Group Actions

Chapter 2Group ActionsThe purpose of this chapter is to explain what it means for a group to ‘act’on a set. There are two main reasons for studying group actions.(i) This is the main way that group theory applies to other branches ofmathematics as well as to computer science and the physical sciences:an action turns a group into a set of symmetries of an object.(ii) This gives us a useful set of terminology and technology for referringto the behaviour of a group. For example, if we can say that a finitegroup G acts on its set of Sylow p-subgroups, then all the methodsand results of this section can be applied to deduce information aboutthe original group G. Thiswillbethemainwayinwhichwewillusethis chapter later on in the course.2.1 Definition of an actionDefinition 2.1 Let G be a group, and Ω a set. An action of G on Ω is amapsuch thatµ: Ω× G → Ω(ω,x) ↦→ ω x(i) (ω x ) y = ω xy for all ω ∈ Ωandx, y ∈ G,(ii) ω 1 = ω for all ω ∈ Ω.We say that G acts on Ω.We think of an action as a way of making the group G move the pointsin Ω around. Thus the first condition states that applying two elementsx and y in sequence has the same effect as applying the product xy, while21

the second condition is that the identity element fixes every point, and soacts in the same way as the identity permutation of Ω.We will spend some time developing the theory of group actions. First,here are some examples which illustrate the concept’s usefulness and allowus to recall some standard groups at the same time.Example 2.2 (i) Let Ω = {1, 2,...,n}. Recallthatthesymmetric groupof degree n is denoted by S n and consists of all permutations of Ω,under the operation of composition.Then S n acts on Ω by(i, σ) ↦→ iσ(the effect of applying the permutation σ to the number i ∈ Ω). Thetwo conditions of Definition 2.1 hold immediately: the first followssince multiplication in S n is composition and the second holds sincethe identity element in S n is the identity permutation (the map whichfixes all points of Ω).We call this action the natural action of S n .(ii) Recall that the dihedral group of order 2n is the subgroup of S n generatedby the following two permutations:α =(123... n)( )1 2 3 ... nβ =1 n n− 1 ... 2=(2n)(3 n−1) ....We shall denote this group by D 2n .(Thisisdifferentfromthenotationused in MT4003, unfortunately both notations are standard.) Recallthat this has the following properties:|α| = n, |β| =2, βα = α −1 β.Now consider a regular n-gon with vertices labellled from 1 to n.αnβ123Applying α to the vertices induces a rotation of the regular n-gon. Applyingβ produces a reflection in the axis through vertex 1. Hence bothα and β induce transformations of the regular n-gon and consequentlyany product of them does so also.Conclusion: D 2n acts on the vertices of a regular n-gon.22

(iii) Let V be a vector space of dimension n over a field F . Any lineartransformation T : V → V can be represented by an n × n matrixwith entries from F and the transformation is invertible when thecorresponding matrix is non-singular (i.e., has non-zero determinant).Recall that the general linear group of degree n over F isGL n (F )={ A | A is an n × n matrix over F with det A ≠0}.Then GL n (F )actsonV : a matrix A in GL n (F )movesthevectorv(from V )accordingtothelineartransformationdeterminedbyA.These give us examples of group actions where Ω is a combinatorialobject (a set), a geometric object (an n-gon) and a vector space. We willalso consider lots of examples of groups acting on something related to theirown structure, such as their subgroups or their elements. Let’s first developthe basic theory of group actions, and then consider applications.2.2 OrbitsDefinition 2.3 Let the group G act on the set Ω. The orbit of ω ∈ Ωisω G = { ω x | x ∈ G }⊆Ω.Thus, the orbit of ω consists of all the points of Ω we can get to byapplying elements of the group G to ω.Example 2.4 Consider the dihedral group D 2n from Example 2.2, actingon the vertices of the n-gon. The permutation α rotates the verticesanticlockwise by one vertex. Consider vertex number 1. By actingwith α, wecaneventuallymovethisvertextoeveryothervertex. Hence1 D 2n= {1,...,n}.The basic properties of orbits are as follows.Proposition 2.5 Let the group G act on the set Ω, andletα, β ∈ Ω. Then(i) α ∈ α G ;(ii) either α G = β G or α G ∩ β G = ∅.Thus part (ii) asserts that any two orbits are either disjoint orareequal.The proposition then yields:Corollary 2.6 Let the group G act on the set Ω. Then Ω is the disjointunion of the orbits of G.□23

