof Lemma 2.8. By Lemma 4.3, θ F must cycle through a generating set of size3 for K/Φ(K). The 2 × 2 matrix representing θ F with respect to the basis{vϕ, wϕ} is B = [b i j ] where b i j = a i j (mod 3), so b 1 1 , b 2 2 ∈ {1, −1} andb 1 2 = 0. Therefore det B = b 1 1 b 2 2 = ±1, but B 3 is the identity matrix, sowe must have det B = 1, and therefore b 1 1 = b 2 2 , so( )±1 0B =.b 2 1 ±1Because of the form of B, the 1,1 entry in B 3 is b 3 1 1, so we must have b 1 1 =b 2 2 = 1, and therefore b 2 1 = ±1 since otherwise B would be the identitymatrix. It follows that a 1 1 , a 2 2 ≡ 1 (mod 3), and a 2 1 ≡ ±1 (mod 3).Now let ψ be the natural homomorphism K → K/9K, and let θ ′ denote theautomorphism of K/9K induced by θ. We know that θ ′ must cycle through agenerating set of size 3 and the 2×2 matrix representing θ ′ with respect to thebasis {vψ, wψ} is C = [c i j ] where c i j ≡ a i j (mod 9). From the informationwe have about A we know that c 1 2 = 0, c 1 1 , c 2 2 ≡ 1 (mod 3) and c 2 1 ≢ 0(mod 3). Therefore( ) ( ) ( )C 3 c 1 1 0 c 1 1 0 c 1 1 0=c 2 1 c 2 2 c 2 1 c 2 2 c 2 1 c 2 2( ) ()c 1 1 0 c 2 1 1 0=c 2 1 c 2 2 c 1 1 c 2 1 + c 2 1 c 2 2 c 2 2 2()c 3 1 1 0=c 2 1 1c 2 1 + c 1 1 c 2 1 c 2 2 + c 2 1 c 2 2 2 c 3 2 2and by assumption C 3 = I 3 so we must have c 2 1 1c 2 1 + c 1 1 c 2 1 c 2 2 + c 2 1 c 2 2 2 ≡ 0(mod 9).Now,c 2 1 1c 2 1 + c 1 1 c 2 1 c 2 2 + c 2 1 c 2 2 2 = c 2 1 (c 2 1 1 + c 1 1 c 2 2 + c 2 2 2)and c 2 1 is coprime to 3, so we must havec 2 1 1 + c 1 1 c 2 2 + c 2 2 2 ≡ 0 (mod 9).

100But by completing the square we getc 2 1 1 + c 1 1 c 2 2 + c 2 2 2 ≡ (c 1 1 + 5c 2 2 ) 2 − 6c 2 2 2 (mod 9),so(c 1 1 + 5c 2 2 ) 2 ≡ 6c 2 2 2 (mod 9).Suppose that there exists some x in the group of units of Z 9 such that x 2 ≡6c 2 2 2 (mod 9). Then x = c 2 2 y where y 2 = 6. But then 3 must divide yand hence y 2 ≡ 0 (mod 9), so x ≡ 0 (mod 3), which is a contradiction.Therefore C cannot have order 3. This proves that Z 9 × Z 81 cannot becyclically generated by 3 generators.5.1.1 ExamplesConsider the groupG = Z 14 × Z 196 × Z 1372 .The primary parts of G areG 2 = Z 2 × Z 4 × Z 4andG 7 = Z 7 × Z 49 × Z 343 .If G 2 is written in the form Z a × Z ab × Z abc then b = 2 and c = 1 so G 2satisfies Abért’s conditions for n = 3, and is therefore cyclically generated by3 generators, g 1 , g 2 , g 3 , where o(g i ) = 4.As 7 ≡ 1 (mod 6) we can construct a cyclic automorphism of order 3 for G 7using a cube root of 1 in Z 343 . For example, if α = 18, thenα 3 = 18 3 = 17 × 343 + 1 ≡ 1 (mod 343)and α 2 = 324 ≡ −19 (mod 343) so α + α 2 + 1 ≡ 0 (mod 343). With respectto the usual basis {y 1 , y 2 , y 3 } for G 7 , let θ be the automorphism defined bythe matrix⎛ ⎞18 0 0⎜ ⎟⎝ 0 −19 0⎠ .0 0 1

101θ has order 3 and cycles through the set{y 1 + y 2 + y 3 , 18y 1 − 19y 2 + y 3 , −19y 1 + 18y 2 + y 3 }.Let h 1 = y 1 + y 2 + y 3 , h 2 = 18y 1 − 19y 2 + y 3 , h 3 = −19y 1 + 18y 2 + y 3 , andlet H be the subgroup generated by h 1 , h 2 , h 3 . Adding these generators gives3y 3 and hence y 3 is in H. Moreover,(y 1 + y 2 + y 3 ) + 18(18y 1 − 19y 2 + y 3 ) − 19(−19y 1 + 18y 2 + y 3 ) = 3y 2so y 2 and hence y 1 are in H. Therefore H = G 7 .Any element of G can be written in the form x+y where x ∈ G 2 and y ∈ G 7 .Let ϕ be the automorphism that cycles through the g i and let ψ : G → G bethe map given by (x + y)ψ = xϕ + yθ. Then repeatedly applying ψ to thegroup element g 1 + y 1 + y 2 + y 3 gives the setS = {(g 1 + y 1 + y 2 + y 3 ), (g 2 + 18y 1 − 19y 2 + y 3 ), (g 3 − 19y 1 + 18y 2 + y 3 )}.Clearly multiplying any of these generators by either 343 or 4 gives a cyclicgenerator of G 2 or G 7 respectively, so S is a cyclic generating set for G.As a further example consider the group G = Z 3 ×Z 243 ×Z 2187 . If we write Gin the form Z a × Z ab × Z abc then b = 81 and c = 9, so G 3 does not satisfy theconditions in Theorem 5.1. The factor group G 3 /G 3 (27) is Z 9 × Z 81 , whichwe have seen is not cyclically generated by 3 generators. By Lemma 4.3, itfollows that G is not cyclically generated.5.2 Rank 4 abelian groups with minimal cyclicgenerating setsSearches using GAP suggested the following result for a rank 4 abelian group.Theorem 5.2. Let G = Z a × Z ab × Z abc × Z abcd where a ≥ 2 and let B bethe set of natural numbers that are the products of primes of the form 4λ + 1,and 1. Then G is cyclically generated by 4 generators if and only if either(b, c, d) is in B, or 1 (b, c, d) is in B.2

102In this case the theorem says that if 4 or any prime of the form 4λ−1 divides(b, c, d) then G does not have a minimal cyclic generating set. We shall seewhy the condition is on divisors of (b, c, d) rather than just on the divisors of(c, d).Proof. As with the rank 3 result, we look at the possible forms the p-primaryparts of G can have. However, by Lemma 4.3, we are looking for cyclicgenerating sets of either size 4 or size 2 and Abért notes in his paper that everyfinite abelian group of rank 2 is symmetrically generated by 2 generators.This means that every finite abelian group of rank 2 is cyclically generatedby two generators, and therefore those primary parts of G corresponding toprimes that divide (c, d), but do not divide ab, are cyclically generated by 2generators. This is why the condition in Theorem 5.2 is not a condition on(c, d).The result about rank 2 groups also means that we only need to consider thep-primary parts that have rank 3 or 4. These are:(1) Z p µ × Z p µ × Z p µ(2) Z p µ × Z p µ × Z p µ+σ(3) Z p µ × Z p µ+ν × Z p µ+ν(4) Z p µ × Z p µ+ν × Z p µ+ν+σ(5) Z p λ × Z p λ × Z p λ × Z p λ(6) Z p λ × Z p λ × Z p λ × Z p λ+σ(7) Z p λ × Z p λ × Z p λ+ν × Z p λ+ν(8) Z p λ × Z p λ × Z p λ+ν × Z p λ+ν+σ(9) Z p λ × Z p λ+µ × Z p λ+µ × Z p λ+µ(10) Z p λ × Z p λ+µ × Z p λ+µ × Z p λ+µ+σ(11) Z p λ × Z p λ+µ × Z p λ+µ+ν × Z p λ+µ+ν

103(12) Z p λ × Z p λ+µ × Z p λ+µ+ν × Z p λ+µ+ν+σwhere λ, µ, ν, σ ≥ 1.The existence of a cyclic generating set of size four depends on the choiceof p only in cases (4) and (12). For the other cases, we either use Abért’sTheorem, or we construct a suitable automorphism.In cases (1), (5), (6), and (9), G p satisfies Abért’s conditions for n = 4. Thereforeif G p has one of these forms it is cyclically generated by 4 generators.In cases (2), (3), (7), (8), (10), and (11) we construct an automorphism oforder 4 that cycles through a generating set for G p , with no condition on p.Suppose first that G p is of the form (8). Let {w, x, y, z} be a basis for G pdefined byG p = Z p λ × Z p λ × Z p λ+ν × Z p λ+ν+σ= 〈w〉 × 〈x〉 × 〈y〉 × 〈z〉.Define a map θ, with respect to {w, x, y, z}, by the matrix⎛⎞0 −1 0 01 0 0 0A = ⎜⎟⎝0 1 −1 0⎠ ,that is,0 0 1 1w ↦→ −x, x ↦→ w, y ↦→ x − y, z ↦→ y + z.We show that this map can be extended to an automorphism that cyclesthrough a generating set of size 4 for G p .Note that 0 = p λ (wθ) = p λ (xθ) = p λ+ν (yθ) = p λ+ν+σ (zθ), and θ maps ontoa generating set for G p , so θ can be extended to an automorphism. Now letθ denote this automorphism.Under θ,z ↦→ y + z ↦→ x − y + y + z = x + z ↦→ w + y + z ↦→ −x + x − y + y + z = z,

104andG p = 〈z, y + z, x + z, w + y + z〉 = 〈z, xθ, zθ 2 , zθ 3 〉so G p is cyclically generated by 4 generators.If H p is the factor group G p /p λ+ν G p thenH p = 〈w, x, y, z p λ w, p λ x, p λ+ν y, p λ+ν+σ z,[w, x], [w, y], [w, z], [x, y], [x, z], [y, z],p λ+ν w, p λ+ν x, p λ+ν y, p λ+ν z〉= Z p λ × Z p λ × Z p λ+ν × Z p λ+νwhich is the form of (7), and p λ+ν G p is a characteristic subgroup of G p , soany group of the form (7) is the factor group of a group of the form (8)factored by a characteristic subgroup. As we know that θ as defined abovecycles through a generating set of size 4, Lemma 4.3 tells us that θ mustinduce on H p an automorphism that cycles through a generating set of size4. Therefore any group of the form (7) is cyclically generated by 4 generators.If G p has the form (10), then G p /G p (p λ ) is of the form (2), so by the sameargument as before, we construct an automorphism that produces a cyclicgenerating of size 4 for the group of the form (10) and then this will inducean automorphism of order 4 on G p /G p (p λ ) that cycles through a generatingset.LetG p = Z p λ × Z p λ+µ × Z p λ+µ × Z p λ+µ+σ= 〈w〉 × 〈x〉 × 〈y〉 × 〈z〉.We define a map θ, with respect to {w, x, y, z}, by the matrix⎛⎞−1 0 0 01 0 1 0A = ⎜⎟⎝ 0 −1 0 0⎠ .0 0 1 1As 0 = p λ (wθ) = p λ+µ (xθ) = p λ+µ (yθ) = p λ+µ+σ (zθ) and θ maps onto agenerating set for G p , it can be extended to an automorphism of G p , which

105will be denoted by θ. As before, we have to show that θ cycles through agenerating set of size 4 for G p .Applying successive powers of θ to z we getz ↦→ y+z ↦→ −x+y+z ↦→ −w−y−x+y+z = −w−x+z ↦→ w−w−y+y+z = zand 〈z, y + z, −x + y + z, −w − x + z〉 = G p so G p is cyclically generated by4 generators.Now let G p be of the form (11). Then G p /G p (p λ ) is of the form (3), so wejust need to prove that G p is cyclically generated by 4 generators to provethat a group of the form (2) is also. We write G p asG p = Z p λ × Z p λ+µ × Z p λ+µ+ν × Z p λ+µ+ν= 〈w〉 × 〈x〉 × 〈y〉 × 〈z〉and define a map θ by the matrix⎛⎞1 0 0 01 −1 0 0A = ⎜⎟⎝0 1 0 1⎠ .0 0 −1 0As 0 = p λ (wθ) = p λ+µ (xθ) = p λ+µ+ν (yθ) = p λ+µ+ν (zθ) and θ maps onto agenerating set for G p , it can be extended to an automorphism, which willbe denoted by θ. We now show that θ has order 4 and cycles through agenerating set for G p :z ↦→ −y ↦→ −x − z ↦→ −w + x + y ↦→ −w + w − x + x + z = z,and 〈z, −y, −x − z, −w + x + y〉 = G p , so G p is cyclically generated by 4generators.For the remaining forms of G p , the choice of p affects whether or not G p iscyclically generated by 4 generators. We prove that G p is cyclically generatedby 4 generators if and only if p ≡ 1 (mod 4) or p = 2 and one of µ, ν, σ is atmost 1. We begin with the case that p is odd.

106Let G p be of the form (12) or (4) and let {w, x, y, z} be a basis for G p definedbyG p = Z p λ × Z p λ+µ × Z p λ+µ+ν × Z p λ+µ+ν+σ= 〈w〉 × 〈x〉 × 〈y〉 × 〈z〉.where λ ≥ 0 and if λ = 0, we put w = 0.As p ≡ 1 (mod 4), −1 is a quadratic residue modulo p and hence modulop n for any n, see [16] p.69. Therefore there exists some α in Z p λ+µ+ν+σ suchthat α 2 = −1 and hence α 4 = 1. We use α to construct an automorphismof G p of order 4 that cycles through a generating set. We represent thisautomorphism with respect to the basis {w, x, y, z} by a matrix with entriesover Z p λ+µ+ν+σ. Define a map θ by the matrix⎛⎞α 0 0 00 −1 0 0⎜⎟⎝0 0 −α 0⎠ .0 0 0 1By Lemma 4.6 we know θ is an automorphism, and by choice of α it hasorder 4. It remains to show that it cycles through a generating set for G p .We apply successive powers of θ to w + x + y + z and show that this gives agenerating set for G p . Under θ we havew+x+y+z ↦→ αw−x−αy+z ↦→ −w+x−y+z ↦→ −αw−x+αy+z ↦→ w+x+y+z.LetH = 〈w + x + y + z, αw − x − αy + z, −w + x − y + z, −αw − x + αy + z〉.On adding the first and third generators we get(w + x + y + z) + (−w + x − y + z) = 2x + 2z,so 2x + 2z is in H, and on adding the second and fourth generators we get(αw − x − αy + z) + (−αw − x + αy + z) = −2x + 2z,

107so −2x + 2z is in H. But(2x + 2z) + (−2x + 2z) = 4z,so z is in H and from this it follows that 2x and hence x are in H. Thereforew+y and αw−αy are in H, and so is 2αy because α(w+y)−(αw−αy) = 2αy.As 2α is coprime to p, this means that y is in H and hence w is in H.This completes the proof that if p ≡ 1 (mod 4) and G p is of the form (12)or (4) then it is cyclically generated by 4 generators.Now suppose p ≡ −1 (mod 4) and G p is of the form (4) or (12). If G p has theform (12) then we can factor G p by the characteristic subgroup G p (p λ ) to geta factor group of the form (4). We prove that if p ≡ −1 (mod 4) then a p-group of the form (4) cannot be cyclically generated by 4 generators becauseit has no automorphism of order 4. Then by Theorem 4.4 it will follow thata group of the form (12) cannot be cyclically generated by 4 generators.Let G p be of the form (4). The automorphism group of Z p n is cyclic of orderp n−1 (p − 1) because p is odd, see [30] p.83. As p ≡ −1 (mod 4), the Sylow2-subgroup of each of Aut Z p µ, Aut Z p µ+ν, and Aut Z p µ+ν+σ, is Z 2 . Thereforewe haveZ 2 × Z 2 × Z 2 ≤ Aut Z p µ × Aut Z p µ+ν × Aut Z p µ+ν+σ ≤ Aut G p .But by Corollary 4.7, the size of Aut G p is p r (p − 1) 3 for some r, so the Sylow2-subgroup of Aut G p has size 8 and therefore must beZ 2 × Z 2 × Z 2 .Hence Aut G p has no element of order 4.Finally we consider the case p = 2. If G p is of one of the forms (4) or (12)and one of µ, ν, σ is at most 1 then we define an automorphism of order 4that cycles through a generating set for G p . Conversely, if µ, ν, σ are all atleast 2 then we prove that G p cannot be cyclically generated by 4 generators.Let G 2 be of the form (12). There are three cases to consider: µ = 1, ν = 1,and σ = 1. In each case we construct a suitable automorphism. If µ = 1,

108then define a map θ with respect to the usual basis {w, x, y, z} by the matrix⎛⎞−1 −2 0 01 1 0 0A = ⎜⎟⎝ 0 1 −1 0⎠ .0 0 1 1By Lemma 4.6, θ is an automorphism of G 2 . We show that θ has order 4 andcycles through a generating set for G 2 . As⎛⎞ ⎛⎞ ⎛⎞−1 −2 0 0 −1 −2 0 0 −1 0 0 0A 2 1 1 0 01 1 0 0= ⎜⎟ ⎜⎟⎝ 0 1 −1 0⎠⎝ 0 1 −1 0⎠ = 0 −1 0 0⎜⎟⎝ 1 0 1 0⎠0 0 1 1 0 0 1 1 0 1 0 1we have⎛⎞ ⎛⎞ ⎛ ⎞−1 0 0 0 −1 0 0 0 1 0 0 0A 4 0 −1 0 00 −1 0 0= ⎜⎟ ⎜⎟⎝ 1 0 1 0⎠⎝ 1 0 1 0⎠ = 0 1 0 0⎜ ⎟⎝0 0 1 0⎠0 1 0 1 0 1 0 1 0 0 0 1so o(θ) = 4.We now show that 〈z, zθ, zθ 2 , zθ 3 〉 = G 2 . Applying successive powers of θ toz givesz ↦→ z+y ↦→ z+y−y+x = z+x ↦→ z+y+x+w ↦→ z+y−y+x+x+w−2x−w = z,which is clearly a generating set for G 2 .Now consider the case ν = 1 and define a map θ by the matrix⎛⎞−1 0 0 01 1 −2 0A = ⎜⎟⎝ 0 1 −1 0⎠ .0 0 1 1By Lemma 4.6 the map defined by A is an automorphism of G 2 , but againwe also need to check that θ has order 4 and cycles through a generating set

109for G 2 . As⎛⎞ ⎛⎞ ⎛⎞−1 0 0 0 −1 0 0 0 1 0 0 0A 2 1 1 −2 01 1 −2 0= ⎜⎟ ⎜⎟⎝ 0 1 −1 0⎠⎝ 0 1 −1 0⎠ = 0 −1 0 0⎜⎟⎝1 0 −1 0⎠0 0 1 1 0 0 1 1 0 1 0 1we have⎛⎞ ⎛⎞ ⎛ ⎞1 0 0 0 1 0 0 0 1 0 0 0A 4 0 −1 0 00 −1 0 0= ⎜⎟ ⎜⎟⎝1 0 −1 0⎠⎝1 0 −1 0⎠ = 0 1 0 0⎜ ⎟⎝0 0 1 0⎠0 1 0 1 0 1 0 1 0 0 0 1so o(θ) = 4. Applying θ to z givesz ↦→ z + y ↦→ z + y − y + x = z + x↦→ z + y − 2y + x + w = z − y + x + w↦→ z + y + y − x − 2y + x + w − w = z,so G 2 is cyclically generated by 4 generators.Finally we consider the case σ = 1 and define a map θ by the matrix⎛⎞−1 0 0 01 1 0 0A = ⎜⎟⎝ 0 1 −1 −2⎠ .0 0 1 1Again θ is an automorphism by Lemma 4.6, and we check now that θ hasorder 4 and cycles through a generating set for G 2 . As⎛⎞ ⎛⎞ ⎛⎞−1 0 0 0 −1 0 0 0 1 0 0 0A 2 1 1 0 01 1 0 0= ⎜⎟ ⎜⎟⎝ 0 1 −1 −2⎠⎝ 0 1 −1 −2⎠ = 0 1 0 0⎜⎟⎝1 0 −1 0 ⎠0 0 1 1 0 0 1 1 0 1 0 −1we have⎛⎞ ⎛⎞ ⎛ ⎞1 0 0 0 1 0 0 0 1 0 0 0A 4 0 1 0 00 1 0 0= ⎜⎟ ⎜⎟⎝1 0 −1 0 ⎠ ⎝1 0 −1 0 ⎠ = 0 1 0 0⎜ ⎟⎝0 0 1 0⎠0 1 0 −1 0 1 0 −1 0 0 0 1