Thus “being in the same orbit” is an equivalence relation on the pointsin Ω. Note that the orbits (equivalence classes) do not all need to be thesame size.Proof of Proposition 2.5: (i) By Definition 2.1(ii) of a group action,α = α 1 ,soα ∈ α G .(ii) Suppose γ ∈ α G ∩ β G . Hence there exist x, y ∈ G such that γ =α x = β y .Applyy −1 :Now, for all g ∈ G:α xy−1 =(α x ) y−1 =(β y ) y−1 = β yy−1 = β 1 = β.β g =(α xy−1 ) g = α xy−1g ∈ α Gand we deduce β G ⊆ α G .Similarly, from β y = α x ,wededuceα g = β yx−1 gfor all g ∈ Gand hence α G ⊆ β G .Hence if α G ∩ β G ≠ ∅, thenα G = β G .□Definition 2.7 AgroupG acts transitively on a set Ω if the whole of Ωforms a single orbit under this action.Thus G acts transitively on Ω if and only if for all α, β ∈ Ωthereexistsx ∈ G such that β = α x .(Forthisiswhatitmeansforβ to lie in the orbitcontaining α.)Example 2.8 We saw in Example 2.4 that 1 D 2n= {1,...,n}, soD 2n actstransitively on the vertices of the regular n-gon.The 0 vector is mapped to itself by all elements of GL n (F ), so the actionof GL n (F )onthevectorsofV is intransitive.2.3 StabilisersDefinition 2.9 Let the group G act on the set Ω, and let ω ∈ Ω.stabiliser of ω in G isTheG ω = { x ∈ G | ω x = ω }.Thus the stabiliser of ω is the set of all group elements which fix ω.24

Example 2.10 Let G =GL n (F )actonV ,asinExample2.2. ForallA ∈ G, theimageofthe0vectorunderA is 0, so the stabiliser of 0 is thewhole of G.If v ∈ V is nonzero, then vA = v if and only if v is an eigenvector forA with eigenvalue 1. So the stabiliser of v is the set of all matrices witheigenvector v and eigenvalue 1.Lemma 2.11 Let the group G act on the set Ω, and let ω ∈ Ω.stabiliser G ω of ω is a subgroup of G.TheProof: Firstly, ω 1 = ω, byDefinition2.1,so1∈ G ω and hence G ω isnonempty. Suppose x, y ∈ G ω .Thenso xy ∈ G ω ,whileω xy =(ω x ) y = ω y = ωω x−1 =(ω x ) x−1 = ω xx−1 = ω 1 = ωso x −1 ∈ G ω .HenceG ω is a subgroup of G.□One crucial reason why stabilisers help us is the following:Theorem 2.12 (Orbit-Stabiliser Theorem) Let the group G act on theset Ω, andletω ∈ Ω. Then|ω G | = |G : G ω |.Thus the size of an orbit equals the index of the corresponding stabiliser.Proof: We demonstrate the existence of a bijection from the set of cosetsof the stabiliser G ω in G to the orbit of ω. Defineφ: G ω x ↦→ ω x .We first check that this is well-defined.x and y. Thenxy −1 ∈ G ω ,soSuppose G ω x = G ω y for someω xy−1 = ω.Apply y:But also:Therefore:ω xy−1y = ω y .ω x(y−1 y) = ω x .ω x = ω y .25

Hence φ is well-defined.Next we show that φ is one-to-one. Suppose that (G ω x)φ =(G ω y)φ forsome x, y ∈ G. Thenω x = ω y .Therefore, upon applying y −1 ,ω xy−1 = ω yy−1 = ω 1 = ω,so xy −1 ∈ G ω and we deduce G ω x = G ω y.Thusφ is one-to-one.Finally, if α ∈ ω G then there exists x ∈ G such that α = ω x .Therefore,(G ω x)φ = α, soφ is onto.Hence φ: G ω x ↦→ ω x defines a bijection from the set of cosets of G ω tothe orbit of ω, andtheresultfollows.□One thing to consider is the following observation: Suppose G acts on Ωand that α, β ∈ Ωaretwopointsthatlieinthesameorbit. Weknowthatorbits are either disjoint or equal, soα G = β G .Hence, by the Orbit-Stabiliser Theorem,|G : G α | = |G : G β |.In particular, if G is a finite group, we can deduce immediately that |G α | =|G β |.Example 2.13 (i) Consider the natural action of S n ,asinExample2.2(i).This action is transitive, so n Sn = {1,...,n}. The stabiliser in S n ofthe point n consists of all permutations that fix the point n, andisisomorphic to S n−1 .Wecheck:|S n : S n−1 | =(ii) Consider the dihedral groupn!(n − 1)! = n = |nSn |.D 2n = 〈α, β〉 = 〈(1 2 ...n), (2 n)(3(n − 1)) ···〉,acting on the n-gon as in Example 2.2(ii). We saw in Example 2.4that 1 D 2n= {1,...,n}, sothestabiliserof1hasorder|D 2n ||1 D 2n |= 2n n =2.Now, the element β fixes 1 and has order 2, so we conclude(D 2n ) 1 = 〈β〉.26