110so o(θ) = 4.By applying θ to z we getz ↦→ z + y ↦→ z + y − 2z − y + x = −z + x↦→ −z − y + x + w ↦→ −z − y + 2z + y − x + x + w − w = zso G 2 is cyclically generated by 4 generators.By Lemma 4.3 this means that G 2 is also cyclically generated by 4 generatorsif G 2 is of the form (4) and one of µ, ν, σ is at most 1.Now we come to the proof that G 2 cannot be cyclically generated by 4 generatorsif µ, ν, σ ≥ 2.As µ ≥ 2, we can write µ = µ ′ + 2, soG 2 = Z 2 λ × Z 2 λ+µ ′ +2 × Z 2 λ+µ ′ +ν+2 × Z 2 λ+µ ′ +ν+σ+2.By Lemma 4.3, if G 2 is cyclically generated by 4 generators then so isG 2 /G 2 (2 λ+µ′ ). We prove that G 2 cannot be cyclically generated by 4 generatorsby proving that G 2 /G 2 (2 λ+µ′ ) is not. The proof follows the methodof a proof that Z 4 × Z 16 × Z 64 is not cyclically generated by 4 generatorsprovided by Anton Evseev. A proof of this result was also provided by PabloSpiga and employed a similar argument, but the details presented here arebased on the proof by Evseev.Let H = G 2 /G 2 (2 λ+µ′ ), soH = Z 2 2 × Z 2 ν+2 × Z 2 ν+σ+2.Suppose, for a contradiction, that H is cyclically generated by 4 generators,i.e. there exists an automorphism θ of order 4 that cycles through a generatingset for H. Let A be the matrix representing θ with respect to the usualbasis for H. By Lemma 4.6, A must be of the form⎛⎞a 1 1 a 1 2 a 1 3⎜⎟⎝a 2 1 a 2 2 a 2 3 ⎠a 3 1 a 3 2 a 3 3

111where a 1 1 , a 2 2 , a 3 3 must be odd and 2 ν |a 1 2 , 2 ν+σ |a 1 3 , and 2 σ |a 2 3 .The Frattini subgroup of H is characteristic and the factor group H/Φ(H)is isomorphic to Z 2 × Z 2 × Z 2 . The automorphism θ induced on H/Φ(H) byθ is given, with respect to the usual basis, by a matrix, B, whose entries arethe result of reducing the entries in A modulo 2, i.e. by a matrix of the form⎛ ⎞1 0 0⎜ ⎟⎝b 2 1 1 0⎠b 3 1 b 3 2 1where b 2 1 , b 3 1 and b 3 2 are in Z 2 . The matrix B lies in a copy of D 8 , whichsits inside Aut (H/Φ(H)), and consists of lower triangular matrices with alldiagonal entries 1 and entries below the diagonal either 0 or 1. As θ cyclesthrough a generating set for H/Φ(H), B must have order 4. Hence it is oneof two elements of D 8 . Either⎛ ⎞⎛ ⎞1 0 01 0 0⎜ ⎟⎜ ⎟B = ⎝1 1 0⎠ or B = ⎝1 1 0⎠ .0 1 11 1 1However, since these are mutual inverses, we can assume that B has the firstform without any loss of generality.Now let C be the matrix produced when the entries of A are reduced modulo4. This is the matrix defining the automorphism induced by θ on the factorgroup H/4H. We know that 4|a 1 2 , 4|a 1 3 , and 4|a 2 3 because ν, σ ≥ 2 andhence c 1 2 = c 1 3 = c 2 3 = 0. Moreover, we know that the diagonal entriesof C must be ±1, and from the form of B we know that c 3 1 ≡ 0 (mod 2).Therefore⎛⎞c 1 1 0 0⎜⎟C = ⎝±1 c 2 2 0 ⎠ .2u ±1 c 3 3We can conjugate C by diagonal matrices with entries ±1 to bring C intothe form,⎛⎞c 1 1 0 0⎜⎟Ĉ = ⎝ 1 c 2 2 0 ⎠ ,2u 1 c 3 3

112because⎛ ⎞ ⎛⎞ ⎛ ⎞ ⎛⎞1 0 0 c 1 1 0 0 1 0 0 c 1 1 0 0⎜ ⎟ ⎜⎟ ⎜ ⎟ ⎜⎟⎝0 −1 0⎠⎝−1 c 2 2 0 ⎠ ⎝0 −1 0⎠ = ⎝ 1 c 2 2 0 ⎠ ,0 0 1 2u −1 c 3 3 0 0 1 2u 1 c 3 3⎛ ⎞ ⎛⎞ ⎛ ⎞ ⎛⎞1 0 0 c 1 1 0 0 1 0 0 c 1 1 0 0⎜ ⎟ ⎜⎟ ⎜ ⎟ ⎜⎟⎝0 1 0 ⎠ ⎝ 1 c 2 2 0 ⎠ ⎝0 1 0 ⎠ = ⎝ 1 c 2 2 0 ⎠ ,0 0 −1 2u −1 c 3 3 0 0 −1 −2u 1 c 3 3and ⎛ ⎞ ⎛⎞ ⎛ ⎞ ⎛⎞−1 0 0 c 1 1 0 0 −1 0 0 c 1 1 0 0⎜ ⎟ ⎜⎟ ⎜ ⎟ ⎜⎟⎝ 0 1 0⎠⎝−1 c 2 2 0 ⎠ ⎝ 0 1 0⎠ = ⎝ 1 c 2 2 0 ⎠ ,0 0 1 2u 1 c 3 3 0 0 1 −2u 1 c 3 3and −2u ≡ 2u (mod 4). On squaring Ĉ we get⎛⎞c 2 1 1 0 0Ĉ 2 ⎜= ⎝ c 1 1 + c 2 2 c 2 ⎟2 2 0 ⎠ .1 + 2u(c 1 1 + c 3 3 ) c 2 2 + c 3 3 c 2 3 3Note that c 2 i i = 1, and either c i i = c j j , in which case c i i + c j j = ±2, orc i i = −c j j , in which case c i i + c j j = 0. Therefore⎛ ⎞1 0 0Ĉ 2 ⎜ ⎟= ⎝2v 1 0⎠1 2w 1for some v, w. It follows that⎛⎞ ⎛ ⎞1 0 0 1 0 0Ĉ 4 ⎜⎟ ⎜ ⎟= ⎝ 4v 1 0⎠ = ⎝0 1 0⎠ ≠ I 4 .1 + 4vw + 1 4w 1 2 0 1Therefore C is conjugate to a matrix whose order is not 4, so C cannothave order 4, which is a contradiction. Therefore Z 2 2 × Z 2 ν+2 × Z 2 ν+σ+2not cyclically generated by four generators if ν, σ ≥ 2, so by Theorem 4.4,Z 2 λ × Z 2 λ+µ × Z 2 λ+µ+ν × Z 2 λ+µ+ν+σ cannot be cyclically generated by fourgenerators if µ, ν, σ ≥ 2. This completes the proof for cases (4) and (12), andhence the proof of Theorem 5.2.is

1135.2.1 ExampleLet G = Z 5 ×Z 75 ×Z 2,250 ×Z 90,000 . Then the primary parts of G are as followsG 2 = Z 2 × Z 16G 3 = Z 3 × Z 9 × Z 9G 5 = Z 5 × Z 25 × Z 125 × Z 625 .As G 2 has rank 2 it is cyclically generated by 2 generators by the notefollowing Theorem 1 (Abért’s Theorem) in [1].The subgroup G 3 is of the form (3), so if {y 1 , y 2 , y 3 } is the usual basis for G 3then the map defined byy 1 ↦→ −y 1 , y 2 ↦→ y 1 + y 3 , y 3 ↦→ −y 2is a cyclic automorphism of order 4.Finally, if {z 1 , z 2 , z 3 , z 4 } is the usual basis for G 5 then we define an automorphismθ with respect to the z i by the matrix⎛⎞182 0 0 00 −1 0 0⎜⎟⎝ 0 0 −182 0⎠ .0 0 0 1Applying θ to z 1 + z 2 + z 3 + z 4 we getz 1 + z 2 + z 3 + z 4 ↦→ 182z 1 − z 2 − 182z 3 + z 4 ↦→ −z 1 + z 2 − z 3 + z 4↦→ −182z 1 − z 2 + 182z 3 + z 4 ↦→ z 1 + z 2 + z 3 + z 4 .Let H be the subgroup generated by this set. Adding the first and thirdgenerators shows that 2z 2 + 2z 4 is in H, and adding the second and fourthshows that −2z 2 +2z 4 is in H, so H contains 4z 4 and hence z 4 , because z 4 hasorder 625. From this it follows that z 2 lies in H, and hence H also containsz 1 + z 3 and 182z 1 − 182z 3 . Therefore H contains182(z 1 + z 3 ) − (182z 1 − 182z 3 ) = 364z 3 .

114As 364 and 125 are coprime, it follows that z 3 and hence z 1 are in H.As each of the primary parts of G p are cyclically generated by 2 or 4 generators,the automorphism cycling through a generating set for each G p canbe composed in the natural way to give an automorphism of G of order 4.Applying this automorphism to the elementx 1 + y 1 + z 1 + z 2 + z 3 + z 4gives a generating set because we can multiply each element of this set by5625(= 9 × 625), 2500(= 4 × 625), or 36, to get the generators of G 2 , G 3 ,and G 5 respectively.

Chapter 6Rank 5 abelian groupsIt might be natural to suppose that necessary and sufficient conditions for anarbitrary rank 5 abelian group Z a ×Z ab ×Z abc ×Z abcd ×Z abcde to be cyclicallygenerated by 5 generators would be that the greatest common divisor of b, c, d,and e is 1 or a product of 5 and primes of the form 10λ + 1. We shall seethat if a group satisfies this condition then the primary parts correspondingto primes of the form 10λ + 1 do have cyclic generating sets of size 5, butas the following example shows, there are groups satisfying this condition on(b, c, d, e) that are not cyclically generated by 5 generators. Let G be thegroup Z 5 × Z 5 × Z 5 × Z 10 × Z 100 , so (b, c, d, e) = 1. The primary part G 5satisfies Abért’s conditions for n = 5 and is therefore cyclically generatedby 5 generators, but G 2 = Z 2 × Z 4 has no automorphism of order 5, soG is not cyclically generated by 5 generators. This suggests that a generaltheorem for rank 5 groups would require a further condition involving (d, e),but these two conditions are still not sufficient. For example, searches usingGAP proved that the group H = Z 19 × Z 19 2 × Z 19 3 × Z 19 3does not have acyclic generating set of size 5 while the group K = Z 19 × Z 19 × Z 19 3 × Z 19 3does, even though in each case 19 does not divide (b, c, d, e) or (d, e). Wegive a classification of rank 5 groups of order coprime to 5 that have minimalcyclic generating sets and also give some partial results for 5-groups.115

1166.0.2 Why is the rank 5 case different from ranks 3and 4?Theorem 5.1 says that if G is a finite abelian group of rank 3 such that,for some prime p of the form 6λ − 1, G p is not cyclic, then G p is cyclicallygenerated by 3 generators if and only if G p satisfies Abért’s conditions forn = 3. The analogous statement for finite abelian groups of rank 5 and primesof the form 10λ − 1 is false in general. In other words, there are examples ofnoncyclic finite abelian p-groups where p ≡ −1 (mod 10) and the group iscyclically generated by 5 generators even though it does not satisfy Abért’sconditions for n = 5. For example, the group Z 19 × Z 19 × Z 19 2 × Z 19 2 iscyclically generated by 5 generators. In general, however, it is also false thatif p is a prime of the form 10λ − 1 then G p is always cyclically generated by 5generators, as Z 19 × Z 19 2 × Z 19 3 × Z 19 3 is not. These examples also show thatconditions involving only (b, c, d, e) and (d, e) are still insufficient because inboth cases (b, c, d, e) = 1 and (d, e) = 1 if we express the groups in the formZ a × Z ab × Z abc × Z abcd × Z abcde . We need to look more closely at the primedivisors of each of a, b, c, d, and e.The task of determining whether a given p-primary part of a groupG = Z a1 × Z a1 a 2× Z a1 a 2 a 3× · · · × Z a1 a 2 a 3···a n−1× Z a1 a 2 a 3···a n−1 a nhas a cyclic generating sets of size 5 requires much more analysis of thestructure of G p than the rank 3 and 4 cases. One reason for this is thatthere are many possible forms for the primary parts of a group of rank 5that do not satisfy the first condition in Abért’s Theorem for n = 5 or correspondto primes dividing (b, c, d, e). The primary parts having the structurein Abért’s Theorem and the primary parts corresponding to prime divisorsof (a 1 , a 2 , . . . , a n−1 , a n ) can be thought of as being at opposing ends of therange of possibilities for the structure of the primary parts of G, and only forgroups of rank at most 3 do these structures coincide. We need to use differentresults to deal with the primary parts lying between these two extremes.A further reason why the problem is more complicated is that unlike in thecase of 6 or 4, if we work modulo 10 then we cannot write all primes ≥ 5

117in the form 10k ± 1. We have to consider primes of the form 10k ± 1 andprimes of the form 10k ± 3. We shall see that primary parts corresponding toprimes of the form 10k + 1 are always cyclically generated by 5 generators,while those corresponding to primes of the form 10k ± 3 are only cyclicallygenerated by 5 generators if they satisfy Abért’s conditions for n = 5. Wehave seen already that for primes of the form 10k − 1 the results are morecomplicated. In the rank 3 case we can think of primes as ‘good’ if theyare or of the form 6k + 1 and ‘bad’ if they are of the form 6k − 1, andprimary parts corresponding to good primes are always cyclically generatedby 3 generators, whereas primary parts corresponding to bad primes areonly cyclically generated by 3 generators if they satisfy Abért’s conditionsfor n = 3. In the same way, for the rank 4 result, we can say primes are goodif they are of the form 4k + 1 and bad if they are of the form 4k − 1, and onlyprimary parts corresponding to good primes are always cyclically generatedby 4 generators. The splitting into good and bad primes corresponds tothe way primes split modulo 6 according to whether certain numbers arequadratic residues. For example, writing primes modulo 6, −3 is a quadraticresidue modulo powers of good primes and a quadratic non-residue modulopowers of bad primes. Similarly, if write primes modulo 4, −1 is a quadraticresidue modulo powers of good primes and a quadratic non-residue modulopowers of bad primes. Writing primes modulo 10 we can still say that primesare good if 5 is a quadratic residue and bad if it is a quadratic non-residue,but now the primary part corresponding to a good prime is not necessarilycyclically generated by 5 generators.6.0.3 A note on computational resultsWhile GAP was used to carry out some limited searches for p-groups withcyclic automorphisms of order 5, this produced considerably less data thanin the case of rank 3 and 4 groups. This is partly because of the size ofthe groups and partly because of the size of primes of the form 10λ − 1.Other methods, such as considering the orders of elements of GL(2, p) andGL(3, p), provided much more information. GAP was especially useful herebecause it can find conjugacy class representatives of general linear groups

118very quickly. Generalizing and simplifying some of these representatives byhand gave possible candidates for cyclic automorphisms of order 5, and thesewere applied to suitable group elements. ACE was used to check whether thisactually produced generating sets.6.1 Preliminary ideasWe will make use of the following results.(i) If G p is of the formZ p s 1 × Z p s 2 × . . . Z p snwhere s 1 < s 2 < · · · < s n , then |AutG| = p s (p − 1) n . See Corollary 4.7.(ii) If G p has rank n > 1 then G p /Φ(G p ) = Z p × Z p × . . . × Z p , (the directproduct of n copies of Z p ), which has automorphism group GL(n, p).(iii) If G = Z p m × Z p m × · · · × Z p m (n copies of Z p m), then Aut G =GL(n, Z p m), the group of n × n matrices with entries from the ringZ p mand determinant coprime to p, see [28] p.30.(iv) The size of the general linear group GL(n, p) is Π n−1i=0 (pn − p i ). In particular,|GL(2, p)| = p(p + 1)(p − 1) 2and|GL(3, p)| = p 3 (p 2 + p + 1)(p + 1)(p − 1) 3 .(v) For any group G, if G has a characteristic subgroup H such that G/His not cyclic and is not cyclically generated by 5 generators, then G isnot cyclically generated by 5 generators; see Theorem 4.4.(vi) An automorphism θ cycles through a generating set for a group G ifand only if every automorphism conjugate to θ in Aut G cycles througha generating set for G; see Theorem 2.4.We also need the following Lemmas.

119Lemma 6.1. Let p be a prime and letG = Z p λ × Z p λ × Z p λ × Z p λ+µ= 〈w〉 × 〈x〉 × 〈y〉 × 〈z〉,where λ, µ ≥ 1. The matrix⎛⎞a 1 1 a 1 2 a 1 3 a 1 4a 2 1 a 2 2 a 2 3 a 2 4A = ⎜⎟⎝a 3 1 a 3 2 a 3 3 a 3 4 ⎠a 4 1 a 4 2 a 4 3 a 4 4represents an automorphism of G (with respect to {w, x, y, z}) if and only if(i) the submatrix A ′ = [a i j ] 3 i,j=1 is in GL(3, Z p λ),(ii) p µ |a i 4 for i = 1, 2, 3, and(iii) (a 4 4 , p) = 1.Proof. Let θ be the map defined by A. We show first that if A satisfiesconditions (i),(ii), and (iii) then θ is an automorphism of G.The group G has presentation〈w, x, y, z p λ w, p λ x, p λ y, p λ+µ z, [w, x], [w, y], [w, z], [x, y], [x, z], [y, z]〉.Let φ : Z p λ × Z p λ × Z p λ → Z p λ × Z p λ × Z p λ be the map represented (withrespect to {w, x, y}) by the matrix A ′ . As Aut(Z p λ ×Z p λ ×Z p λ) = GL(3, Z p λ),we know that φ is an automorphism. To show that θ is a homomorphismwe just need to show that for each non-commutator relator r = r(w, x, y, z)in the above presentation for G we have r(wθ, xθ, yθ, zθ) = 0. As φ is anautomorphism we havep λ (wθ) = p λ (a 1 1 w + a 1 2 x + a 1 3 y + a 1 4 z)= p λ (a 1 1 w + a 1 2 x + a 1 3 y) + p λ a 1 4 z= p λ wφ + p λ+µ bz for some b, by (ii)= 0 + 0 because p λ+µ z = 0.

120Similarly p λ xθ = p λ yθ = 0, and as p λ+µ is the exponent of G, we also havep λ+µ (zθ) = 0, so θ is a homomorphism.Note that the map represented by A is invertible if and only if (det A, p) = 1.By expanding det A about the last column and reducing modulo p we getdet A ≡ a 4 4 det A ′ (mod p), by (ii)and (det A ′ , p) = 1, by (i), so (a 4 4 det A ′ , p) = 1 by (iii).automorphism.Hence θ is anConversely, suppose that θ is an automorphism. We show that (i),(ii), and(iii) must hold. Suppose first that (ii) does not hold, i.e. p µ does not dividea i 4 for one of i = 1, 2, 3. Consider first the case i = 1, so a 1 4 = p α b whereα < µ and (b, p) = 1. Then, by the assumption that θ is a homomorphism,we have0 = p λ wθ= p λ a 1 1 w + p λ a 1 2 x + p λ a 1 3 y + p λ a 1 4 z= 0 + 0 + 0 + p λ+α bz≠ 0.Similarly p µ must divide a 2 4 and a 3 4 , so (ii) must hold if θ is a homomorphism.As (ii) holds, we know that det A ≡ a 4 4 det A ′ (mod p). But if either (i) or(iii) does not hold, a 4 4 det A ′ ≡ 0 (mod p) and hence det A is not a unit inZ p , which is a contradiction because θ is an automorphism so det A must beinvertible.Lemma 6.2. Let p be a prime and letG = Z p λ × Z p λ+µ × Z p λ+µ × Z p λ+µ= 〈w〉 × 〈x〉 × 〈y〉 × 〈z〉.

121where λ, µ ≥ 1. The matrix⎛⎞a 1 1 a 1 2 a 1 3 a 1 4a 2 1 a 2 2 a 2 3 a 2 4A = ⎜⎟⎝a 3 1 a 3 2 a 3 3 a 3 4 ⎠a 4 1 a 4 2 a 4 3 a 4 4represents an automorphism of G (with respect to {w, x, y, z}) if and only if(i) the submatrix A ′ = [a i j ] 4 i,j=2 is in GL(3, Z p λ+µ),(ii) p µ |a 1 j for j = 2, 3, 4, and(iii) (a 1 1 , p) = 1.Proof. The proof follows a similar method to the previous result. Let θ bethe map defined by A. We show first that if A satisfies conditions (i),(ii), and(iii) then θ is an automorphism of G.The group G has presentation〈w, x, y, z p λ w, p λ+µ x, p λ+µ y, p λ+µ z, [w, x], [w, y], [w, z], [x, y], [x, z], [y, z]〉.Let θ be the map represented by the matrix A. To show that θ is a homomorphismwe need to show that r(wθ, xθ, yθ, zθ) = 0 for each non-commutatorrelator r = r(w, x, y, z) in the presentation above. Note that by (ii) we havep λ (wθ) = p λ (a 1 1 w + a 1 2 x + a 1 3 y + a 1 4 z)= p λ a 1 1 w + p λ (a 1 2 x + a 1 3 y + a 1 4 z)= 0 + p λ (p µ bx + p µ cy + p µ dz) for some b, c, d= 0,and also p λ+µ xθ = p λ+µ yθ = p λ+µ zθ = 0 because p λ+µ is the exponent of G,so θ is a homomorphism.The map represented by A is invertible if and only (det A, p) = 1. Expandingdet A about the first row and reducing modulo p we get (by (ii))det A ≡ a 1 1 det A ′(mod p),

122and by (i), (det A ′ , p) = 1, so (a 1 1 det A ′ , p) = 1 by (iii).automorphism.Hence θ is anConversely, suppose θ is an automorphism. We show that (i),(ii), and (iii)must hold. Suppose first that (ii) does not hold, i.e. p µ does not divide a 1 jfor one of j = 2, 3, 4. Consider the case j = 2, so a 1 2 = p α b where α < µ and(b, p) = 1. As θ is a homomorphism, p λ wθ = 0, but then0 = p λ wθ= p λ a 1 1 w + p λ a 1 2 x + p λ a 1 3 y + p λ a 1 4 z= 0 + p λ a 1 2 x + p λ a 1 3 y + p λ a 1 4 z= 0 + p λ+α bx + p λ a 1 3 y + p λ a 1 4 z≠ 0,because p λ+α bx ≠ 0 and w, x, y, z are independent.Similarly p µ must divide a 1 3 and a 1 4 , so (ii) must hold if θ is a homomorphism.As (ii) holds, we know that det A ≡ a 1 1 det A ′ (mod p). But if either (i) or(iii) does not hold, a 1 1 det A ′ ≡ 0 (mod p) and hence det A is not a unit inZ p λ+µ, which is a contradiction because θ is an automorphism so A must beinvertible.Lemma 6.3. If p ≡ −1 (mod 10) then, for any n ∈ N there exist distinct jand k satisfying the congruencex 2 + x − 1 ≡ 0 (mod p n ).Proof. As p ≡ −1 (mod 10), 5 is a quadratic residue modulo p, (see [16],p.76, Theorem 97) and hence modulo p n for any n. Therefore, for any n,there exists r such that r 2 ≡ 5 (mod p n ). Letj = −1 + r2andk = −1 − r ,2

123so j and k are distinct modulo p n . Thenand similarly,j 2 (−1 + r)2+ j − 1 = + −1 + r − 14 2= [(1 − 2r + r2 ) − 2 + 2r − 4]41 − 2r + 5 − 2 + 2r − 4≡ (mod p n )4≡ 0 (mod p n ),k 2 (−1 − r)2+ k − 1 = + −1 − r − 14 2= [(1 + 2r + r2 ) − 2 − 2r − 4]41 + 2r + 5 − 2 − 2r − 4≡ (mod p n )4≡ 0 (mod p n )so j and k satisfy x 2 + x − 1 ≡ 0 (mod p n ) as required.Lemma 6.4. Let p ≡ −1 (mod 10) and let j, k be two distinct solutions ofx 2 + x − 1 ≡ 0 (mod p). Let⎛ ⎞⎛ ⎞1 0 01 0 0⎜ ⎟⎜ ⎟v j = ⎝0 0 −1⎠ , and let v k = ⎝0 0 −1⎠ .0 1 j0 1 kAny element of order 5 in GL(3, p) is conjugate either to v j or to v k .Proof. Let j and k be defined as in the proof of Lemma 6.3.We provethat v j and v k have order 5 in GL(3, p), but include only the details of theproof for v j . The fact that j 2 + j − 1 ≡ 0 (mod p) from Lemma 6.3 is usedrepeatedly.⎛ ⎞ ⎛ ⎞ ⎛⎞ ⎛ ⎞1 0 0 1 0 0 1 0 0 1 0 0vj 2 ⎜ ⎟ ⎜ ⎟ ⎜⎟ ⎜ ⎟= ⎝0 0 −1⎠⎝0 0 −1⎠ = ⎝0 −1 −j ⎠ = ⎝0 −1 −j⎠ ,0 1 j 0 1 j 0 j j 2 − 1 0 j −j

124so⎛ ⎞ ⎛ ⎞ ⎛⎞ ⎛ ⎞1 0 0 1 0 0 1 0 0 1 0 0vj 4 ⎜ ⎟ ⎜ ⎟ ⎜= ⎝0 −1 −j⎠⎝0 −1 −j⎠ = ⎝0 1 − j 2 j + j 2 ⎟ ⎜ ⎟⎠ = ⎝0 j 1⎠ .0 j −j 0 j −j 0 −j − j 2 0 0 −1 0ThereforeAsand⎛ ⎞ ⎛ ⎞ ⎛ ⎞1 0 0 1 0 0 1 0 0vj 5 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟= ⎝0 0 −1⎠⎝0 j 1⎠ = ⎝0 1 0⎠ .0 1 j 0 −1 0 0 0 1|GL(3, p)| = p 3 (p 2 + p + 1)(p + 1)(p − 1) 3|GL(2, p)| = p(p + 1)(p − 1) 2 ,and p ≡ −1 (mod 10), the only factor of the order of GL(3, p) that is divisibleby 5 is (p + 1), which is also a factor of |GL(2, p)|. Therefore, if P 5 andP ′ 5denote Sylow 5-subgroups of GL(3, p) and GL(2, p) respectively, then|P ′ 5| = |P 5 |.Let K be the subgroup of GL(3, p) generated by the matrices of the form⎛⎞1 0 0⎜⎟⎝0 x 1 1 x 1 2 ⎠ ,0 x 2 1 x 2 2where X = [x i j ] 2 i,j=1 is in P 5. ′ Then K is isomorphic to P 5, ′ and therefore hassize equal to the size of P 5 , so K is a Sylow 5-subgroup of GL(3, p), and henceit is conjugate to P 5 . Therefore any matrix Y in GL(3, p) of order a power of5 is conjugate to a matrix in K, and the corresponding X must have orderequal to the order of Y. In particular, this means that any matrix of order 5is conjugate to a matrix of the form⎛⎞1 0 0⎜⎟⎝0 x 1 1 x 1 2 ⎠ ,0 x 2 1 x 2 2

125where X has order 5 in GL(2, p).Therefore we can identify the conjugacy classes of the elements of order 5in GL(3, p) by identifying the conjugacy classes of elements of order 5 inGL(2, p). The following table, which gives conjugacy class representativesand the orders of their centralizers in GL(2, p), is taken from [19], p.326.Class rep g( ) ( ) ( ) ()s 0 s 1 s 0 0 10 s 0 s 0 t −t 1+p t + t ps ∈ Z ∗ p s ∈ Z ∗ p s, t ∈ Z ∗ p t ∈ F p 2 \ F p|C GL(2,p) (g)| (p 2 − 1)(p 2 − p) (p − 1)p (p − 1) 2 p 2 − 1As an element must lie in its own centralizer, it follows from the orders of thecentralizers that the only conjugacy classes that can have elements of order5 must have representatives of the form()0 1m t =−t 1+p t + t pwhere t ∈ F p 2 \ F p .Now consider SL(2, p). This has size|GL(2, p)|,p − 1and SL(2, p) is normal in GL(2, p), so the factor group GL(2, p)/SL(2, p) hassize p − 1. This is not divisible by 5, so the Sylow 5-subgroups of GL(2, p)are all contained in SL(2, p), and hence any element of order a power of 5 inGL(2, p) must be in SL(2, p). In particular, this means that m t is in SL(2, p),and hence det m t = 1. Therefore −t 1+p = −1, so in fact m t has order 5 ifand only ifm t =()0 1−1z( ) 50 1for some z such that= I 2 .−1 z

126As( ) ( ) ( )m 5 0 1 −1 z −1 zt =−1 z −z z 2 − 1 −z z 2 − 1( ) ()0 1 1 − z 2 z 3 − 2z=−1 z 2z − z 3 (z 2 − 1) 2 − z 2()2z − z 3 (z 2 − 1) 2 − z 2=z 2 − 1 + 2z 2 − z 4 2z − z 3 + z(z 2 − 1) 2 − z 3()z(2 − z 2 ) z 4 − 3z 2 + 1=,−z 4 + 3z 2 − 1 z(2 − z 2 ) + z(z 4 − 3z 2 + 1)z must satisfy the two equationsButso eitherz 4 − 3z 2 + 1 = 0 and z(2 − z 2 ) = 1.z 4 − 3z 2 + 1 = (z 2 + z − 1)(z 2 − z − 1),z 2 + z − 1 = 0 or z 2 − z − 1 = 0.As p ≡ −1 (mod 10), both of these quadratics have solutions in Z p because5 is a quadratic residue modulo p. The solutions arez = −1 ± r or z = 1 ± r22where r 2 = 5, i.e. either z = j, or z = k, orz = 1 ± r2 .If z = j or z = k then( ) ( ( ))−1 ± r 1 ∓ 2r + rz(2 − z 2 2) =2 −24( ) ( )−1 ± r 8 − 6 ± 2r=24( ) ( )−1 ± r 1 ± r=2 2−1 + r2=4= 1

127but ifthenz = 1 ± r2( ) ( ( ))1 ± r 1 ± 2r + rz(2 − z 2 2) =2 −24( ) ( )1 ± r 8 − 6 ∓ 2r=2 4( ) ( )1 ± r 1 ∓ r=2 2= 1 − r24= −1so in these cases z does not satisfy both the equations, so m t does not haveorder 5.Therefore there are only two conjugacy classes in GL(2, p) that contain elementsof order 5, one with representative( )0 1−1and the other with representative( )0 1.−1 kNote that conjugating each of these matrices by the matrix( )1 0j0 −1gives ( ) ( )0 1 0 −1∼−1 j 1 jand

128( ) ( )0 1 0 −1∼ .−1 k 1 kWe have shown that any element of order 5 in GL(3, p) is conjugate to anelement of K, so this shows that an element of K of order 5 is conjugate (inGL(3, p)) either to v j or v k , and therefore an element of order 5 in GL(3, p)is conjugate either to v j or v k . It remains to show that v j is not conjugate tov k .Suppose, for a contradiction, that there exists some y in GL(3, p) such thatv j y = yv k . If⎛ ⎞a b c⎜ ⎟y = ⎝d e f⎠g h ithen we have⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞1 0 0 a b c a b c 1 0 0⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝0 0 −1⎠⎝d e f⎠ = ⎝d e f⎠⎝0 0 −1⎠0 1 j g h i g h i 0 1 ki.e.⎛⎞ ⎛⎞a b c a c kc − b⎜⎟ ⎜⎟⎝ −g −h −i ⎠ = ⎝d f kf − e⎠ . (6.1)d + jg e + jh f + ji g i ki − hComparing the 2,1 entries gives d = −g, and substituting this into the equationobtained by comparing the 3,1 entries gives g(j − 1) = g, so g(j − 2) = 0and as j ≠ 2, we must have g = d = 0. Similarly, comparing the 1,2 entriesand the 1,3 entries gives b = c = 0, so we can substitute these in (6.1) to get⎛⎞ ⎛⎞a 0 0 a 0 0⎜⎟ ⎜⎟⎝0 −h −i ⎠ = ⎝0 f kf − e⎠ .0 e + jh f + ji 0 i ki − hFrom the 2,2 and 3,3 entries we have the equations f = −h and f+ji = ki−h,so from these it follows that f + ji = ki + f and hence i(j − k) = 0, so i = 0.

129The 2,3 entry then gives e = kf and the 3,2 entry gives e + jh = 0, sokf = −jh, but f = −h, so hk = jh, and hence h(j − k) = 0, so we musthave h = 0 and f = 0. It follows that e = 0 and therefore⎛ ⎞a 0 0⎜ ⎟y = ⎝0 0 0⎠ ,0 0 0which is a contradiction because we assumed y ∈ GL(3, p). Therefore v j andv k are not conjugate in GL(3, p).6.2 Rank 5 resultTheorem 6.5. Let G be a finite abelian group of rank 5 and order coprime to5. Then G has a minimal cyclic generating set if and only if, for each primep dividing |G| with p ≢ 1 (mod 10), G p is cyclic, or G p satisfies Abért’sconditions for n = 5, or, if p ≡ −1 (mod 10), there exist integers α, β ≥ 0and γ ≥ 1 such that G p = Z p α × Z p β × Z p β × Z p γ × Z p γ.By Lemma 4.5 we know that the group G is cyclically generated by 5 generatorsif and only if each of its noncyclic p-primary parts is cyclically generatedby 5 generators. We split the primes into 3 types and for each type wepresent a theorem that gives necessary and sufficient conditions for a noncyclicabelian p-group of rank at most 5 to be cyclically generated by 5generators. By Lemma 4.5, this proves Theorem 6.5.Let G be a noncyclic abelian p-group of rank at most 5. There are 30 possibilitiesfor the form of G and these are listed below. Whether or not sucha p-group is cyclically generated by 5 generators depends on both the choiceof the prime p and the form of the group.With λ, µ, ν, σ, τ ≥ 1, the 30 possible forms for G are:(1) Z p σ × Z p σ(2) Z p σ × Z p σ+τ

130(3) Z p ν × Z p ν × Z p ν(4) Z p ν × Z p ν × Z p ν+τ(5) Z p ν × Z p ν+σ × Z p ν+σ(6) Z p ν × Z p ν+σ × Z p ν+σ+τ(7) Z p µ × Z p µ × Z p µ × Z p µ(8) Z p µ × Z p µ × Z p µ × Z p µ+τ(9) Z p µ × Z p µ × Z p µ+σ × Z p µ+σ(10) Z p µ × Z p µ × Z p µ+σ × Z p µ+σ+τ(11) Z p µ × Z p µ+ν × Z p µ+ν × Z p µ+ν(12) Z p µ × Z p µ+ν × Z p µ+ν × Z p µ+ν+τ(13) Z p µ × Z p µ+ν × Z p µ+ν+σ × Z p µ+ν+σ(14) Z p µ × Z p µ+ν × Z p µ+ν+σ × Z p µ+ν+σ+τ(15) Z p λ × Z p λ × Z p λ × Z p λ × Z p λ(16) Z p λ × Z p λ × Z p λ × Z p λ × Z p λ+τ(17) Z p λ × Z p λ × Z p λ × Z p λ+σ × Z p λ+σ(18) Z p λ × Z p λ × Z p λ × Z p λ+σ × Z p λ+σ+τ(19) Z p λ × Z p λ × Z p λ+ν × Z p λ+ν × Z p λ+ν(20) Z p λ × Z p λ × Z p λ+ν × Z p λ+ν × Z p λ+ν+τ(21) Z p λ × Z p λ × Z p λ+ν × Z p λ+ν+σ × Z p λ+ν+σ(22) Z p λ × Z p λ × Z p λ+ν × Z p λ+ν+σ × Z p λ+ν+σ+τ(23) Z p λ × Z p λ+µ × Z p λ+µ × Z p λ+µ × Z p λ+µ(24) Z p λ × Z p λ+µ × Z p λ+µ × Z p λ+µ × Z p λ+µ+τ

131(25) Z p λ × Z p λ+µ × Z p λ+µ × Z p λ+µ+σ × Z p λ+µ+σ(26) Z p λ × Z p λ+µ × Z p λ+µ × Z p λ+µ+σ × Z p λ+µ+σ+τ(27) Z p λ × Z p λ+µ × Z p λ+µ+ν × Z p λ+µ+ν × Z p λ+µ+ν(28) Z p λ × Z p λ+µ × Z p λ+µ+ν × Z p λ+µ+ν × Z p λ+µ+ν+τ(29) Z p λ × Z p λ+µ × Z p λ+µ+ν × Z p λ+µ+ν+σ × Z p λ+µ+ν+σ(30) Z p λ × Z p λ+µ × Z p λ+µ+ν × Z p λ+µ+ν+σ × Z p λ+µ+ν+σ+τ .Note that if G is of any of the forms (7), (15), (16), or (23) then G satisfiesAbért’s conditions for n = 5, so G is cyclically generated by 5 generators forany prime p.Theorem 6.6. Let G be a noncyclic finite abelian p-group of rank at most5. If p ≡ 1 (mod 10) then G is cyclically generated by 5 generators.Proof. The proof follows a similar method to part of the proof of Theorem5.1. LetG = Z p λ × Z p λ+µ × Z p λ+µ+ν × Z p λ+µ+ν+σ × Z p λ+µ+ν+σ+τ= 〈v〉 × 〈 w 〉 × 〈 x 〉 × 〈 y 〉 × 〈 z 〉(6.2)where λ, µ, ν, σ, τ ≥ 0 and p ≡ 1 (mod 10). The group of units of Z p λ+µ+ν+σ+τhas p λ+µ+ν+σ+τ−1 (p − 1) elements, so it has an element α of order 5. Notethatα + α 2 + α 3 + α 4 + 1 ≡ 0 (mod p λ+µ+ν+σ+τ ),by Lemma 7.4 in the proof of Theorem 7.1. Define a map θ with respect to{v, w, x, y, z} by the matrix⎛⎞α 0 0 0 00 α 2 0 0 00 0 α 3 0 0.⎜⎝0 0 0 α 4 ⎟0⎠0 0 0 0 1

132By Corollary 4.8, θ is an automorphism and clearly it has order 5. Notethat if the rank of G is less than 5, G can be written in the form (6.2) byextending it by trivial factors. For example, if λ = µ = 0, then let w = v = 0and then the matrix above still defines an automorphism of G.Applying successive powers of θ to v + w + x + y + z givesv + w + x+y + z ↦→ αv + α 2 w + α 3 x + α 4 y + z↦→ α 2 v + α 4 w + αx + α 3 y + z ↦→ α 3 v + αw + α 4 x + α 2 y + z↦→ α 4 v + α 3 w + α 2 x + αy + z ↦→ v + w + x + y + z.LetA = v + w + x + y + z,B = αv + α 2 w + α 3 x + α 4 y + z,C = α 2 v + α 4 w + αx + α 3 y + z,D = α 3 v + αw + α 4 x + α 2 y + z,E = α 4 v + α 3 w + α 2 x + αy + z,and let H be the subgroup generated by A, B, C, D, E. We want to provethat H = G.By adding the generators we getA + B + C + D + E= (v + w + x + y + z) + (αv + α 2 w + α 3 x + α 4 y + z)+ (α 2 v + α 4 w + αx + α 3 y + z) + (α 3 v + αw + α 4 x + α 2 y + z)+ (α 4 v + α 3 w + α 2 x + αy + z)= 5zand as (5, p) = 1, z is in H. Now we can take the following linear combinationsof A, B, C, D, and E to show that H also contains y, x, w, and hence v :

133A + αB + α 2 C + α 3 D + α 4 E= (v + w + x + y + z) + α(αv + α 2 w + α 3 x + α 4 y + z)+ α 2 (α 2 v + α 4 w + αx + α 3 y + z) + α 3 (α 3 v + αw + α 4 x + α 2 y + z)+ α 4 (α 4 v + α 3 w + α 2 x + αy + z)= 5yA + α 2 B + α 4 C + αD + α 3 E= (v + w + x + y + z) + α 2 (αv + α 2 w + α 3 x + α 4 y + z)+ α 4 (α 2 v + α 4 w + αx + α 3 y + z) + α(α 3 v + αw + α 4 x + α 2 y + z)+ α 3 (α 4 v + α 3 w + α 2 x + αy + z)= 5xA + α 4 B + α 3 C + α 2 D + αE= (v + w + x + y + z) + α 4 (αv + α 2 w + α 3 x + α 4 y + z)+ α 3 (α 2 v + α 4 w + αx + α 3 y + z) + α 2 (α 3 v + αw + α 4 x + α 2 y + z)+ α(α 4 v + α 3 w + α 2 x + αy + z)= 5wso w, x, and y all lie in H. Therefore H = G as required.For example, if we take p = 11 and let G = Z 11 × Z 11 × Z 121 then we can usea 5th root of 1 in Z 121 and construct a cyclic automorphism of order 5. Firstnote thatG 11 = Z 11 × Z 11 × Z 121= Z 1 × Z 1 × Z 11 × Z 11 × Z 121= 〈v〉 × 〈 w 〉 × 〈 x 〉 × 〈 y 〉 × 〈 z 〉.Now note that 3 is a 5th root of 1 in Z 121 because 3 5 = 243 ≡ 1 (mod 121).