Proposition 2.14 Let the group G act on the set Ω, andletα, β ∈ Ω. Ifα and β lie in the same orbit of G on Ω, thenthestabilisersG α and G β areconjugate subgroups of G. Inparticular,G α x =(G α ) x .Proof: Since β lies in the orbit of α, thereexistsx ∈ G such that β = α x .We will show thatG β =(G α ) x = x −1 G α x.Let g ∈ G α ,sothatx −1 gx ∈ (G α ) x .Thenβ x−1 gx =(α x ) x−1 gx= α xx−1 gx= α gx= α x (as g ∈ G α )= β.Hence x −1 gx ∈ G β ;thatis,(G α ) x ⊆ G β .For the reverse inclusion, note first that from β = α x ,wededuceβ x−1 = α xx−1 = α,so, reasoning in the same way as we did from the equation α x = β, wegetthat is,(G β ) x−1 ⊆ G α ;xG β x −1 ⊆ G α .Multiply on the left by x −1 and on the right by x:G β ⊆ x −1 G α x =(G α ) x .Thus G β =(G α ) x ,asrequired.□Example 2.15 Let G = S n ,initsnaturalaction. Wechoosex =(1n) asan element mapping n to 1 (there are plenty of other choices!). We haveseen thatG n = {σ ∈ S n : nσ = n} ∼ = S n−1 .It follows from the previous proposition thatG 1 = {σ ∈ S n : 1σ =1}=(G n )(1 n)= {σ (1 n) : nσ = n}We will continue to develop the theory of group actions later,butlet’sfirst consider a couple of examples which illustrate how we can applythistheory to the study of groups.27

2.4 ConjugationIn this section we study what is probably the second most important waythat a group can act on itself.Example 2.16 (Conjugation Action) Let G be a group. We define anaction of G on itself byG × G → G(g, x) ↦→ x −1 gx = g x ,the conjugate of g by x. Let’sfirstcheckthatthisisindeedagroupaction.(i) Let g, x, y be any elements of G. Then (g x ) y = y −1 (x −1 gx)y =y −1 x −1 gxy =(xy) −1 g(xy) =g xy .(ii) Let g ∈ G. Theng 1 =1 −1 g1=1g1 =g.Now let’s calculate the orbits and stabilisers for the conjugation action.If g ∈ G, thentheorbitofg is the conjugacy class of g (in G):The stabiliser of g is:g G = { g x | x ∈ G } = { x −1 gx | x ∈ G }.G g = { x ∈ G | g x = g }= { x ∈ G | x −1 gx = g }= { x ∈ G | gx = xg };i.e., under the action of conjugation, the stabiliser of g consists of the set ofelements of G which commute with g.Definition 2.17 If G is a group and g ∈ G, thenthecentraliser of g (in G)isC G (g) ={ x ∈ G | gx = xg }.Let’s now apply our results from Section 2.3 to the conjugation action.Proposition 2.18 Let G be a group. Then(i) G is the disjoint union of its conjugacy classes;(ii) the centraliser of an element g is a subgroup of G;(iii) the number of conjugates of an element g equals the index of its centraliser;28

(iv) if g, x ∈ G thenC G (g x )=C G (g) x .Proof: (i) Immediate from Corollary 2.6: a set is the disjoint union of theorbits in a group action.(ii) Immediate from Lemma 2.11: a stabiliser is a subgroup.(iii) Immediate from the Orbit-Stabiliser Theorem (Theorem 2.12): thesize of an orbit equals the index of the corresponding stabiliser.(iv) Immediate from Proposition 2.14.□(This material also appeared in MT4003, but it can be deduced muchmore naturally by using group actions.)Let G be a finite group. Write G as the disjoint union of its conjugacyclasses:G = C 1 ∪ C 2 ∪···∪C k .Hence|G| = |C 1 | + |C 2 | + ···+ |C k |k∑= |G :C G (g i )|,i=1where g i is a representative for the conjugacy class C i .Supposethat|C i | =1for 1 i l and |C i | > 1fori>l.ThenNotek∑|G| = l + |G :C G (g i )|.i=l+1|C i | =1 ifandonlyif g x i = g i for all x ∈ Gif and only if g i x = xg i for all x ∈ G.Definition 2.19 If G is a group, the centre of G isZ(G) ={ g ∈ G | gx = xg for all x ∈ G };that is, the set of elements in G which commute with all elements of G.The centre of G can easily be seen to be a subgroup of G. [Proof[Omitted in lectures]: 1x = x = x1 forallx ∈ G, so1∈ Z(G) andhence Z(G) ≠ ∅. If g, h ∈ Z(G), then (gh)x = ghx = gxh = xgh = x(gh)for all x ∈ G, sogh ∈ Z(G). Multiply the equation gx = xg on the left andon the right by g −1 ,togetxg −1 = g −1 (gx)g −1 = g −1 (xg)g −1 = g −1 x for allx ∈ G, henceg −1 ∈ Z(G). Therefore Z(G) isasubgroupofG.]Our discussion above now establishes:29

Theorem 2.20 (Class Equation) Let G be a finite group. Then|G| = |Z(G)| +k∑|G :C G (x i )|i=l+1where x l+1 ,...,x k are representatives for the conjugacy classes of lengthgreater than 1.□Example 2.21 Let’s look at D 2×4 = {1,α,α 2 =(13)(24),α 3 =(1432),β,αβ =(14)(23),α 2 β =(13),α 3 β =(12)(34)}. The element α 2 commuteswith both α and β, soliesinZ(D 2×4 ). It is clear that α, α −1 ∉Z(D 2×4 ), since α and β don’t commute. It’s easy to check that α i β does notcommute with α for i ∈{0, 1, 2, 3}, soZ(D 2×4 )=〈α 2 〉 ∼ = Z 2 .It is then straightforward to check that all other conjugacy classes have size2(andconsistofthetwoelementswiththesamecyclestructure). Thus allelements of D 2×4 that do not lie in Z(D 2×4 )havecentralisersoforder4.2.5 Conjugation on subgroupsExample 2.22 (Conjugation action on subsets and subgroups)Let G be a group and let P(G) denotethesetofallsubsetsofG (the powerset of G). We define an action of G on P(G) byP(G) × G → P(G)(A, x) ↦→ A x = x −1 Ax = { x −1 ax | a ∈ A }.AsimilarargumenttoExample2.16checksthatthisisindeedan action.The orbit of A is the set of all conjugates of A and the stabiliser is the‘normaliser’ of A:Definition 2.23 If G is a group and A is a subset of G, thenormaliserof A (in G) isN G (A) ={ x ∈ G | A x = A }.Since this is a stabiliser, it is a subgroup of G (by Lemma 2.11). Weshall be most interested in the case when A is also a subgroup.Let G be a group and let x ∈ G. Write τ x for the permutation of theelements of G which is induced by conjugation by x, sothatτ x (g) =g x .30

Observations:(i) Let g, h ∈ G. Then (gh)τ x = x −1 ghx = x −1 gx · x −1 hx =(gτ x )(hτ x ).That is, τ x is a homomorphism.(ii) Let g ∈ G. Then gτ x τ x −1 = x(x −1 gx)x −1 = g, sothatτ x τ x −1 =id G .Similarly τ x −1τ x =id G . Hence τ x is an bijection (it has τ x −1 as itsinverse).Definition 2.24 Let G be a group. An automorphism of G is a map G → Gwhich is an isomorphism. An automorphism φ is inner if there exists x ∈ Gsuch that gφ = gτ x ,forallg ∈ G.Example 2.25 Consider A 4 ,thealternatinggroupon4points. Theelementx =(123)∈ A 4 ,soconjugationby(123)isaninnerautomorphismof A 4 .Theelementy =(12)belongstoS 4 but not to A 4 ,soconjugationofA 4 by y is an automorphism of A 4 that is not inner. Conversely, conjugationof S 4 by y is an inner automorphism of S 4 .Since τ x is an automorphism of G, ifH is a subgroup of G, thenitsimage under τ x is also a subgroup. Hence the conjugateH x = { x −1 hx | h ∈ H }is a subgroup of G. Furthermore, the Orbit-Stabiliser Theorem (Theorem2.12) tells us that the number of conjugates of H equals the indexin G of the normaliser of H, becausethestabiliserofH in this action is thenormaliser of H.Summarizing, we have proved:Proposition 2.26 Let G be a group.(i) If x ∈ G, thentheconjugationmapτ x : g ↦→ g x is an automorphismof G.(ii) If H is a subgroup of G and x ∈ G, thentheconjugateH x is a subgroupof G.(iii) If H is a subgroup of G, thenthenormaliserN G (H) of H in G is asubgroup of G.(iv) If H is a subgroup of G, thenthenumberofconjugatesofH in Gequals the index |G :N G (H)| of the normaliser of H in G. □31