134Let θ be defined by the matrix⎛⎞3 0 0 0 00 9 0 0 00 0 27 0 0.⎜⎟⎝0 0 0 81 0⎠0 0 0 0 1As v = w = 0 we just need to apply θ to x + y + z. This givesx+y+z ↦→ 27x+81y+z ↦→ 3x+27y+z ↦→ 81x+9y+z ↦→ 9x+3y+z ↦→ x+y+z.The following three linear combinations of these elements give 3z, 3y, and 3xrespectively:(x+y+z)+(27x+81y+z)+(3x+27y+z)+(81x+9y+z)+(9x+3y+z) = 3z(x+y+z)+3(27x+81y+z)+9(3x+27y+z)+27(81x+9y+z)+81(9x+3y+z) = 3y(x+y+z)+9(27x+81y+z)+81(3x+27y+z)+3(81x+9y+z)+27(9x+3y+z) = 3x.It follows immediately that G is cyclically generated by 5 generators.For certain primes p the next result uses Abért’s Theorem and the order ofcertain general linear groups to prove necessary and sufficient conditions onthe form of G for it to have a cyclic generating set of size 5.Theorem 6.7. Let G be a noncyclic finite abelian p-group of rank at most 5.If p = 2 or p ≡ ±3 (mod 10) then G is cyclically generated by 5 generatorsif and only if G satisfies Abért’s conditions for n = 5.Proof. If G satisfies Abért’s conditions for n = 5 then it is cyclically generatedby 5 generators, and G satisfies these conditions if and only if it is ofone of the forms (7), (15), (16), or (23). We show that if G is of any of theother forms then it is not cyclically generated by 5 generators.As p = 2 or p ≡ ±3 (mod 10), we know that 5 does not divide |GL(2, p)|or |GL(3, p)|, so if G is of any of the forms (1),. . . ,(6), then G/Φ(G) is notcyclically generated by 5 generators, and it follows from Theorem 4.4 that G

135is not. Moreover, if G is of the form (9), ..., (14), (17),...,(22), (25), ..., (30)then G has a characteristic subgroup H (as listed in the following table) suchthat G/H is of one of the forms (1),. . .,(6), so G is not cyclically generatedby 5 generators. Note that in the table, λ, µ, ν, σ, τ ≥ 1.G H G/H(9) Z p µ×Z p µ×Z p µ+σ×Z p µ+σ G(p µ ) Z p σ×Z p σ(10) Z p µ×Z p µ×Z p µ+σ×Z p µ+σ+τ G(p µ ) Z p σ×Z p σ+τ(11) Z p µ×Z p µ+ν×Z p µ+ν×Z p µ+ν G(p µ ) Z p ν×Z p ν×Z p ν(12) Z p µ×Z p µ+ν×Z p µ+ν×Z p µ+ν+τ G(p µ ) Z p ν×Z p ν×Z p ν+τ(13) Z p µ×Z p µ+ν×Z p µ+ν+σ×Z p µ+ν+σ G(p µ ) Z p ν×Z p ν+σ×Z p ν+σ(14) Z p µ×Z p µ+ν×Z p µ+ν+σ×Z p µ+ν+σ+τ G(p µ ) Z p ν×Z p ν+σ×Z p ν+σ+τ(17) Z p λ×Z p λ×Z p λ×Z p λ+σ×Z p λ+σ G(p λ ) Z p σ×Z p σ(18) Z p λ×Z p λ×Z p λ×Z p λ+σ×Z p λ+σ+τ G(p λ ) Z p σ×Z p σ+τ(19) Z p λ×Z p λ×Z p λ+ν×Z p λ+ν×Z p λ+ν G(p λ ) Z p ν×Z p ν×Z p ν(20) Z p λ×Z p λ×Z p λ+ν×Z p λ+ν×Z p λ+ν+τ G(p λ ) Z p ν×Z p ν×Z p ν+τ(21) Z p λ×Z p λ×Z p λ+ν×Z p λ+ν+σ×Z p λ+ν+σ G(p λ ) Z p ν×Z p ν+σ×Z p ν+σ(22) Z p λ×Z p λ×Z p λ+ν×Z p λ+ν+σ×Z p λ+ν+σ+τ G(p λ ) Z p ν×Z p ν+σ×Z p ν+σ+τ(25) Z p λ×Z p λ+µ×Z p λ+µ×Z p λ+µ+σ×Z p λ+µ+σ G(p λ+µ ) Z p σ×Z p σ(26) Z p λ×Z p λ+µ×Z p λ+µ×Z p λ+µ+σ×Z p λ+µ+σ+τ G(p λ+µ ) Z p σ×Z p σ+τ(27) Z p λ×Z p λ+µ×Z p λ+µ+ν×Z p λ+µ+ν×Z p λ+µ+ν G(p λ+µ ) Z p ν×Z p ν×Z p ν(28) Z p λ×Z p λ+µ×Z p λ+µ+ν×Z p λ+µ+ν×Z p λ+µ+ν+τ G(p λ+µ ) Z p ν×Z p ν×Z p ν+τ(29) Z p λ×Z p λ+µ×Z p λ+µ+ν×Z p λ+µ+ν+σ×Z p λ+µ+ν+σ G(p λ+µ ) Z p ν×Z p ν+σ×Z p ν+σ(30) Z p λ×Z p λ+µ×Z p λ+µ+ν×Z p λ+µ+ν+σ×Z p λ+µ+ν+σ+τ G(p λ+µ ) Z p ν×Z p ν+σ×Z p ν+σ+τThe remaining forms to consider are (8) and (24), where G is of the formZ p µ × Z p µ × Z p µ × Z p µ+τorZ p λ × Z p λ+µ × Z p λ+µ × Z p λ+µ × Z p λ+µ+τ .

136We prove that G is not cyclically generated by 5 generators if p = 2 or p ≡ ±3(mod 10).Note that if G is of the form (8), thenG/G(p µ−1 ) = Z p × Z p × Z p × Z p 1+τand if G has the form (24) and K = G/G(p λ ) thenK = Z p µ × Z p µ × Z p µ × Z p µ+τandK/K(p µ−1 ) = Z p × Z p × Z p × Z p 1+τ .By Lemma 6.1, ifN = Z p × Z p × Z p × Z p 1+τ= 〈w〉 × 〈x〉 × 〈y〉 × 〈z〉,then a matrix representing an automorphism of N with respect to the basis{w, x, y, z} is of the form⎛⎜⎝C ′⎞p τ ap τ b⎟p τ c⎠d e f gwhere (g, p) = 1 and C ′ ∈ GL(3, p). It follows that|Aut N| = |GL(3, p)|p α ϕ(p) = |GL(3, p)|p α (p − 1) for some α,and as p = 2 or p ≡ ±3 (mod 10), 5 does not divide |GL(3, p)|. ThereforeN has no automorphism of order 5 and hence, by Theorem 4.4, G cannot becyclically generated by 5 generators.Finally we come to primes of the form 10λ−1. As mentioned in the introductionto this chapter, searches using GAP found certain p-groups with p ≡ −1(mod 10) that do not satisfy Abért’s conditions for n = 5 and yet are cyclicallygenerated by 5 generators. Working modulo powers of these primes,

1375 is a quadratic residue, so in certain circumstances we can use this factto construct an automorphism of order 5 that cycles through a generatingset. Where no such automorphism exists, we use Lemmas 6.1 and 6.2 and ageneralization of part of the proof of Theorem 6.7 to prove the result. Thismethod of proof was suggested by the results obtained from GAP by lookingat 19-groups. For the existence side, after finding a matrix that representeda suitable automorphism for a 19-group of rank 3, the process of trying togeneralize this to a general prime led to Lemmas 6.3 and the matrices inLemma 6.4. To prove that there is no such automorphism for certain othergroups, we used the example of a particular 19-group of rank 4. Lemmas6.1 and 6.2 gave a possible form for a matrix representing a suitable automorphism,and an exhaustive search over the remaining unknowns suggestedthat no such matrix exists. Arguments involving conjugate automorphismscompleted the proof for this particular group and generalizing these argumentsand applying Theorem 4.4 gave a proof of the result for the generalcase.Theorem 6.8. Let G be a noncyclic finite abelian p-group of rank at most 5.If p ≡ −1 (mod 10) then G is cyclically generated by 5 generators if and onlyif G satisfies Abért’s conditions for n = 5 or there exist integers α, β ≥ 0and γ ≥ 1 such that G = Z p α × Z p β × Z p β × Z p γ × Z p γ.Proof. Note first that if G satisfies Abért’s conditions for n = 5 then G iscyclically generated by 5 generators and is of one of the forms (7), (15), (16),or (23).If G is of the form (1), (3), (4), (5), (9), (17), (19), (20), (21), or (25) then weuse Lemma 6.3 and work modulo the exponent of G to find roots j and k ofthe equation x 2 +x−1 ≡ 0. We use these to construct a cyclic automorphismof G of order 5.Let G be of the form (1), soG = Z p σ × Z p σ= 〈y〉 × 〈z〉.

138Defining j as in the proof of Lemma 6.3 working modulo p σ , let θ : G → Gbe the map defined, with respect to {y, z}, by the matrix T where( )0 −1T = .1 jNote that Aut G = GL(2, Z p σ) and det T = 1, so θ ∈ AutG.We apply successive powers of θ to y :y ↦→ −z↦→ −y − jz↦→ z − jy − j 2 z = −jy + (1 − j 2 )x = −jy + jz↦→ jz + jy + j 2 z = jy + (j + j 2 )z = jy + z↦→ −jz + y + jz = yso θ cycles through a generating set of order 5.Now let G be of one of the forms (3), (4) or (5). If necessary we can rearrangethe cyclic factors of G so that it has the formZ p α × Z p β × Z p β.Let j be defined as in Lemma 6.3 and let {x, y, z} be the basis given byG = Z p α × Z p β × Z p β= 〈x〉 × 〈y〉 × 〈z〉.Let θ be the map of G defined (with respect to {x, y, z}) by⎛ ⎞1 0 0⎜ ⎟T = ⎝0 0 −1⎠ .0 1 jAs p β (yθ) = 0 = p β (zθ), θ is a homomorphism, and det T = 1, so θ is inAut G.

139We apply successive powers of θ to x + y :x + y ↦→ x − z↦→ x − y − jz↦→ x + z − jy − j 2 z = x − jy + (1 − j 2 )x = x − jy + jz↦→ x + jz + jy + j 2 z = x + jy + (j + j 2 )z = x + jy + z↦→ x − jz + y + jz = x + y.We need to show that θ cycles through a generating set. Let H be thesubgroup generated by x + y, (x + y)θ, (x + y)θ 2 , (x + y)θ 3 , and (x + y)θ 4 , soH = 〈x + y, x − z, x − y − jz, x − jy + jz, x + jy + z〉.On adding the generators of H we get(x + y) + (x − z) + (x − y − jz) + (x − jy + jz) + (x + jy + z) = 5x,so x is in H because (p, 5) = 1. It follows immediately that y and z are in Hand hence H = G as required.Now we consider the following rank 4 and rank 5 forms of G :(9) Z p µ × Z p µ × Z p µ+σ × Z p µ+σ(17) Z p λ × Z p λ × Z p λ × Z p λ+σ × Z p λ+σ(19) Z p λ × Z p λ × Z p λ+ν × Z p λ+ν × Z p λ+ν(20) Z p λ × Z p λ × Z p λ+ν × Z p λ+ν × Z p λ+ν+τ(21) Z p λ × Z p λ × Z p λ+ν × Z p λ+ν+σ × Z p λ+ν+σ(25) Z p λ × Z p λ+µ × Z p λ+µ × Z p λ+µ+σ × Z p λ+µ+σ.We prove that in each case G is cyclically generated by 5 generators. Notethat in each case G has two pairs of repeated cyclic factors so we can rearrangethe factors of G so that it has the formZ p α × Z p β × Z p β × Z p γ × Z p γ,

140where, if necessary, α = β, or in the case of (9), α = 0. The rearrangementsare as follows:(9) G ↦→ Z p µ × Z p µ × Z p µ+σ × Z p µ+σ(17) G ↦→ Z p λ × Z p λ × Z p λ × Z p λ+σ × Z p λ+σ(19) G ↦→ Z p λ+ν × Z p λ+ν × Z p λ+ν × Z p λ × Z p λ(20) G ↦→ Z p λ+ν+τ × Z p λ × Z p λ × Z p λ+ν × Z p λ+ν(21) G ↦→ Z p λ+ν × Z p λ × Z p λ × Z p λ+ν+σ × Z p λ+ν+σ(25) G ↦→ Z p λ × Z p λ+µ × Z p λ+µ × Z p λ+µ+σ × Z p λ+µ+σ.We will now work with G in this new form, with basis {v, w, x, y, z} whereG = Z p α × Z p β × Z p β × Z p γ × Z p γ= 〈v〉 × 〈w〉 × 〈x〉 × 〈y〉 × 〈z〉.Let θ : G → G be the map defined (with respect to {v, w, x, y, z}) by⎛⎞1 0 0 0 00 0 −1 0 0T =0 1 j 0 0.⎜⎟⎝0 0 0 0 −1⎠0 0 0 1 kNote that θ is an automorphism because0 = p β (wθ) = p β (xθ) = p γ (yθ) = p γ (zθ)and det T = 1.

141We apply successive powers of θ to v + w + y :v + w + y ↦→ v − x − z↦→ v − w − jx − y − kz↦→ v + x − jw − j 2 x + z − ky − k 2 z= v − jw + (1 − j 2 )x − ky + (1 − k 2 )z= v − jw + jx − ky + kz↦→ v + jx + jw + j 2 x + kz + ky + k 2 z= v + jw + (j + j 2 )x + ky + (k + k 2 )z= v + jw + x + ky + z↦→ v − jx + w + jx − kz + y + kz= v + w + y.Let H be the subgroup generated by the setso{(v + w + y), (v + w + y)θ, (v + w + y)θ 2 , (v + w + y)θ 3 , (v + w + y)θ 4 },H = 〈v + w + y, v − x − z, v − w − jx − y − kz,On adding the generators of H we getv − jw + jx − ky + kz, v + jw + x + ky + z〉.(v + w + y)+(v − x − z) + (v − w − jx − y − kz)+ (v − jw + jx − ky + kz) + (v + jw + x + ky + z) = 5v,so v is in H. Therefore w + y, −x − z, and −w − jx − y − kz are in H,and so (w + y) + (−w − jx − y − kz) = −jx − kz is in H, and hencek(−x − z) − (−jx − kz) = (j − k)x is in H. By choice of j and k,( ) ( )−1 + r −1 − rj − k =−= 2r22 2 = r,and (r, p) = 1, so x is in H, and hence z is in H.As we know v, x, z are in H, if follows from the last generator of H thatjw + ky is in H. Therefore H contains jw + ky − k(w + y), but this is

142(j − k)w, so w and hence y are in H. Therefore H = G and this proves thatif p ≡ −1 (mod 10) and G is of any of the forms (9),(17),(19),(20),(21) or(25), then G is cyclically generated by 5 generators.Now we prove that if G is not of one of the forms listed in the statement ofthe Theorem then there is no automorphism of order 5 that cycles througha generating set for G.If G is of the form (2) or (6), then by Corollary 4.7, |Aut G| = p s (p − 1) rfor some r and s, so G has no automorphism of order 5 and hence G cannotbe cyclically generated by 5 generators. If G is of the form (10), (14), (18),(22), (26), or (30) then G has a characteristic subgroup H such that G/H ∼ =Z p σ × Z p σ+τ , which again has no automorphism of order 5 by Corollary 4.7.Therefore G cannot be cyclically generated by 5 generators.The remaining forms to consider are (8), (11), (12), (13), (24), (27), (28), and(29). If G is of one of these forms and has rank 5 then G/G(p λ ) is isomorphicto one of the rank 4 groups listed, so if we can show that groups of the form(8),(11),(12), or (13) are not cyclically generated by 5 generators then thefact that the rank 5 groups are not will follow by Theorem 4.4.Let G have either of forms (12) or (13). ThenG/p µ+ν G ∼ = Z p µ × Z p µ+ν × Z p µ+ν × Z p µ+ν,which is of form (11). Therefore, if a group of form (11) is not cyclicallygenerated by 5 generators then neither are groups of forms (12) or (13), sowe only need to prove that groups of the form (8) and (11) are not cyclicallygenerated by 5 generators.Let G be a group either of the form (8) and (11) and let {w, x, y, z} be theusual basis for G. In both cases G has a subgroup of the form Z p α ×Z p α ×Z p α,so if B ′ = {b 1 , b 2 , b 3 } is a basis for this subgroup then B ′ can be extended toa basis B for G in the following way. In the first case, where G is of the form

143(8), we haveG = Z p µ × Z p µ × Z p µ × Z p µ+τ= Z p µ × Z p µ × Z p µ × 〈z〉= 〈b 1 〉 × 〈b 2 〉 × 〈b 3 〉 × 〈z〉,so B = B ′ ∪ {z} is a basis for G, and in the second case, where G is of theform (11), we haveG = Z p µ × Z p µ+τ × Z p µ+τ × Z p µ+τ= 〈w〉 × Z p µ+τ × Z p µ+τ × Z p µ+τ= 〈w〉 × 〈b 1 〉 × 〈b 2 〉 × 〈b 3 〉,so B = {w} ∪ B ′ is a basis for G. By Lemma 6.1, if G is of the form (8) thenan automorphism θ of G is represented with respect to B by the matrix⎛⎞a 1 1 a 1 2 a 1 3 a 1 4a 2 1 a 2 2 a 2 3 a 2 4A = ⎜⎟⎝a 3 1 a 3 2 a 3 3 a 3 4 ⎠a 4 1 a 4 2 a 4 3 a 4 4if and only if(i) the submatrix A ′ = [a i j ] 3 i,j=1 is in GL(3, Z p µ),(ii) p τ |a i 4 for i = 1, 2, 3, and(iii) (a 4 4 , p) = 1and by Lemma 6.2, if G is of the form (11) then an automorphism θ of G isrepresented with respect to B by the matrix⎛⎞a 1 1 a 1 2 a 1 3 a 1 4a 2 1 a 2 2 a 2 3 a 2 4A = ⎜⎟⎝a 3 1 a 3 2 a 3 3 a 3 4 ⎠a 4 1 a 4 2 a 4 3 a 4 4if and only if