2.6 Permutation representationsThe Orbit-Stabiliser Theorem tells us about the link between agroupactionand the indices of certain subgroups: the point stabilisers. We will nowconstruct a homomorphism associated to the group action. The imageofthe homomorphism is a subgroup of a symmetric group.Definition 2.27 Let Ω be any set. The set of all permutations of Ω undercomposition of maps is called the symmetric group on Ωandisdenotedby Sym(Ω). We recover S n by considering the special case of Ω ={1, 2,...,n}.Let the group G act on the set Ω, and let x ∈ G. Thenx induces a mapρ x from Ω to itself by ωρ x = ω x .Nowandωρ x ρ x −1 =(ω x ) x−1 = ω xx−1 = ω 1 = ωωρ x −1ρ x =(ω x−1 ) x = ω x−1x = ω 1 = ω.Hence ρ x ρ x −1 = ρ x −1ρ x =1,soρ x is a bijection and thereforeρ x ∈ Sym(Ω) for all x ∈ G.So to each element x of G we associate a permutation, ρ x ,ofthesetΩ.We therefore have a constructed a map ρ : G → Sym(Ω), xρ = ρ x .Nowfor all ω ∈ Ωandx, y ∈ G, soThusωρ x ρ y =(ω x ) y = ω xy = ωρ xyρ x ρ y = ρ xy for all x, y ∈ G.(xρ)(yρ) =(xy)ρ for all x, y ∈ G.That is, ρ : G → Sym(Ω) is a homomorphism.following theorem.We record this in theTheorem 2.28 Let G be a group, let Ω be a set and let G act on Ω. Foreach x ∈ G, themapis a permutation of Ω. Themapis a homomorphism.ρ x :Ω→ Ωω ↦→ ω xρ: G → Sym(Ω)x ↦→ ρ x32

Definition 2.29 A permutation representation of G is a homomorphismfrom G to Sym(Ω), for some set Ω. The kernel of ρ is the kernel of theaction. Thiskernelconsistsoftheelementsx of G such thatωρ x = ω for all x ∈ G;i.e., the elements of G which fix all points in Ω.First, an easy example:Example 2.30 Consider the group (Z 6 , +). We can make this group acton {0, 1, 2} by addition modulo 3:a + b = a + b mod 3 for a ∈{0, 1, 2},b ∈ Z 6. This action will give a permutation representation ρ of Z 6 as a subgroupof Sym(Ω), where Ω = {0, 1, 2}. Wecalculate:0ρ =()=3ρ, 1ρ =(0, 1, 2) = 4ρ, 2ρ =(0, 2, 1) = 5ρ.Note that the kernel of the action is {0, 3} Z 6 .Now we consider an important family of examples.Example 2.31 (Right Regular Action) Let G be a group. We define amap from G × G to G by(g, x) ↦→ gx.Wefirstcheckthatthismapisagroupaction:(i) (gx)y = g(xy) forallg, x, y ∈ G (by associativity),(ii) g1 =g for all g ∈ G.This action is the right regular action of G.Theorem 2.28 provides us with a homomorphism ρ: G → Sym(G). Whatis the kernel of ρ?x ∈ ker ρ if and only if ρ x =1if and only if gρ x = g for all g ∈ Gif and only if gx = g for all g ∈ Gif and only if x =1.Hence ker ρ = 1 and so ρ is one-to-one. It follows that G is isomorphicto im ρ and we have proved Cayley’s Theorem.Theorem 2.32 (Cayley’s Theorem) Every group is isomorphic to a subgroupof a symmetric group.□33