144(i) the submatrix A ′ = [a i j ] 4 i,j=2 is in GL(3, Z p µ+τ ),(ii) p τ |a 1 j for j = 2, 3, 4, and(iii) (a 1 1 , p) = 1.Suppose that G has an automorphism θ of order 5 that cycles through agenerating set for G. Then by Lemma 4.3, θ induces an automorphism θ F onG/Φ(G) that also has order 5 and cycles through a generating set for G/Φ(G).Let ϕ : G → G/Φ(G) be the homomorphism given by gϕ = g + Φ(G). Thenθ F is is represented (with respect to the basis Bϕ) by the matrix C = [c i j ]where c i j ≡ a i j (mod p). Because of the form A must have, this means thatin the first case θ F is represented by the matrix⎛⎞c 1 1 c 1 2 c 1 3 0c 2 1 c 2 2 c 2 3 0C = ⎜⎟⎝c 3 1 c 3 2 c 3 3 0 ⎠c 4 1 c 4 2 c 4 3 c 4 4where (c 4 4 , p) = 1 and the submatrix C ′ = [c i j ] 3 i,j=1 is in GL(3, p), and in thesecond case θ F is represented by the matrix⎛⎞c 1 1 0 0 0c 2 1 c 2 2 c 2 3 c 2 4C = ⎜⎟⎝c 3 1 c 3 2 c 3 3 c 3 4 ⎠c 4 1 c 4 2 c 4 3 c 4 4where (c 1 1 , p) = 1 and the submatrix C ′ = [c i j ] 4 i,j=2 is in GL(3, p).Now note that in each case C ′ must have order 5 because in the first case C nhas the form⎛⎞0(C ′ ) n 0⎜⎟⎝0 ⎠α β γ c n 4 4

145and in the second it has the form⎛⎞c n 1 1 0 0 0α⎜⎟⎝ β (C ′ ) n ⎠γfor some α, β, γ.Therefore C ′ lies in one of the two conjugacy classes of GL(3, p) containingelements of order 5. One has representative v j and the other has representativev k , where v j and v k are as defined previously. Without loss of generalitywe can assume that C ′ lies in the class represented by v j , and a suitablechoice of basis B ′ means that we can assume that C ′ = v j .Now we prove that in each case, C is conjugate to the matrix T where⎛⎞1 0 0 00 1 0 0T = ⎜⎟⎝0 0 0 −1⎠ .0 0 1 jThe details are different for the two cases so we consider each case separately.In the first case we can assume that⎛⎞1 0 0 00 0 −1 0C = ⎜⎟⎝ 0 1 j 0 ⎠c 4 1 c 4 2 c 4 3 c 4 4and we know that C 5 is the identity matrix, so this provides further informationabout the c 4 j s. Immediately we have that c 5 4 4 = 1, so c 4 4 = 1. Forease of notation, we write r, s, t, for c 4 1 , c 4 2 , and c 4 3 respectively. Using thefact that j 2 + j − 1 ≡ 0 (mod p) we get⎛⎞1 0 0 0C 2 0 −1 −j 0= ⎜⎟⎝ 0 j −j 0⎠2r t + s t(1 + j) − s 1

146so⎛⎞1 0 0 0C 4 0 1 − j 2 j + j 2 0= ⎜⎟⎝ 0 −j − j 2 0 0⎠4r j(t(1+j)−s) −j(t+s)−j(t(1+j)−s)+t(1+j)−s 1⎛⎞1 0 0 00 j 1 0= ⎜⎟⎝ 0 −1 0 0⎠ .4r t − js −s 1Therefore⎛⎞ ⎛⎞1 0 0 0 1 0 0 0C 5 0 j 1 00 0 −1 0= ⎜⎟ ⎜⎟⎝ 0 −1 0 0⎠⎝0 1 j 0⎠4r t − js −s 1 r s t 1⎛ ⎞1 0 0 00 1 0 0= ⎜ ⎟⎝ 0 0 1 0⎠5r 0 0 1so r = 0 and there is no restriction on s and t. Now we show that C isconjugate to the matrix T.Letx =s(1 − j) + tj − 2andy = t − sj − 2and let Y be the matrix⎛ ⎞1 0 0 00 0 1 0Y = ⎜ ⎟⎝0 0 0 1⎠ .0 1 x y

147Then det Y = 1, so Y is in GL(4, p), and⎛⎞ ⎛ ⎞1 0 0 0 1 0 0 00 0 −1 00 0 1 0CY = ⎜⎟ ⎜ ⎟⎝0 1 j 0⎠⎝0 0 0 1⎠0 s t 1 0 1 x y⎛⎞1 0 0 00 0 0 −1= ⎜⎟⎝0 0 1 j ⎠0 1 s + x t + y⎛⎞1 0 0 0=0 0 0 −1⎜⎝0 0 1 j ⎟⎠0 1 s + (1−j)s+t t + t−sj−2j−2⎛⎞1 0 0 0=0 0 0 −1⎜⎝0 0 1 j ⎟⎠t−s jt−t−s0 1j−2 j−2⎛1 0 0 0=0 0 0 −1⎜⎝0 0 1 jt−s0 1j−2 j−2⎛⎞1 0 0 00 0 0 −1= ⎜⎟⎝0 0 1 j ⎠jt−js+js−t−s⎞⎟⎠0 1 y jy − x⎛ ⎞ ⎛⎞1 0 0 0 1 0 0 00 0 1 00 1 0 0= ⎜ ⎟ ⎜⎟⎝0 0 0 1⎠⎝0 0 0 −1⎠= Y T0 1 x y0 0 1 j

148so C is conjugate to T.Now consider the second case. Again we prove that C is conjugate to T. Weknow that the basis B ′ can be chosen so that C is of the form⎛⎞c 1 1 0 0 0c 2 1 1 0 0C = ⎜⎟⎝c 3 1 0 0 −1⎠c 4 1 0 1 jand by the same argument as before, c 1 1 = 1. Again, for ease of notation wewrite r, s, t for c 2 1 , c 3 1 , and c 4 1 respectively.As⎛⎞ ⎛⎞1 0 0 0 1 0 0 0C 2 s 1 0 0s 1 0 0= ⎜⎟ ⎜⎟⎝t 0 0 −1⎠⎝t 0 0 −1⎠u 0 1 j u 0 1 j⎛⎞1 0 0 02s 1 0 0= ⎜⎟⎝ t − u 0 −1 −j ⎠u(1 + j) + t 0 j j 2 − 1⎛⎞1 0 0 02s 1 0 0= ⎜⎟⎝ t − u 0 −1 −j⎠u(1 + j) + t 0 j −jwe have

149⎛⎞ ⎛⎞1 0 0 0 1 0 0 0C 4 2s 1 0 02s 1 0 0= ⎜⎟ ⎜⎟⎝ t − u 0 −1 −j⎠⎝ t − u 0 −1 −j⎠u(1 + j) + t 0 j −j u(1 + j) + t 0 j −j⎛⎞1 0 0 04s 1 0 0= ⎜⎟⎝−j(u(1 + j) + t) 0 j − j 2 j + j 2 ⎠u(1 + j) + t + j(t − u) − j(u(1 + j) + t) 0 −j − j 2 −j 2 + j 2⎛⎞1 0 0 04s 1 0 0= ⎜⎟⎝−u − jt 0 j 1⎠t 0 −1 0and hence⎛⎞ ⎛⎞1 0 0 0 1 0 0 0C 5 4s 1 0 0s 1 0 0= ⎜⎟ ⎜⎟⎝−u − jt 0 j 1⎠⎝t 0 0 −1⎠t 0 −1 0 u 0 1 j⎛⎞1 0 0 05s 1 0 0= ⎜⎟⎝−u − jt + jt + u 0 1 −j + j⎠t − t 0 0 1so s = 0 and there is no restriction on the choice for t and u.Letand letx =t(1 − j) − u2 − jand⎛ ⎞1 0 0 00 1 0 0Y = ⎜ ⎟⎝x 0 1 0⎠ .y 0 0 1y = t + u2 − j

150Then Y ∈ GL(4, p), and⎛⎞ ⎛ ⎞1 0 0 0 1 0 0 00 1 0 00 1 0 0CY = ⎜⎟ ⎜ ⎟⎝t 0 0 −1⎠⎝x 0 1 0⎠u 0 1 j y 0 0 1⎛⎞1 0 0 00 1 0 0= ⎜⎟⎝ t − y 0 0 −1⎠u + x + jy 0 1 j⎛⎞1 0 0 0=0 1 0 0⎜⎝ t − t+u 0 0 −1⎟2−j ⎠u + t(1−j)−u + j t+u 02−j 2−j⎛⎞1 j1 0 0 0=0 1 0 0⎜ t(1−j)−u⎝ 0 0 −1⎟2−j ⎠t+u0 1 j2−j⎛⎞1 0 0 00 1 0 0= ⎜⎟⎝x 0 0 −1⎠y 0 1 j⎛ ⎞ ⎛⎞1 0 0 0 1 0 0 00 1 0 00 1 0 0= ⎜ ⎟ ⎜⎟⎝x 0 1 0⎠⎝0 0 0 −1⎠y 0 0 1 0 0 1 j= Y Tand hence C is conjugate to T.Therefore, in both cases, the matrix defining θ F is conjugate to T. In otherwords, there is a basis E = {e 1 , e 2 , e 3 , e 4 } for Z p × Z p × Z p × Z p such that T

151is the matrix defining θ F with respect to E. If we can show that an automorphismthat is represented by T cannot cycle through a generating set thenwe will have a contradiction, as required.Suppose that x = t 1 e 1 + t 2 e 2 + t 3 e 3 + t 4 e 4 , soxθ F = t 1 e 1 + t 2 e 2 + t 4 e 3 + (jt 4 − t 3 )e 4xθ 2 F = t 1 e 1 + t 2 e 2 + (jt 4 − t 3 )e 3 − j(t 3 + t 4 )e 4xθ 3 F = t 1 e 1 + t 2 e 2 − j(t 3 + t 4 )e 3 + (jt 3 − t 4 )e 4xθ 4 F = t 1 e 1 + t 2 e 2 + (jt 3 − t 4 )e 3 + t 3 e 4and suppose x is such that {x, xθ F , . . . , xθ 4 F } generates Z p × Z p × Z p × Z p .As 〈e 1 〉, 〈e 2 〉, and 〈e 3 , e 4 〉 are θ F invariant, we cannot have t 1 = 0, t 2 = 0, ort 3 = t 4 = 0. By assumption, the set {x, xθ F , . . . , xθF 4 } is a generating set, soin particular, e 1 ∈ 〈x, xθ F , . . . , xθF 4 〉, i.e. there exist λ 0, λ 1 , . . . , λ 4 in Z p suchthatBut thene 1 = λ 0 x + λ 1 xθ F + λ 2 xθ 2 F + λ 3 xθ 3 F + λ 4 xθ 4 F .e 1 = λ 0 (t 1 e 1 + t 2 e 2 + t 3 e 3 + t 4 e 4 ) + λ 1 (t 1 e 1 + t 2 e 2 + t 4 e 3 + (jt 4 − t 3 )e 4 )+ λ 2 (t 1 e 1 + t 2 e 2 + (jt 4 − t 3 )e 3 − j(t 3 + t 4 )e 4 )+ λ 3 (t 1 e 1 + t 2 e 2 − j(t 3 + t 4 )e 3 + (jt 3 − t 4 )e 4 )+ λ 4 (t 1 e 1 + t 2 e 2 + (jt 3 − t 4 )e 3 + t 3 e 4 )= t 1 (λ 0 + λ 1 + λ 2 + λ 3 + λ 4 )e 1 + t 2 (λ 0 + λ 1 + λ 2 + λ 3 + λ 4 )e 2+ (λ 0 t 3 + λ 1 t 4 + λ 2 (jt 4 − t 3 ) − λ 3 j(t 3 + t 4 ) + λ 4 (jt 3 − t 4 ))e 3+ (λ 0 t 4 + λ 1 (jt 4 − t 3 ) − λ 2 j(t 3 + t 4 ) + λ 3 (jt 3 − t 4 ) + λ 4 t 3 )e 4so t 1 (λ 0 +λ 1 +λ 2 +λ 3 +λ 4 ) = 1 and in particular, t 2 (λ 0 +λ 1 +λ 2 +λ 3 +λ 4 ) = 0.The first of these conditions proves that λ 0 + λ 1 + λ 2 + λ 3 + λ 4 must be a unitin Z p , but as t 2 ≠ 0, we must have λ 0 + λ 1 + λ 2 + λ 3 + λ 4 = 0 from the second

152equation, which is a contradiction. Therefore θ F does not cycle through agenerating set of size 5 and hence G/Φ(G) is not cyclically generated.It follows by Theorem 4.4 that if G is of the form (8) or (11), then G is notcyclically generated by 5 generators, and hence, if G is of the form (12) or(13), then G is not cyclically generated by 5 generators.As an example of the automorphism constructed in the proof of this result,consider the groupG = Z 29 × Z 29 × Z 29 2 × Z 29 3 × Z 29 3.First we need to rearrange the cyclic factors of G and label the basis elements.G = Z 29 2 × Z 29 × Z 29 × Z 29 3 × Z 29 3= 〈v〉 × 〈w〉 × 〈x〉 × 〈y〉 × 〈z〉.We work modulo the exponent of G. Note that 3978 2 +3978−1 ≡ 0 (mod 29 3 )and 20410 2 + 20410 − 1 ≡ 0 (mod 29 3 ), so we put j = 3978 and k = 20410in the proof of Theorem 6.8 and let θ be defined by the matrix⎛⎞1 0 0 0 00 0 −1 0 00 1 3978 0 0.⎜⎟⎝0 0 0 0 −1 ⎠0 0 0 1 20410

153Repeatedly applying θ to w + w + y givesv + w + y ↦→ v − x − z↦→ v − w − 3978x − y − 20410z↦→ v + x − 3978w − 3978 2 x + z − 20410y − 20410 2 z= v − 3978w + (1 − 3978 2 )x − 20410y + (1 − 20410 2 )z= v − 3978w + 3978x − 20410y + 20410z↦→ v + 3978x + 3978w + 3978 2 x + 20410z + 20410y + 20410 2 z= v + 3978w + (3978 + 3978 2 )x + 20410y + (20410 + 20410 2 )z= v + 3978w + x + 20410y + z↦→ v − 3978x + w + 3978x − 20410z + y + 20410z= v + w + yso it remains to show that the setS = {v + w + y, v − x − z, v − w − 3978x − y − 20410z,v − 3978w + 3978x − 20410y + 20410z, v + 3978w + x + 20410y + z}generates G. Let H be the subgroup generated by S. Adding the generatorsof H gives(v + w + y)+(v − x − z) + (v − w − 3978x − y − 20410z)+ (v − 3978w + 3978x − 20410y + 20410z)+ (v + 3978w + x + 20410y + z) = 5v,so v is in H. It follows that w +y, −x−z, and −w −3978x−y −20410z mustbe in H, and therefore (w+y)+(−w−3978x−y−20410z) = −3978x−20410zis in H, and hence H contains20410(−x − z) − (−3978x − 20410z) = (3978 − 20410)x = −16432x.But (−16432) 2 ≡ 5 (mod 29 3 ), and therefore 29 and −16432 are coprime, sox is in H, and hence z is in H. From the last generator of H, we know thatH contains 3978w + 20410y and we already know that H contains w + y, soH also contains3978(w + y) − (3978w + 20410y) = (3978 − 20410)y,

154and by the argument above it follows that y and hence w are in H, so H = Gas required.Theorem 6.5 follows by combining Theorems 6.6, 6.7, and 6.8 with Lemma4.5.6.2.1 A note on rotationsAs an aside we mention here that there is a connection between the matricesdefining cyclic automorphisms of order 5 in certain abelian p-groups andthe matrices describing certain rotations in the plane. Clearly the equationx 2 + x − 1 = 0 has roots in the real numbers, and we prove that if j and kare defined using the the usual square root of 5 in R, rather than the rootsof x 2 = 5 in some Z p n, then the matrices v j and v k are conjugate to realmatrices defining rotations in the plane through angles of 2π/5 and 4π/5respectively. This is perhaps not surprising as the automorphism group canbe thought of as a measure of the symmetry of a group, so an automorphismthat cycles through a generating set highlights a certain cyclic or rotationalsymmetry of the generators, and hence of the group.Let A be the matrix representing a rotation through 2π/5 in R 2 .()cos 2π/5 sin 2π/5A =,− sin 2π/5 cos 2π/5so A has characteristic polynomialbutandχ A (x) = det (xI − A) = (x − cos 2π/5) 2 + sin 2 2π/5= x 2 − 2x cos 2π/5 + cos 2 2π/5 + sin 2 2π/5= x 2 − 2x cos 2π/5 + 1= x 2 − x(2 cos 2π/5) + 1,2 cos 2π/5 = 4 cos 2 π/5 − 22 cos π/5 = φ = 1 + √ 5, the golden ratio, see [7] p.112,2(6.3)

155so cos π/5 = φ/2 and therefore 4 cos 2 π/5 = φ 2 , soBut φ 2 − φ − 1 = 0, soSubstituting in (6.3) then gives2 cos 2π/5 = φ 2 − 2.φ 2 − 2 = 1 + φ − 2 = φ − 1.χ A (x) = x 2 − (φ − 1)x + 1.However, this is the same as the characteristic polynomial of the matrix( )0 −1B =,1 φ − 1so A and B have the same characteristic polynomial. This quadratic is not aperfect square, so it must also be the minimum polynomial of A and B, andhence A and B are conjugate.As φ satisfies the equation x 2 − x − 1, we have(φ − 1) 2 + (φ − 1) − 1 = φ 2 − 2φ + 1 + φ − 1 − 1 = φ 2 − φ − 1 = 0,so φ − 1 satisfies x 2 + x − 1 = 0. If we work in R rather than in the ring Z p α,thensoj = −1 + √ 52and k = −1 − √ 5,2φ − 1 = 1 + √ 5− 1 = −1 + √ 5= j,22i.e. B = v j , so v j is conjugate to a matrix representing a rotation through2π/5. Now note thatsok = −1 − √ 5= −φ,2( )0 −1v k = ,1 −φ

156which has characteristic polynomial χ(v k ) = x(x + φ) + 1 = x 2 + φx + 1.As A is conjugate to v j , we know that A 2 is conjugate to v 2 j , butv 2 j =(0 −11 φ − 1) (0 −11 φ − 1)=(which has characteristic polynomial−1(x + 1)(x − (1 − φ))−(φ − 1)(1 − φ)1 − φφ − 1 (φ − 1) 2 − 1)=(−1φ − 1= x 2 − (1 − φ)x + x − (1 − φ) + (φ − 1) 2= x 2 − (1 − φ)x + x − (1 − φ) + φ 2 − 2φ + 1= x 2 + φx − 1 + φ + φ 2 − 2φ + 1= x 2 + φx − φ + φ 2= x 2 + φx − φ + 1 + φ (as φ 2 = 1 + φ)= x 2 + φx + 1= χ(v k ),so v k is conjugate to the matrix representing a rotation by 4π/5.)1 − φ,1 − φ

1576.3 Abelian 5-groups with cyclic generatingsets of size 5We now consider the case where 5 divides the order of a rank 5 group G andthis means that G 5 is a 5-group of rank 5 or less, i.e.G 5 = Z 5 λ × Z 5 λ+µ × Z 5 λ+µ+ν × Z 5 λ+µ+ν+σ × Z 5 λ+µ+ν+σ+τ ,where λ, µ, ν, σ, τ ≥ 0.Clearly if G 5 is cyclic then there is nothing to prove, and if G 5 is not cyclic,but it satisfies Abért’s conditions for n = 5 then we know that it is cyclicallygenerated by 5 generators.As with the other primes, whether or not G 5 is cyclically generated by 5generators depends on the form of G 5 , i.e. it depends on the occurrence ofrepeated cyclic direct factors. The situation is more complicated, however,because we are looking for automorphisms of order p for a p-group, and thereforewe also need to look at whether these powers are ≥ 2. This was also thecase for 3-groups of rank 3 and 2-groups of rank 4, where the requirement wasthat at least one of the the relevant powers must be at most 1. In some caseswe can use results from earlier chapters to construct cyclic automorphisms oforder 5, while in other cases we can prove that no such automorphism exists.Some results were proved using a combination of GAP results and Lemmasproved previously. These results cannot be combined to give a classificationof 5-groups with cyclic automorphisms of order 5, but the following table indicatesto what extent the question has been partially answered. As before,an (A) indicates that the groups satisfies Abért’s conditions for n = 5 and istherefore always cyclically generated by 5 generators. Note that in the tableλ, µ, ν, σ, τ ≥ 1 and CG5 means ‘cyclically generated by 5 generators’.

158G 5 is G 5 is G 5 isForm of G 5 CG5 CG5 CG5⇒ ⇐ ⇔(1) Z 5 σ×Z 5 σ σ = 1(2) Z 5 σ×Z 5 σ+τ σ = 1(3) Z 5 ν×Z 5 ν×Z 5 ν ν = 1(4) Z 5 ν×Z 5 ν×Z 5 ν+τ ν = 1(5) Z 5 ν×Z 5 ν+σ×Z 5 ν+σ νσ = 1(6) Z 5 ν×Z 5 ν+σ×Z 5 ν+σ+τ νσ = 1(7) Z 5 µ×Z 5 µ×Z 5 µ×Z 5 µ (A) (A) (A)(8) Z 5 µ×Z 5 µ×Z 5 µ×Z 5 µ+τ µ = 1 or τ = 1(9) Z 5 µ×Z 5 µ×Z 5 µ+σ×Z 5 µ+σ σ = 1(10) Z 5 µ×Z 5 µ×Z 5 µ+σ×Z 5 µ+σ+τ σ = 1(11) Z 5 µ×Z 5 µ+ν×Z 5 µ+ν×Z 5 µ+ν ν = 1(12) Z 5 µ×Z 5 µ+ν×Z 5 µ+ν×Z 5 µ+ν+τ ν = 1(13) Z 5 µ×Z 5 µ+ν×Z 5 µ+ν+σ×Z 5 µ+ν+σ νσ = 1(14) Z 5 µ×Z 5 µ+ν×Z 5 µ+ν+σ×Z 5 µ+ν+σ+τ νσ = 1(15) Z 5 λ×Z 5 λ×Z 5 λ×Z 5 λ×Z 5 λ (A) (A) (A)(16) Z 5 λ×Z 5 λ×Z 5 λ×Z 5 λ×Z 5 λ+τ (A) (A) (A)(17) Z 5 λ×Z 5 λ×Z 5 λ×Z 5 λ+σ×Z 5 λ+σ σ = 1(18) Z 5 λ×Z 5 λ×Z 5 λ×Z 5 λ+σ×Z 5 λ+σ+τ σ = 1(19) Z 5 λ×Z 5 λ×Z 5 λ+ν×Z 5 λ+ν×Z 5 λ+ν ν = 1(20) Z 5 λ×Z 5 λ×Z 5 λ+ν×Z 5 λ+ν×Z 5 λ+ν+τ ν = 1(21) Z 5 λ×Z 5 λ×Z 5 λ+ν×Z 5 λ+ν+σ×Z 5 λ+ν+σ νσ = 1(22) Z 5 λ×Z 5 λ×Z 5 λ+ν×Z 5 λ+ν+σ×Z 5 λ+ν+σ+τ νσ = 1(23) Z 5 λ×Z 5 λ+µ×Z 5 λ+µ×Z 5 λ+µ×Z 5 λ+µ (A) (A) (A)(24) Z 5 λ×Z 5 λ+µ×Z 5 λ+µ×Z 5 λ+µ×Z 5 λ+µ+τ µ = 1 or τ = 1(25) Z 5 λ×Z 5 λ+µ×Z 5 λ+µ×Z 5 λ+µ+σ×Z 5 λ+µ+σ σ = 1(26) Z 5 λ×Z 5 λ+µ×Z 5 λ+µ×Z 5 λ+µ+σ×Z 5 λ+µ+σ+τ σ = 1(27) Z 5 λ×Z 5 λ+µ×Z 5 λ+µ+ν×Z 5 λ+µ+ν×Z 5 λ+µ+ν ν = 1(28) Z 5 λ×Z 5 λ+µ×Z 5 λ+µ+ν×Z 5 λ+µ+ν×Z 5 λ+µ+ν+τ ν = 1(29) Z 5 λ×Z 5 λ+µ×Z 5 λ+µ+ν×Z 5 λ+µ+ν+σ×Z 5 λ+µ+ν+σ νσ = 1(30) Z 5 λ×Z 5 λ+µ×Z 5 λ+µ+ν×Z 5 λ+µ+ν+σ×Z 5 λ+µ+ν+σ+τ νσ = 1

159We begin with two results concerning 5-groups that are cyclically generatedby 5 generators.Theorem 6.9. LetG 5 = Z 5 λ × Z 5 λ+µ × Z 5 λ+µ+ν × Z 5 λ+µ+ν+σ × Z 5 λ+µ+ν+σ+τwhere λ, µ, ν, σ, τ ≥ 0 and G 5 is not cyclic. If λ, µ, ν, σ and τ satisfy oneof the following conditions then G 5 is cyclically generated by 5 generators:(i) µ = ν = σ = 0,(ii) ν = σ = τ = 0,(iii) ν = σ = 0 and either µ = 1 or τ = 1.Proof. In each case G 5 satisfies Abért’s conditions for n = 5.We already knew that when G 5 is of one of the forms (7), (15), (16), or(23), then it is cyclically generated by 5 generators by Abért’s Theorem, butTheorem 6.9 adds partial results if G 5 is of the form (8) or (24) and µ = 1or τ = 1.LetG = Z a1 × Z a1 a 2× · · · × Z a1 a 2···a n= 〈x 1 〉 × 〈x 2 〉 × · · · × 〈x n 〉.Using the additive notation, the cyclic automorphism defined in the proof ofTheorem 2.2 is induced by the mapx 1 ↦→ x 1 , x 2 ↦→ x 1 + x 2 , . . . , x n−1 ↦→ x n−2 + x n−1 , x n ↦→ x n−1 + x n .We already know that the set {x n , x n θ, x n θ 2 , . . .} generates G, and in theproof of the next result we show that if G is of a certain form then thisautomorphism has order 5.Theorem 6.10. For any β ≥ 1, groups of each of the following forms arecyclically generated by 5 generators:

160(i) Z 5 × Z 5 β,(ii) Z 5 × Z 5 × Z 5 β,(iii) Z 5 × Z 5 × Z 5 × Z 5 β,(iv) Z 5 × Z 5 × Z 5 × Z 5 × Z 5 β.Proof. We use the same argument as that used in part of the proof of Lemma3.19.For each of the forms of group in the statement of the theorem, let G 5 be thegroup and let θ be the automorphism of G 5 defined above. In each of the 4cases we have(i) G 5 = Z 5 × Z 5 β with usual basis {x 1 , x 2 } and x 2 θ 5 = 5x 1 + x 2 = x 2 ,(ii) G 5 = Z 5 × Z 5 × Z 5 βwith usual basis {x 1 , x 2 , x 3 } andx 3 θ 5 = 10x 1 + 5x 2 + x 3 = x 3 ,(iii) G 5 = Z 5 × Z 5 × Z 5 × Z 5 βwith usual basis {x 1 , x 2 , x 3 , x 4 } andx 4 θ 5 = 10x 1 + 10x 2 + 5x 3 + x 4 = x 4 ,(iv) G 5 = Z 5 × Z 5 × Z 5 × Z 5 × Z 5 βwith usual basis {x 1 , x 2 , x 3 , x 4 , x 5 } andx 5 θ 5 = 5x 1 + 10x 2 + 10x 3 + 5x 4 + x 5 = x 5 .While the fact that θ 5 fixes the generator of the largest cyclic factor is notsufficient to prove that θ has order 5, we can use the fact that θ is alsoknown to cycle through a generating set containing this generator to provethe result. Let G 5 be one of the groups given above, and let x n generate thelargest factor of G 5 . Then(x n θ i )θ 5 = x n θ i+5 = x n θ 5 θ i = x n θ ifor each i. As {x n , x n θ, . . .} generates G 5 and θ 5 fixes each of these generators,θ must have order 5.

161Note that groups of the form (iii) or (iv) also satisfy Abért’s conditions forn = 5, but this result shows that if G 5 is of the form (1) or (2) and σ = 1then G 5 is cyclically generated by 5 generators, and if G 5 is of the form (3)or (4) and ν = 1 then G 5 is cyclically generated by 5 generators.The next results give conditions on σ and ν for G 5generated by 5 generators.not to be cyclicallyTheorem 6.11. Let G 5 = Z 5 λ × Z 5 λ+µ × Z 5 λ+µ+ν × Z 5 λ+µ+ν+σ × Z 5 λ+µ+ν+σ+τ .If σ ≥ 2 then G 5 is not cyclically generated.Proof. Note that if σ ≥ 2, then we can write σ = 2 + σ ′ for some σ ′ ≥ 0,and letLetH = G 5 /G 5 (5 λ+µ+ν+σ′ ) ≡ Z 5 2 × Z 5 2+τ .K = H/5 2 H = Z 25 × Z 25 .We can prove that Z 25 × Z 25 is not cyclically generated by 5 generators usingGAP and the result follows from Theorem 4.4.GAP can either be used to prove directly that Z 25 ×Z 25 , does not have a cyclicautomorphism of order 5, or it can be used in combination with Lemma 4.3and Theorem 4.4 give a more transparent proof. A summary of the secondproof is given here.We used GAP to find representatives for the conjugacy classes of GL(2, Z 25 )that contain elements of order 5. The form of these representatives suggesteda reason for the non-existence of a cyclic automorphism of order 5. Recallthat if there were a cyclic automorphism, θ, of order 5, then it would inducea cyclic automorphism θ F on Z 5 × Z 5 , because this is K/Φ(K). However,by using GAP we were able to see that the conjugacy classes of GL(2, Z 25 )containing elements of order 5 all have representatives of the form( )a 5bwhere a, d ≡ 1 (mod 5).5c dBy the above argument, the automorphism represented by this matrix inducesan automorphism of order 5 on Z 5 × Z 5 , but this is a contradiction

162because this induced automorphism must be represented by the matrix withits entries reduced modulo 5, i.e. by the identity matrix. In other words,any automorphism of order 5 of Z 25 × Z 25 is conjugate to an automorphismthat induces only the identity automorphism on Z 5 × Z 5 . This proves thatthere is no automorphism of order 5 that cycles through a generating set forZ 25 × Z 25 .The details of this proof can be generalized to show that Z p 2 × Z p 2 is notcyclically generated by p generators for any prime p ≥ 5. A proof of thisresult is given in the next chapter.The combination of GAP results, Lemma 4.3 and Theorem 4.4 used to provethat Z 25 × Z 25 is not cyclically generated by 5 generators can also be usedto prove that Z 25 × Z 25 × Z 25 is not. Any cyclic automorphism θ of order5 of Z 25 × Z 25 × Z 25 would induce a cyclic automorphism of order 5 onZ 5 × Z 5 × Z 5 . However, by using GAP to find conjugacy class representativesin GL(3, 5), we can see that such an automorphism is conjugate to one thatcan be represented by the matrix⎛ ⎞0 0 1⎜ ⎟⎝1 0 2⎠0 1 3and therefore we can assume θ is represented by the matrix⎛⎞5a 5b 1 + 5c⎜⎟A = ⎝1 + 5d 5e 2 + 5f⎠5g 1 + 5h 3 + 5kfor some a, b, c, d, e, f, g, h, k in Z 5 . Using GAP to run through all possibilitiesfor a, b, c, d, e, f, g, h, and k we see that A does not have order 5 in GL(3, Z 25 )for any choice of a, b, c, d, e, f, g, h, k. Therefore we have the following result.Theorem 6.12. Let G 5 = Z 5 λ × Z 5 λ+µ × Z 5 λ+µ+ν × Z 5 λ+µ+ν+σ × Z 5 λ+µ+ν+σ+τ .If ν ≥ 2 then G 5 is not cyclically generated.By combining Theorem 4.4 with either Theorem 6.11 or Theorem 6.12 wecomplete the results for 5-groups of the form (1), (2), (3), and (4), and obtain

163the remaining partial results for groups that do not satisfy Abért’s conditionsfor n = 5 (or cannot be made to satisfy Abért’s conditions for n = 5 by asuitable choice of µ or τ). For example, if G 5 = Z 5 × Z 5 × Z 125 × Z 625 × Z 625then G 5 is of the form (21) where λ = σ = 1, and ν = 2. If H = G 5 (5)then G 5 /H = Z 25 × Z 125 × Z 125 and if we let K denote this group thenK/25K = Z 25 × Z 25 × Z 25 , which does not have a cyclic generating set ofsize 5 by Theorem 6.12. It follows by Theorem 4.4 that G 5 is not cyclicallygenerated by 5 generators.Further investigations using GAP might provide more information to help findnecessary and sufficient conditions for a 5-group to have a cyclic generatingset of size 5, but perhaps a more interesting problem is the question of howthe results obtained in this chapter, and for rank 3 and rank 4 groups, canbe generalized to groups of higher rank. Some of these more general resultsare contained in the next chapter.

164

Chapter 7Abelian groups of higher rankSome of the results from previous chapters can be generalized to obtainpartial answers to the question of when a finite abelian group of rank n iscyclically generated by a minimal generating set. In a particular case wegive necessary and sufficient conditions for a finite abelian p-group of primerank to have a minimal cyclic generating set. The question for finite abeliangroups of composite rank is more complicated because we have a choice overthe size of the cyclic generating sets that we require each primary part tohave. For example, if G is a finite abelian group of rank 6, then we onlyrequire each primary part to be cyclically generated by 2, 3, or 6 generatorsto conclude that G is cyclically generated by 6 generators, and conversely, inorder to prove that G is not cyclically generated by 6 generators we need toprove that at least one of its primary parts is not cyclically generated by 2,3, or 6 generators. This means that results already obtained about the nonexistenceof cyclic generating sets of certain sizes cannot necessarily be used.For example, if G has rank 6 and G 3 = Z 3 × Z 27 × Z 243 , then we cannot usethe fact that G 3 is not cyclically generated by 3 generators to conclude thatG is not cyclically generated by 6 generators. In fact, GAP can be used toshow that G 3 is generated by 6 generators. If the methods from the precedingchapters were to be used then any attempt at finding a classification resultfor a group of arbitrary rank would require at least some examination ofall the possible sizes of cyclic generating sets for a finite abelian p-group ofarbitrary rank. We concentrate here on the specific question of when a finite165

166abelian p-group is cyclically generated by a prime number of generators andthis leads to a result about cyclic generating sets of square free size and aclassification of rank 2 abelian p-groups with cyclic generating sets of size p.7.1 Abelian p-groups with cyclic generatingsets of prime size qThe method used to prove Theorem 6.6 can be generalized to a result thatproves the existence of a cyclic generating set of prime size q for certain finiteabelian p-groups G. Provided the rank of G is small enough, the existenceof the cyclic generating set depends only on the relationship between p andq and not on the structure of G. If, however, G is of a certain form, thiscondition becomes both necessary and sufficient.Theorem 7.1. Let q be a prime. A finite abelian p-group of rank ≤ q iscyclically generated by q generators if p ≡ 1 (mod q).Proof. The result follows from a series of lemmas.Let p be prime with p ≡ 1 (mod q) and let G be an abelian p-group withrank G ≤ q. By extending G by trivial groups if necessary, we can writeG = Z p λ 1 × Z p λ 1 +λ 2 × · · · × Z p λ 1 +λ 2 +···+λqwhere λ i ≥ 0 for i = 1, 2, 3, . . . , q. We use the result of the following lemmato construct an automorphism of G.Lemma 7.2. For any λ ≥ 1 there is a q-th root of unity in the group of unitsof Z p λ.Proof. Let U be the group of units of Z p λ. Then|U| = ϕ(p λ ) = p λ−1 (p − 1)and as p ≡ 1 (mod q), U must have an element α of order q, and henceα is a q-th root of 1 in Z p λ. Note too that since q is prime, the elementsα 2 , α 3 , . . . , α q−1 are also q-th roots of 1.

167Let Z p λ 1 +λ 2 +···+λ i be generated by x i and let α ≠ 1 be a q-th root of 1 inZ p λ 1 +λ 2 +···+λq . For ease of notation let λ = λ 1 + λ 2 + · · · + λ q , the exponent ofG. Define a map θ : G → G, with respect to the basis x 1 , x 2 , . . . , x q of G, bythe matrix⎛⎞αα 2 . . .,⎜⎝α q−1 ⎟⎠1i.e. x i ↦→ α i x i . As (α, p) = 1, θ is an automorphism of G, by Lemma 4.6, andclearly o(θ) = q.Let x = x 1 + x 2 + · · · + x q and let H = 〈x, xθ, xθ 2 , . . . , xθ q−1 〉. We prove thatH = G by showing that each x i lies in H.Lemma 7.3. For i = 1, 2, . . . , q − 1 the set of coefficients of x i in the set{x, xθ, xθ 2 , . . . , xθ q−1 } is {1, α, α 2 , . . . , α q−1 }.Proof. Fix i with 1 ≤ i ≤ q − 1. The coefficient of x i in xθ j is (α i ) j = α ij .Suppose that the coefficient of x i in xθ j 1is the same as that in xθ j 2for somej 1 , j 2 with 0 ≤ j 1 , j 2 ≤ q − 1 and j 1 ≠ j 2 . Then α ij 1≡ α ij 2(mod p λ ), soij 1 ≡ ij 2 (mod q), i.e. i(j 1 − j 2 ) ≡ 0 (mod q), and hence j 1 = j 2 , which isa contradiction. As these q coefficients are powers of α and are all distinct,they must be 1, α, α 2 , . . . , α q−1 , as required.Lemma 7.4. If α ≠ 1 is a q-th root of 1 in Z p λthen1 + α + α 2 + · · · + α q−1 ≡ 0 (mod p λ ).Proof. First we note that(α − 1)(1 + α + α 2 + . . . + α q−1 ) = α q − 1,and α q ≡ 1 (mod p λ ), so(α − 1)(1 + α + α 2 + · · · + α q−1 ) ≡ 0 (mod p λ ). (7.1)

168If α − 1 is a unit in Z p λprove that α − 1 is coprime to p.then the result will follow. We therefore need toSuppose p divides α − 1. Then for some k not necessarily coprime to p,α = 1 + kp, and for any n ≥ λ,Let∑λ−1( ) nα n = (1 + kp) n ≡ 1 + (kp) i (mod p λ ). (7.2)ii=1µ = maxi=1,2,...,λ−1 {r : pr |i!},so for each i in the range 1, 2, . . . , λ − 1, we can write i! = p µ−ν im i whereν i ≥ 0 and (m i , p) = 1.If we let n = p λ+µ , we have( ni)= pλ+µ (p λ+µ − 1)!i!(p λ+µ − i)!= pλ+µ (p λ+µ − 1) · · · (p λ+µ − i + 1)i!= p λ+µ−µ+ν i(p λ+µ − 1) · · · (p λ+µ − i + 1)m i≡ 0 (mod p λ ),so from (7.2) we have α n ≡ 1 (mod p λ ). Therefore either α = 1, which is acontradiction, or the order of α is a power of p, which is also a contradiction,and hence α − 1 is a unit in Z p λ. By (7.1) it now follows that1 + α + α 2 + · · · + α q−1 ≡ 0 (mod p λ )as required.Returning to the proof that each x i is in H, note that for j = 0, 1, 2, . . . , q−1,the coefficient of x q in xθ j is 1, and by Lemma 7.3 we know that, for i =1, 2, . . . , q − 1, the set of coefficients of x i in the set {x, xθ, xθ 2 , . . . , xθ q−1 } isthe set {1, α, α 2 , . . . , α q−1 }. Therefore, by Lemma 7.4 we have

169q−1 q−1q−1∑ ∑∑xθ j = (1 + α + α 2 + · · · + α q−1 )x k +j=0 k=1j=0q−1 q−1∑ ∑≡ 0x k + x q (mod p λ )k=1 j=0= qx qx qand as (q, p) = 1, x q must lie in H.Now fix t with 1 ≤ t ≤ q − 1. We show that x t is in H. For j = 0, 1, . . . , q − 1,let y j = α q−jt (xθ j ). Then as the coefficient of x t in xθ j is α jt , the coefficientof x t in y j is α q−jt α jt = α q ≡ 1 (mod p λ ).Lemma 7.5. For s with 1 ≤ s ≤ q and s ≠ t, if we define C s to be theset of coefficients of x s in the set {y j : j = 0, 1, . . . , q − 1}, then C s ={1, α, α 2 , . . . , α q−1 }.Proof. Fix s with 1 ≤ s ≤ q and s ≠ t. The coefficient of x s in y j isα q−jt α js = α q+j(s−t) , so it remains to show that these coefficients are distinctfor distinct j.Suppose that we have α q+j1(s−t) ≡ α q+j2(s−t) (mod p λ ) for some j 1 , j 2 with0 ≤ j 1 , j 2 ≤ q − 1 and j 1 ≠ j 2 . Then q + j 1 (s − t) ≡ q + j 2 (s − t) (mod q), soj 1 (s − t) ≡ j 2 (s − t)(mod q),and hence (j 1 − j 2 )(s − t) ≡ 0 (mod q), which is a contradiction by choice ofs, t, j 1 , and j 2 .Now, by adding the y j we get