Example 2.33 Consider D 2×3∼ = S3 .Let’slisttheelementsinthefollowingorder:x 1 =1,x 2 = α, x 3 = α 2 ,x 4 = β,x 5 = αβ, x 6 = α 2 β.Then the regular representation sendsα ↦→ (x 1 x 2 x 3 )(x 4 x 6 x 5 ),β ↦→ (x 1 x 4 )(x 2 x 5 )(x 3 x 6 ).(The images of the other elements of D 2×3 can be worked out from these.)Our final general example is extremely important: itwilloccurthroughoutthe course.Example 2.34 (Action on Cosets) Let G be a group and let H be asubgroup of G. Let Ω = { Hg | g ∈ G }, thesetofcosetsofH in G. Wedefine an action of G on Ω as follows:Ω × G → Ω(Hg,x) ↦→ Hgx.This can be checked to be a well-defined action of G on Ω which is transitive(i.e., there is exactly one orbit). The stabiliser of the cosetHg is theconjugate H g of the original subgroup H. Tothisactiontherecorrespondsapermutationrepresentationρ: G → Sym(Ω). The basic properties of ρare summarised as follows:Theorem 2.35 Let H be a subgroup of G, letΩ={ Hx | x ∈ G }, andletG act on Ω by right multiplication. Let ρ: G → Sym(Ω) be the associatedpermutation representation.(i) If H

2.7 p-GroupsWe finish our chapter on group actions by establishing some tools concerningactions of p-groups. We then apply them to deduce results about thestructure of such groups.Definition 2.38 Let p ∈ N be prime. A finite group G is called a p-groupif its order is a power of p.An important tool relating to p-groups is the following:Lemma 2.39 Let G be a finite p-group and let G act on the finite set Ω.DefineFix G (Ω) = { ω ∈ Ω | ω x = ω for all x ∈ G },the set of fixed-points of the action. Then|Fix G (Ω)| ≡|Ω|(mod p).Proof: Express Ω as a disjoint union of orbits:Ω=Ω 1 ∪ Ω 2 ∪···∪Ω k .Suppose (without loss of generality) that |Ω i | =1fori =1,2,...,l, andthat |Ω i | > 1fori = l +1,...,k. SoFix G (Ω) = Ω 1 ∪ Ω 2 ∪···∪Ω l and |Fix G (Ω)| = l.By the Orbit-Stabiliser Theorem (Theorem 2.12),|Ω i | = |G : G ωi | = |G|/|G ωi |for all i, whereω i ∈ Ω i . Now |G| is a power of the prime p and henceeach |Ω i | is also a power of p. Thereforep divides |Ω i |for i = l +1,...,k.Hence|Ω| = |Ω 1 | + |Ω 2 | + ···+ |Ω k |k∑= |Fix G (Ω)| + |Ω i |i=l+1≡|Fix G (Ω)| (mod p).□Here is an application:35

Proposition 2.40 Let G be a finite p-group. If N is a non-trivial normalsubgroup of G, thenN ∩ Z(G) ≠ 1.In particular, since G is a normal subgroup of itself, we obtain:Corollary 2.41 Let G be a finite p-group. Then the centre of G is nontrivial.□Proof of Proposition 2.40: Since N G, itfollowsthatg x = x −1 gx ∈N for all g ∈ N and x ∈ G. We may therefore define an action of G on NbyN × G → N(g, x) ↦→ g x = x −1 gx.(The conditions for an action follow as in Example 2.16. What is special hereis that normality ensures that when we apply an element of G to an elementof N we end up back inside N.) Now G is a p-group, so by Lemma 2.39:(since |N| = p r for some r 1). Note|Fix G (N)| ≡|N| ≡0 (mod p)Fix G (N) ={ g ∈ N | x −1 gx = g for all x ∈ G }= { g ∈ N | gx = xg for all x ∈ G }= { g ∈ N | g ∈ Z(G) }= N ∩ Z(G).Thusso|N ∩ Z(G)| ≡0 (mod p),N ∩ Z(G) ≠ 1.□Example 2.42 Let’s continue our analysis of D 2×4 from Example 2.21.Lets think about the normal subgroups of D 2×4 .WeknowthatZ(D 2×4 )={1,α 2 }. Thus any proper normal subgroup must have order 2 (and be thecentre) or order 4 (and contain the center), by Proposition 2.40. Since anormal subgroup must be a union of conjugacy classes, and any subgroupmust be closed under multiplication and inversion, it follows easily that theset of normal subgroups of D 2×4 of order 4 is:〈α〉, {1,α 2 ,β,βα 2 }, {1,α 2 ,αβ,α 3 β}.We shall meet several further applications of Lemma 2.39 in the nextchapter.36