170q−1 q−1∑ ∑y j = α q−jt (xθ j )j=0=j=0q−1∑α jk x k)( q∑α q−jtj=0 k=1q−1∑ q∑= α q−jt α jk x k=j=0 k=1q−1∑j=0 k=1⎛q−1∑⎜= ⎝==≡≡j=0q−1∑j=0q∑k=1k≠tq∑α q+j(k−t) x k⎞q∑(α q+j(k−t) x k ) + α q ⎟x t ⎠k=1k≠tq−1q∑∑α q+j(k−t) x k + α q x tk=1k≠tj=0( ∑q−1)q−1∑α q+j(k−t) x k +j=0j=0α q x t(q∑ ∑q−1∑c)x k + α q x t (mod p λ )c∈C kk=1k≠tj=0q−1q∑ ∑0x k + α q x t (mod p λ )k=1k≠tj=0q−1∑≡ α q x t (mod p λ )j=0q−1∑≡ x t (mod p λ )j=0= qx tso x t is in H, as required. Therefore H = G and we have proved that G iscyclically generated by q generators. This completes the proof of Theorem

1717.1.For example, the groups Z 43 × Z 43 × Z 43 × Z 43 2 × Z 43 3 × Z 43 3 × Z 43 4 andZ 29 × Z 29 × Z 29 × Z 29 2 × Z 29 4 are both cyclically generated by 7 generators.Alternatively, Theorem 7.1 could be restated as An abelian p-group G iscyclically generated by q generators for any prime q such that q is at least aslarge as the rank of G and q divides p − 1. For example, Z 43 × Z 43 2 has cyclicgenerating sets of sizes 2, 3, and 7.In the special case that rank G = q, we have proved that G has a minimalcyclic generating set.Combining Theorem 7.1 with Theorem 4.4 and Lemma 4.6 gives us the followingresult.Theorem 7.6. Let q be prime and for p ≠ q, let G be a finite abelian p-group,with rank G ≤ q and the 3 largest cyclic direct factors of G are all distinct.Then G is cyclically generated by q generators if and only if p ≡ 1 (mod q).Proof. If p ≡ 1 (mod q) then by Theorem 7.1 we know that any finiteabelian p-group of rank ≤ q is cyclically generated by q generators. Nowsuppose that p ≢ 1 (mod q) and p ≠ q. Then since G can be written in theformG = Z p λ 1 × Z p λ 1 +λ 2 × · · · × Z pλ 1 +λ 2 +···+λ q−2 +λ q−1 × Z pλ 1 +λ 2 +···+λ q−2 +λ q−1 +λqwhere λ i ≥ 0 for i = 1, 2, . . . , q − 2, and λ q−1 , λ q ≥ 1. LetH = G/G(p λ 1+λ 2 +···+λ q−2) = Z pλ q−1 × Z pλ q−1 +λq .Then since |Aut H| = p r (p − 1) 2 for some r, by Corollary 4.7, we know thatH is not cyclically generated by q generators and hence G is not, by Theorem4.4.For example, if G = Z 23 × Z 23 × Z 23 2 × Z 23 3 then G is cyclically generatedby 11 generators, but it is not cyclically generated by 5 generators.Applying Theorem 7.6 to a group of prime rank we have the following result.

172Theorem 7.7. Let G be a finite abelian p-group of prime rank q ≠ p. If the 3largest cyclic direct factors of G are all distinct then G has a minimal cyclicgenerating set if and only if p ≡ 1 (mod q).For example,Z 23 × Z 23 × Z 23 × Z 23 × Z 23 × Z 23 × Z 23 × Z 23 × Z 23 × Z 23 2 × Z 23 3andZ 29 × Z 29 × Z 29 × Z 29 2 × Z 29 2 × Z 29 4 × Z 29 5both have minimal cyclic generating sets, butZ 13 × Z 13 × Z 13 × Z 13 2 × Z 13 2 × Z 13 4 × Z 13 5does not.7.2 Cyclic generating sets of square free sizeThe following result is a generalization of Theorem 7.1.Theorem 7.8. Let n be a square free integer and let p be a prime withp ≡ 1 (mod n). Then any finite abelian p-group G of rank ≤ n is cyclicallygenerated by n generators.Proof. Let n = q 1 q 2 · · · q r and let G be an abelian p-group with exponent p λand rank ≤ n. Suppose p satisfies p ≡ 1 (mod n). Then p ≡ 1 (mod q i ) fori = 1, 2, . . . , r, so the group of units of Z p λ contains an element α i of orderq i for each i. Let α = α 1 α 2 · · · α r . Then α has order n, so α n ≡ 1 (mod p λ ).We can use α to construct a cyclic automorphism of G.If the rank of G is less than n we can extend G to the direct product of ncyclic subgroups by adding copies of the trivial group, soG = Z p λ 1 × Z p λ 1 +λ 2 × · · · × Z p λ 1 +λ 2 +···+λn= 〈x 1 〉 × 〈x 2 〉 × · · · × 〈x n 〉

173where λ 1 , λ 2 , . . . , λ n ≥ 0 and λ 1 + λ 2 + · · · + λ n = λ. Define a map θ : G → Gby x i ↦→ α i x i , so θ is represented, with respect to the basis {x 1 , x 2 , . . . , x n },by the matrix⎛⎞αα 2 α 3 . . ..α n−2 ⎜⎝α n−1 ⎟⎠1As α is coprime to p, the map θ is an automorphism, and it has order n bythe definition of α. We need to show that θ cycles through a generating setfor G.Let x = x 1 + x 2 + · · · + x n . We want to show that the set produced when θis repeatedly applied to x is indeed a generating set. For any j in the rangej = 0, 1, . . . , n − 1, we have( n∑)xθ j = x i θ j =i=1Adding all the xθ j s givesn∑x i θ j =i=1n∑α ij x i .i=1∑n−1xθ j =j=0∑n−1j=0n∑α ij x ii=1(n∑ n−1) ∑= x i α iji=1 j=0(∑n−1n−1) ∑ ∑n−1= x i α ij + x ni=1 j=0j=0α nj(∑n−1n−1) ∑≡ x i α ij + nx n (mod p λ ).i=1 j=0(7.3)

174We show that for i = 1, 2, . . . , n − 1,∑n−1α ij ≡ 0 (mod p λ ),j=0but in order to do this we need the following two lemmas.Lemma 7.9. For i = 1, 2, . . . , n − 1,where m i is the order of α i .1 + α i + α 2i + · · · + α i(m i−1) ≡ 0 (mod p λ )Proof. Let β i denote α i , and let m i denote the order of β i , which isβ m ii− 1 = (β i − 1)(1 + β i + βi 2 + · · · + β m i−1i ) and β m iin. As (i,n)≡ 1 (mod p λ ), we have(β i − 1)(1 + β i + βi 2 + · · · + β m i−1i ) = β m ii − 1 ≡ 0 (mod p λ ). (7.4)If β i −1 is a unit in Z p λ, then the result follows. It therefore remains to provethat p does not divide β i − 1. The method used is the same as the methodused to prove Lemma 7.4.If we suppose that β i = 1 + kp then, by a suitable choice of µ, we can provethat β pλ+µi ≡ 1 (mod p λ ).LetThenandµ = maxj=1,2,...,λ−1 {r : pr |j!}.∑λ−1β pλ+µi = (1 + kp) pλ+µ ≡ 1 +( ) pλ+µjj=1( pλ+µj)(kp) j (mod p λ )= pλ+µ (p λ+µ − 1) · · · (p λ+µ − j + 1),j!so by the choice of µ, any power of p dividing j! divides p µ , and therefore( ) pλ+µ≡ 0 (mod p λ ),j

175so β pλ+µi ≡ 1 (mod p λ ). It follows that the order of β i must divide p λ+µ , whichis a contradiction because p ≡ 1 (mod q i ) for each i and is therefore coprimeto the order of β i . As this shows that β i − 1 is a unit in Z p λ, it follows from(7.4) that1 + β i + β 2 i + · · · + β m i−1i ≡ 0 (mod p λ ).Lemma 7.10. For i = 1, 2, . . . , n − 1,∑n−1α ij ≡ (i, n)j=0n(i,n) −1∑j=0α ij ≡ 0 (mod p λ ).Proof. Let β i = α i . If (i, n) = 1 then the result follows by Lemma 7.9. If(i, n) > 1, then without loss of generality we can assume that there is some k isuch that (i, n) = q 1 q 2 · · · q ki . Let q = q 1 q 2 · · · q ki and let ˆq = q ki +1q ki +2 · · · q r .Then i = lq for some l, and β i has ordern(i,n)1 + β i + · · · + β ˆq−1i ≡ 0 (mod p λ ).We also know that if j ≡ j ′ (mod ˆq) then β j i ≡ βj′ ias required∑n−1α ij =j=0q−1∑m=0mˆq−1∑j=0α ijˆq−1∑≡ q α ij (mod p λ )j=0ˆq−1∑= qj=0β j i= ˆq, so by Lemma 7.9,≡ 0 (mod p λ ), by Lemma 7.9From Lemma 7.10 and equation (7.3), we now have(mod p λ ). Therefore,(∑n−1∑n−1n−1) ∑xθ j ≡ x i α ij + nx n (mod p λ )j=0 i=1 j=0n∑≡ 0 + nx n (mod p λ )i=1= nx n

176and as (n, p) = 1, the proves that x n is in the subgroup generated by thexθ j s. We use a similar method to show that the other x i s are also in thissubgroup.Fix t with 1 ≤ t ≤ n − 1. We show that x t lies in the subgroup generated bythe xθ j s.For each j in the range 1, 2, . . . , n, letn∑y j = α n−tj (xθ j ) = α n−tj+ij x ii=1= x t + ∑ α n+j(i−t) x i ≡ x t + ∑i≠ti≠tα j(i−t) x i (mod p λ ).We show that the sum of all the y j s is congruent to nx t modulo p λ . Note firstthat(∑n−1∑n−1y j = x t + ∑ )α j(i−t) x ij=0 j=0 i≠t∑n−1∑n−1∑= x t + α j(i−t) x ij=0j=0i≠t∑n−1∑= nx t + α j(i−t) x ij=0= nx t + ∑ i≠ti≠t( n−1) ∑x i α j(i−t) .j=0(7.5)But for any i ≠ t, if we reduce i − t modulo n we get i − t = 1, 2, . . . , n − 1because t ≠ n, and therefore, if we replace i by i − t in Lemmas 7.9 and 7.10,then we get∑n−1(α (i−t) ) j ≡ (i − t, n)j=0n(i−t,n) −1∑j=0It therefore follows from equations (7.5) and (7.6) that∑n−1y j = nx t + ∑j=0i≠tα (i−t)j ≡ 0 (mod p λ ). (7.6)( n−1) ∑x i α j(i−t) ≡ nx t (mod p λ )j=0

177and hence x t is in the subgroup generated by the xθ j s. This completes theproof of Theorem 7.8.For example, suppose n = 6 and G = Z 7 × Z 7 × Z 7 × Z 7 2. We produce acyclic generating set of size 6 as follows.We need to define a basis for G, but first we extend G by copies of the trivialgroup to obtainG = Z 1 × Z 1 × Z 7 × Z 7 × Z 7 × Z 7 2= 〈x 1 〉 × 〈x 2 〉 × 〈x 3 〉 × 〈x 4 〉 × 〈x 5 〉 × 〈x 6 〉where x 1 = x 2 = 0. As −1 is a square root of 1 in Z 49 and 30 is a cuberoot of 1, we let α = −30 = 19. Then we define θ, with respect to the basis{x 1 , x 2 , . . . , x 6 }, by the matrix⎛⎞19 0 0 0 0 00 18 0 0 0 00 0 48 0 0 00 0 0 30 0 0.⎜⎟⎝ 0 0 0 0 31 0⎠0 0 0 0 0 1Let x = x 1 + x 2 + x 3 + x 4 + x 5 + x 6 . Applying θ repeatedly to x we getx 1 + x 2 + x 3 + x 4 + x 5 + x 6↦→ 19x 1 + 18x 2 + 48x 3 + 30x 4 + 31x 5 + x 6↦→ 18x 1 + 30x 2 + x 3 + 18x 4 + 30x 5 + x 6↦→ 48x 1 + x 2 + 48x 3 + x 4 + 48x 5 + x 6↦→ 30x 1 + 18x 2 + x 3 + 30x 4 + 18x 5 + x 6↦→ 31x 1 + 30x 2 + 48x 3 + 18x 4 + 19x 5 + x 6↦→ x 1 + x 2 + x 3 + x 4 + x 5 + x 6 .As 19+18+48+30+31+1 = 147 ≡ 0 (mod 49), and clearly 18+30+1 ≡ 0

178(mod 49) and 48 + 1 ≡ 0 (mod 49), adding the xθ j s givesx+xθ + xθ 2 + xθ 3 + xθ 4 + xθ 5= (1 + 19 + 18 + 48 + 30 + 31)x 1 + 2(1 + 18)x 2 + 3(1 + 48)x 3+ 2(1 + 30 + 18)x 4 + (1 + 31 + 30 + 48 + 18 + 19)x 5 + 6x 6≡ 6x 6 (mod 49).so x 6 is in the subgroup generated by the xθ j s.It remains to show that x 5 , x 4 , and x 3 are in the subgroup generated by thexθ j s. As an illustration we show that x 3 is in the subgroup.For j = 0, 1, . . . , 5 lety j = 19 6−3j xθ j =6∑19 6+(i−3)j x i .i=1Theny 0 = x 1 + x 2 + x 3 + x 4 + x 5 + x 6y 1 = 30x 1 + 31x 2 + x 3 + 19x 4 + 18x 5 + 48x 6y 2 = 18x 1 + 30x 2 + x 3 + 18x 4 + 30x 5 + x 6y 3 = x 1 + 48x 2 + x 3 + 48x 4 + x 5 + 48x 6y 4 = 30x 1 + 18x 2 + x 3 + 30x 4 + 18x 5 + x 6y 5 = 18x 1 + 19x 2 + x 3 + 31x 4 + 30x 5 + 48x 6 ,and thereforey 0 +y 1 + y 2 + y 3 + y 4 + y 5= 2(1 + 30 + 18)x 1 + (1 + 31 + 30 + 48 + 18 + 19)x 2 + 6x 3+ (1 + 19 + 18 + 48 + 30 + 31)x 4 + 2(1 + 18 + 30)x 5 + 3(1 + 48)x 6≡ 6x 3 (mod 49)as required.Theorem 7.8 can be used to give a condition for an arbitrary abelian groupof square free rank to have a minimum cyclic generating set.

179Theorem 7.11. Let n be a square free integer and let G be a finite abeliangroup of rank n. Then G has a minimal cyclic generating set if every primep dividing |G| satisfies p ≡ 1 (mod n).This result relies on the fact that, by Theorem 7.8, if every prime p dividing|G| satisfies p ≡ 1 (mod n) then G p has a cyclic generating set of size n. Thisdoes not, however, tell us anything about a group whose order is divisible byprimes less than its rank. For example, if G is a group of rank 14, then wecannot use Theorem 7.11 if any of 2, 3, 5, 7, 11, or 13, divides |G|. As weactually only need each primary part to have a cyclic generating set of sizedividing n, we can do better than this.Theorem 7.12. Let n be a square free integer. A group G of rank n iscyclically generated by n generators if, for each prime p dividing |G|, thereis some m p dividing n such that m p ≥ rank G p , and p ≡ 1 (mod m p ).For example, if n = 30 and G has rank 30 and |G| is divisible by 31 and 19,whereG 19 = Z 19 × Z 19 × Z 19 2 × Z 19 3 × Z 19 4 × Z 19 5,then we cannot use Theorem 7.11 to say anything about G because 19 ≢ 1(mod 30). We can, however, use Theorem 7.12 to say that G 19 is cyclicallygenerated by 6 generators, since 19 ≡ 1 (mod 6). We know by Theorem 7.8that G 31 is cyclically generated by 30 generators, and therefore G is cyclicallygenerated by 30 generators.7.3 Abelian p-groups cyclically generated byp generatorsThe final section concerns the existence of cyclic generating sets of size p fora finite abelian p-group of rank 2. Clearly Z 4 ×Z 4 is cyclically generated by 2generators, and from Theorem 5.1 we know that Z 9 ×Z 9 is cyclically generatedby 3 generators. The first result generalizes, to the case of a general primep, the question of whether Z 25 × Z 25 is cyclically generated by 5 generators.As mentioned in the previous chapter, the result for p = 5 was first obtained

180using GAP and further computations showed that the group Z p 2 × Z p 2 is notcyclically generated by p generators if p = 7 or p = 11. This theorem leadsto a classification of rank 2 abelian p-groups with cyclic generating sets ofsize p.Theorem 7.13. Let p be a prime. The group Z p 2 ×Z p 2 is cyclically generatedby p generators if and only if p = 2 or p = 3.Proof. If p = 2 thenG = Z 4 × Z 4= 〈x〉 × 〈y〉and the map x ↦→ y ↦→ x is a cyclic automorphism. If p = 3 thenG = Z 9 × Z 9= 〈x〉 × 〈y〉and we define a map θ with respect to {x, y} by the matrix( )1 1.6 1This matrix defines an automorphism because it is onto and 9(xθ) = 0 and9(yθ) = 0.On applying powers of θ to x we getx ↦→ x + y ↦→ 7x + 2y ↦→ 7x + 7y + 12x + 2y = x,so θ has order 3 and G is cyclically generated by 3 generators.Now let p ≥ 5 and let G = Z p 2 × Z p 2. By Lemma 4.3, if G has an automorphismθ of order p that cycles through a generating set for G, then θ inducesa cyclic automorphism θ ′ of order p on G/Φ(G), and if A = [a i j ] is a matrixrepresenting θ then (with respect to the relevant basis for G/Φ(G)) the inducedautomorphism θ ′ is represented by the matrix A ′ = [a i j (mod p)]. By

181the table in the proof of Lemma 6.4, the only conjugacy classes of GL(2, p)that contain elements of order p are those with representatives of the form( )s 1u s = where s ∈ Z p .0 sAs ( ) p ( )s 1 s p ps p−1= ,0 s 0 s pa matrix of this form will have order p if and only if s = 1. Therefore A ′is conjugate to the matrix u 1 , and without loss of generality we can assumethat the basis of G was chosen so that A ′ = u 1 .It follows that A must be of the form()1 + ip 1 + jpkp1 + lpfor some i, j, k, l ∈ Z p . We prove that if A has this form then A p is not theidentity matrix, and therefore θ does not have order p. This proves that G isnot cyclically generated by p generators.( )a bThe nth power of a 2 × 2 matrix is given by Theorem 1 of [25] inc dterms of the trace and determinant of the matrix as follows: if T and D arethe trace and determinant respectively, and if, for any n, y n is defined by theequationthen (aand thereforeA p =cy n =⌊n/2⌋∑s=0( n − ss)T n−2s (−D) s , (7.7)) n ()b y n − dy n−1 by n−1= ,d cy n−1 y n − ay n−1()y p − (1 + lp)y p−1 (1 + jp)y p−1.kpy p−1 y p − (1 + ip)y p−1

182Putting n = p − 1 in (7.7) we gety p−1 =⌊(p−1)/2⌋∑s=0( p − 1 − ss)T p−1−2s (−D) s .We prove that for p ≥ 5, there are no i, j, k, l such that(1 + jp)y p−1 ≡ 0 (mod p 2 )and therefore A does not have order p.As A ∈ GL(2, Z p 2) we reduce modulo p 2 as required, and therefore the traceand determinant of A areT ≡ 2 + (i + l)p (mod p 2 )andD ≡ 1 + (i + l − k)p (mod p 2 )respectively. Note that for any λ, µ ≥ 1 we haveT λ = (2 + (i + l)p) λ ≡ 2 λ + 2 λ−1 λ(i + l)p (mod p 2 )and(−D) µ = (−1 − (i + l − k)p) µ ≡ (−1) µ − (−1) µ−1 µ(i + l − k)p (mod p 2 )and thereforeT λ (−D) µ≡ [2 λ + 2 λ−1 λ(i + l)p][(−1) µ − (−1) µ−1 µ(i + l − k)p] (mod p 2 )≡ 2 λ (−1) µ + [−2 λ (−1) µ−1 µ(i + l − k) + (−1) µ 2 λ−1 λ(i + l)]p (mod p 2 ).Putting λ = p − 1 − 2s and µ = s we getwhereT p−1−2s (−D) s = 2 p−1−2s (−1) s + Zp

183Z = −2 p−1−2s (−1) s−1 s(i + l − k) + (−1) s 2 p−2−2s (p − 1 − 2s)(i + l)= −2 p−1−2s (−1) s−1 s(i + l − k)+ (p − 1)(−1) s 2 p−2−2s (i + l) − 2s(−1) s 2 p−2−2s (i + l)= 2 p−1−2s (−1) s s(i + l) + 2 p−1−2s (−1) s−1 sk+ (p − 1)(−1) s 2 p−2−2s (i + l) − s(−1) s 2 p−1−2s (i + l)= 2 p−1−2s (−1) s−1 sk + (p − 1)(−1) s 2 p−2−2s (i + l)= 2 p−1−2s (−1) s−1 (p − 1)sk + (−1) s 2 p−1−2s (i + l)2= 2 p−1−2s (−1) s (p − 1)(−sk) + (−1) s 2 p−1−2s (i + l)[ 2 ]= 2 p−1−2s (−1) s (p − 1)−sk + (i + l) ,2so()]T p−1−2s (−D) s = 2 p−1−2s (−1) s +[2 p−1−2s (−1) s (p − 1)−sk + (i + l) p2() ]= 2 p−1−2s (−1)[1 s (p − 1)+ −sk + (i + l) p .2Substituting for this in y p−1 gives

184y p−1===⌊(p−1)/2⌋∑s=0⌊(p−1)/2⌋∑s=0+ p⌊(p−1)/2⌋∑s=0( p − 1 − ss( p − 1 − ss⌊(p−1)/2⌋∑s=0( p − 1 − s− kps⌊(p−1)/2⌋∑s=0)2 p−1−2s (−1) s [1 +)2 p−1−2s (−1) s( p − 1 − s+s)2 p−1−2s (−1) s( p − 1 − ssp(p − 1)2but by equation (7.1) in [25],and thereforey p−1 = p − kp≡ p − kp⎛= p ⎝1 − k⌊(p−1)/2⌋∑s=0⌊(p−1)/2⌋∑s=0⌊(p−1)/2⌋∑s=0⌊(p−1)/2⌋∑s=0(−sk +)2 p−1−2s (−1) s (−sk +)2 p−1−2s (−1) s s⎛(i + l) ⎝( p − 1 − ss( ) p − 1 − ss∑⌊(p−1)/2⌋s=0) ](p − 1)(i + l) p2( p − 1 − ss)2 p−1−2s (−1) s = p)(p − 1)(i + l)2⎞)2 p−1−2s (−1) s ⎠2 p−1−2s (−1) s s + p2 (p − 1)(i + l)2( p − 1 − ss⎞( ) p − 1 − s2 p−1−2s (−1) s s⎠ .s)2 p−1−2s (−1) s s (mod p 2 )It follows that(1 + jp)y p−1 ≡ y p−1 (mod p 2 ).

185We need to show that for any k in Z p , y p−1 is not congruent to 0 modulo p 2 .Claim LetS =⌊(p−1)/2⌋∑s=0Then for p ≥ 5, S ≡ 0 (mod p).( p − 1 − ss)2 p−1−2s (−1) s s.If this claim is true, then for any k in Z p , we will have(1 + jp)y p−1 ≡ y p−1 ≡ p (mod p 2 ),and as (1 + jp)y p−1 is the 1,2 entry of A p , this proves that A p cannot be theidentity matrix.Proof of Claim Note that the first term in the sum is zero, soS =⌊(p−1)/2⌋∑s=1( p − 1 − ss)2 p−1−2s (−1) s s ≡ 0 (mod p) (7.8)and we will use this to prove the claim. By Corollary 1 on page 4 of [25],( ) ⌊(p−1)/2⌋p − 1 − s∑( ) p t= 2 −p+1+2s . (7.9)s2t + 1)(sExpanding the binomial coefficients on the right hand side of (7.9) we get( ) p t p!t!=2t + 1)(s (2t + 1)!(p − 2t − 1)!s!(t − s)! ,and for 1 ≤ s ≤ t < ⌊(p−1)/2⌋, p does not divide (2t+1)!(p−2t−1)!s!(t−s)!so ( ) p t≡ 02t + 1)(st=s(mod p).Therefore, on reducing modulo p, all the terms in the sum on the right handside of (7.9) are zero except for the one where t = ⌊(p − 1)/2⌋ and hence(7.9) gives( ) ()( )p − 1 − s≡ 2 −p+1+2s p⌊(p − 1)/2⌋s2(⌊(p − 1)/2⌋) + 1 s(mod p),

186and since p is odd, ⌊(p − 1)/2⌋ = (p − 1)/2, so() (pp= = 1.2(⌊(p − 1)/2⌋) + 1 p)If we substitute for the binomial coefficient in (7.8) we now haveS =≡≡⌊(p−1)/2⌋∑s=1⌊(p−1)/2⌋∑s=1⌊(p−1)/2⌋∑s=1( p − 1 − ss( ⌊(p − 1)/2⌋s( ⌊(p − 1)/2⌋s)2 p−1−2s (−1) s s)2 −p+1+2s 2 p−1−2s (−1) s s (mod p))(−1) s sHowever, we can use the identity( ( )n n − 1k = n ,k)k − 1(mod p).see [2], p.23, to change this into a constant multiple of an alternating sum ofbinomial coefficients in the following way:S =⌊(p−1)/2⌋∑s=1= ⌊(p − 1)/2⌋= −⌊(p − 1)/2⌋( ⌊(p − 1)/2⌋s⌊(p−1)/2⌋∑)(−1) s s( )⌊(p − 1)/2⌋ − 1(−1) ss − 1s=1⌊(p−1)/2⌋−1∑( )⌊(p − 1)/2⌋ − 1(−1) s .ss=0} {{ }∆(7.10)The part of (7.10) labelled ∆ is the alternating sum of the entries in row⌊(p − 1)/2⌋ − 1 of Pascal’s Triangle (counting the rows starting at row zero).Note that as p ≥ 5,⌊(p − 1)/2⌋ − 1 ≥ 1,so we are not concerned with row zero. Therefore, by the Binomial Theorem,⌊(p−1)/2⌋−1∑s=0( ⌊(p − 1)/2⌋ − 1s)(−1) s = (1+(−1)) ⌊(p−1)/2⌋−1 = 0,as required.

187This completes the proof of the claim and it follows that (1 + jp)y p−1 ≢ 0(mod p 2 ), and therefore a matrix of the form A does not have order p inGL(2, Z p 2).Note that the matrix defining the cyclic automorphism of order 3 for Z 9 × Z 9is of the form ()1 + ip 1 + jp,kp 1 + lpbut this is not a contradiction because in this case ∆ is the alternating sumof the 0th row of Pascal’s Triangle, and therefore S ≡ −1 (mod 3). It followsthat y 3−1 ≡ 3(1 + 2) ≡ 0 (mod 9).We can use Theorems 7.13 and 4.4, and Lemma 4.5 to prove that a group Gis not cyclically generated by p generators in any situation where factoringby a characteristic subgroup reduces some p-primary part of G to Z p 2 × Z p 2where p ≥ 5. For example, ifG = Z 2 × Z 2 × Z 2 × Z 2 × Z 2.7 × Z 2.343 × Z 2.343 ,then G 7 = Z 7 × Z 343 × Z 343 , so G 7 /G 7 (7) = Z 49 × Z 49 , which is not cyclicallygenerated by 7 generators by Theorem 7.13. Therefore by Theorem 4.4, G 7is not cyclically generated by 7 generators, and hence, by Lemma 4.5, G doesnot have a minimal cyclic generating set.We finish with a classification of rank 2 abelian p-groups.Theorem 7.14. A rank 2 abelian p-group, Z p λ × Z p λ+µ, is cyclically generatedby p generators if and only if λ, µ, and p satisfy one of the followingconditions:(i) p = 2, or(ii) p = 3 and λ = 1 or µ ≤ 1, or(iii) p ≥ 5 and λ = 1.

188Proof. The result follows from a combination of results already proved.By the note following Theorem 1 in Abért’s paper, [1], any rank 2 abelian2-group is cyclically generated by 2 generators.Let G = Z p λ × Z p λ+µ. If p = 3 then G is cyclically generated by 3 generatorsif and only if λ = 1 or µ ≤ 1 by Theorem 5.1. If p = 5 and λ = 1 then G hasa cyclic generating set of size 5 by Theorem 6.10, while if λ ≥ 2 then G/p 2 Gis Z p 2 × Z p 2, which is not cyclically generated by p generators by Theorem7.13, and hence G is not by Theorem 4.4.

Appendix AGAP and ACE codesThe computer codes here are the codes that we used to find and test results.They are included to give an indication of the methods and commands used,and to allow the reader to check certain claims in Chapters 2 and 3, but wedo not claim that the methods are efficient. All of the GAP codes work whenusing version 4.4.5, see [14], but they should also be suitable for use with anyof the versions released since September 2002. The ACE codes are suitablefor use with version 3.000, see [17].A.1 GAP codesA.1.1Naïve code for searching for cyclically generatedgroupsThis code tests whether a group g is cyclically generated by finding its automorphismgroup and then applying every power of each automorphism toeach element of g. We conclude that g is not cyclically generated if found is0 and g is cyclically generated otherwise.g:= ;IsSolvable(g); # GAP has more efficient methods for computing# the automorphism group of soluble groupsA:=AutomorphismGroup(g);189

190found:=0;for x in g dofor t in A dos:=[];i:=0;while i < Order(t) doAdd(s,Image(t^i,x));i:=i+1;od;h:=Subgroup(g,s);if Index(g,h)=1 then found:=found+1; fi;od;od;found;A.1.2Code for finding GSO generating setsThis code establishes for which relevant m the group g is generated by itselements of order m. If there is no output, we conclude that g is a non-GSOgroup.g:=;for m in [2..Size(g)] doorderm:=Filtered(g,x->Order(x)=m);h:=Subgroup(g,orderm); # h is the subgroup generated by# elements of order mif Index(g,h)=1 thenPrint("g is GSO generated by its elements of order ",m,"\n");fi;od;A.1.3Code for ECS groupsThe following code establishes whether every element of a group g lies ina proper characteristic subgroup. It does this by compiling a collection cc

191of all the elements that lie in proper characteristic subgroups of g and thenusing Difference to find the set diff, which consists of the elements of gthat are not in cc. If this set is empty, i.e. if Length(diff) is 0, then g isan ECS group.g:= ;IsSolvable(g);c:=CharacteristicSubgroups(g);cc:=[];for i in [2..Length(c)] do # c[1] is gcc:=Union(cc,c[i]);# collects the elements that# lie in proper characteristic# subgroups of god;diff:=Difference(Elements(g),cc);;Length(diff);A.1.4Less naïve code for searching for cyclically generatedgroupsBy combining the previous codes and using Theorem 2.5 we have the followingcode that determines whether or not a given group g is cyclicallygenerated. In general, the group g is not given as a group from the smallgroups library, i.e. it is not of the form SmallGroup([n,m]) for some m and n,so in order to compute the automorphism group and conjugacy classes moreeasily, the code sets GAP to identify a permutation group G that is isomorphicto g and then search for cyclic generating sets for the group G instead. If gis polycyclic, then it might be quicker to replace IsomorphismPermGroup(g)with IsomorphismPcGroup(g) and then IsSolvable(G) can be removed.g:= ;iso:=IsomorphismPermGroup(g);G:=Image(iso);IsSolvable(G);

192c:=CharacteristicSubgroups(G);; # first tests whether G is ECScc:=[];for i in [2..Length(c)] do cc:=Union(cc,c[i]); od;diff:=Difference(Elements(G),cc);;if Length(diff) 0 thenset:=[];for m in [2..Size(G)] do # finds set of m such that G has# a GSO generating set of elements# of order morderm:=Filtered(G,x->Order(x)=m);h:=Subgroup(G,orderm);if Index(G,h)=1 then Add(set,m); fi;od;A:=AutomorphismGroup(G);cl:=ConjugacyClasses(G);classes:=[];for j in [1..Size(cl)] do # finds conjugacy classes with# representatives of order in ‘set’x:=Representative(cl[j]);if Order(x) in set then Add(classes,j); fi;od;found:=0;while found < 1 dofor j in classes dox:=Representative(cl[j]);for t in A do# tests only one element from each# conjugacy class in ‘classes’; see# Theorem 2.5s:=[];i:=0;while i < Order(t) doAdd(s,Image(t^i,x));

193i:=i+1;od;h:=Subgroup(G,s);if Index(G,h)=1 then found:=found+1; fi;od;od;od;if found 0 thenPrint("g is cyclically generated","\n");fi;fi;If GAP is able to compute the conjugacy classes of the automorphism groupof G, then the following code gives an alternative way of searching for cyclicgenerating sets, which is justified by Theorem 2.4.g:= ;iso:=IsomorphismPermGroup(g);G:=Image(iso);IsSolvable(G);c:=CharacteristicSubgroups(G);; # first tests whether G is ECScc:=[];for i in [2..Length(c)] docc:=Union(cc,c[i]);od;diff:=Difference(Elements(G),cc);;if Length(diff) 0 thenset:=[];for m in [2..Size(G)] do # finds set of m such that G has# a GSO generating set of elements

194# of order morderm:=Filtered(G,x->Order(x)=m);h:=Subgroup(G,orderm);if Index(G,h)=1 then Add(set,m); fi;od;A:=AutomorphismGroup(G);cl:=ConjugacyClasses(A);found:=0;while found < 1 dofor j in [1..Size(cl)] dot:=Representative(cl[j]);for m in set doorderm:=Filtered(G,x->Order(x)=m);for x in orderm do # applies one automorphism from# each conjugacy class of Aut G to# each element of order m for each# m in ‘set’s:=[];i:=0;while i < Order(t) doAdd(s,Image(t^i,x));i:=i+1;od;h:=Subgroup(G,s);if Index(G,h)=1 then found:=found+1; fi;od;od;od;od;if found 0 then

195fi;fi;Print("g is cyclically generated","\n");Both of these codes can easily be modified to search for cyclic generating setsof a particular size. For example, using the second of the codes with a suitablemodification, we can establish whether Z 7 × Z 49 is cyclically generated by 3generators.gap> g:= DirectProduct(CyclicGroup(7), CyclicGroup(49));gap> iso:=IsomorphismPermGroup(g);gap> G:=Image(iso);gap> IsSolvable(G);truegap> c:=CharacteristicSubgroups(G);;gap> cc:=[];[ ]gap> for i in [2..Length(c)] do> cc:=Union(cc,c[i]);> od;gap> diff:=Difference(Elements(G),cc);;gap> if Length(diff) if Length(diff) > 0 then> set:=[];> for m in [2..Size(G)] do> orderm:=Filtered(G,x->Order(x)=m);> h:=Subgroup(G,orderm);> if Index(G,h)=1 then Add(set,m); fi;> od;> A:=AutomorphismGroup(G);> cl:=ConjugacyClasses(A);

196> found:=0;> while found < 1 do> for j in [1..Size(cl)] do> t:=Representative(cl[j]);> if Order(t)=3 then> for m in set do> orderm:=Filtered(G,x->Order(x)=m);> for x in orderm do> s:=[];> i:=0;> while i < Order(t) do> Add(s,Image(t^i,x));> i:=i+1;> od;> h:=Subgroup(G,s);> if Index(G,h)=1 then found:=found+1; fi;> od;> od;> fi;> od;> od;> if found >0> then Print("g is cyclically generated by 3 generators","\n");> fi;> if found >0> then Print("g is cyclically generated by 3 generators","\n");g is cyclically generated by 3 generators> fi;> fi;In practice, however, it is quicker to prove that a given group is cyclically generated,or cyclically generated by a specified number of generators, by usingPseudoRandom to produce a collection of automorphisms and group elementsand then apply all of these automorphisms to each of the group elements

197using part of the naïve code above. To prove that a GSO non-ECS groupg is not cyclically generated, or not cyclically generated by a certain numberof generators, it is usually quicker to use CharacteristicSubgroups(g)to find a factor group of g that we can prove is not cyclically generated ornot cyclically generated by a suitable number of generators, and then useTheorems 2.9 or 4.4 to prove that g does not have this property.A.2 Example of ACE code for testing generatingsetsACE was used to test certain generating sets because of ease of use. Thesetests could also have been carried out using GAP or by using the ACE packageas the coset enumerator within GAP.If G is a group generated by Group Generators and H is a subgroup of Gthat is generated by the elements Subgroup Generators, which are wordsin the Group Generators, then the following code is used to find the indexof H in G and if ACE returns INDEX = 1 then we know that the setSubgroup Generators generates G.wo: 12000000;mess: 1000;Group Name: G;Group Generators: ;Group Relators: ;Subgroup Name: H;Subgroup Generators: ;hard;beg;For example, ifG = Z 19 × Z 19 × Z 19= 〈x〉 × 〈y〉 × 〈z〉

198then we can use the code above to test whether the setS = {x + y, x − z, x − y − 4z, x − 4y + 4z, x + 4y + z}is a generating set for G. (Note that in ACE, we use multiplicative notationand inverses are denoted by capital letters.)wo: 12000000;mess: 1000;Group Name: G;Group Generators:x,y,z ;Group Relators: x^19, y^19, z^19, XYxy, XZxz,YZyz ;Subgroup Name: H;Subgroup Generators: xy, xZ, xYZ^4, xY^4z^4,xy^4z ;hard;beg;#-- ACE 3.000: Run Parameters ---Group Name: G;Group Generators: xyz;Group Relators: XYxy, XZxz, YZyz, (x)^19, (y)^19, (z)^19;Subgroup Name: H;Subgroup Generators: xy, xZ, xYZZZZ, xyyyyz, xYYYYzzzz;Wo:12000000; Max:1999998; Mess:1000; Ti:-1; Ho:-1; Loop:0;As:0; Path:0; Row:1; Mend:0; No:6; Look:0; Com:10;C:1000; R:1; Fi:10; PMod:3; PSiz:256; DMod:4; DSiz:1000;#--------------------------------SG: a=13 r=1 h=1 n=14; l=1 c=+0.00; m=13 t=13RS: a=1 r=2 h=2 n=2; l=2 c=+0.00; m=14 t=14INDEX = 1 (a=1 r=2 h=2 n=2; l=3 c=0.00; m=14 t=14)Therefore S generates G.

Notationo(g)order of the group element gAut Gautomorphism group of GG p〈S〉[x, y]C nZ n {0, 1, 2, . . . , n − 1},Z ∗ nD 2nSmallGroup([n, m])Z(G)Φ(G)p-primary part of the group Gsubgroup generated by the set S−x − y + x + y if the group operation is addition orx −1 y −1 xy if the group operation is multiplicationcyclic group of order n under multiplicationgroup of units of Z ndihedral group of order 2nm-th group of order n in the GAP small groups librarycentre of the group GFrattini subgroup of GmG subgroup generated by {g m : g ∈ G} or {mg : g ∈ G}when the group operation is multiplication or additionrespectivelyG(m)subgroup generated by the set of elements whose orderdivides mF p n⌊x⌋(a, b)field of order p nlargest integer less than or equal to xgreatest common divisor of a and b199

200F (n, r)Fibonacci group defined by the presentation withgenerators x 1 , x 2 , . . . , x n and relators that are all cyclicpermutations of x 1 x 2 · · · x r x −1r+1 under the mapx 1 ↦→ x 2 ↦→ · · · x n ↦→ x 1GL(n, p) group of n × n invertible matrices with entries from Z pGL(n, Z m ) group of n × n invertible matrices with entries from Z mSL(n, p) the n × n matrices with entries from Z p and determinant 1det MC G (g)determinant of the matrix Mcentralizer of g in G

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