Vol. 10 No 3 - Pi Mu Epsilon

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There are other questions that could be asked about similar partitions. Forexample, let 0(m) denote the number of partitions with m parts in which thekth part is less than or equal to m - k + 1 and all parts are odd. Forexample, O(5) = 7 because of the partitions53311 53111 51111 33311 33111 31111 11111.I conjecture that, for m 2 2,andRachele Dembowski did the work that led to this paper while a studentat Seton Hall University, from which she was graduated in May. Heradvisor was Professor Esther Guerin.Xuming Chen (University of Alabama, Tuscaloosa) notes that it seemsas if, given three consecutive odd primes, pn , p +pn +'>, it is always thecase that pn +pn+, > pn+? (3 + 5 > 7, 5 + 7 > 11, ... , 99971 +99989 > 99991, . . .) and wonders if this is easy or hard to prove. Is thereany relation to Bertrand's Theorem that for any positive integer n > 2 thereis a prime between n and 2n7AUTOMORPHISMS OF HASSE SUBGROUP DIAGRAMSFOR CYCLIC GROUPSLars SemeHendrix Collegelit; --.The work presented here extends that of Butt [l] and Woodard [4], whocalculated automorphisms of Hasse subgroup diagrams, Butt for groups ofsmall order and Woodard for the cyclic group C m n , where p and q areprime and m and n are natural numbers. Here we extend their results tocover all finite cyclic groups. The theory of Hasse subgroup diagrams isnot new; the definitive texts are Suzuki [3] and the recently published book[2] by Schmidt. Our results are a special case of Jones' theorem onclassifying the isomorphism classes of Hasse subgroup diagrams associatedto any finite group (see [3], p. 37, Theorem 4.5). However, the work inthis paper was done independently of these references.The Hasse subgroup diagram for a group G is the lattice of subgroupsordered by subgroup containment. The group is the top element of thelattice and the subgroup containing only the identity is the bottom element.Subgroup A is below subgroup B in the lattice if A G B. An edge connectsA and B whenever there are no intermediate subgroups; thus edges impliedby transitivity are suppressed.An automorphism, ip, of a Hasse subgroup diagram, H, is a bijectionfrom H to H that preserves or reverses order. An order-preservingautomorphism has the property that for tyo lattice elements x and y, ifx Â¥ y, then ip(x) Â¥ ~(y) . If x 2 y, then p is an order-reversingautomorphism. Finally, the identity automorphism is the bijection that fixesthe elements of H.nj , where pi is prime and ni and j are positive integers,Let C ,,,PI Pi -pjdenote a finite cyclic group. Since the subgroups of a cyclic group arecyclic, we can give the subgroups the general form < pfl p$ ... p: > ,
SEME, HASSE SUBGROUP DIAGRAMS 217where ki can take on any value between 0 and ni and using the notation< g > to denote the subgroup generated by g. A few simple calculationsshow thatand thatis directly belowin the Hasse subgroup diagram.LEMMA 1. In the Hasse subgroup diagram of C nl nj the subgroupPi P? -Pj< pfl p$ p>jl > has:0 subgroups directly below if ki = ni for i = 1 to j,j subgroups directly below if ki # ni for i = 1 to j,j - m subgroups directly below if m is the number of times ki = nifor i = 1 to j.k kProof. Suppose < pl p22 - p> > is a subgroup of C nl n2 IfPi P. -p;j .ki = ni, then our subgroup is the identity subgroup and can have nosubgroups below it. Suppose ki # ni for i = 1 to j. By adding 1 to anyexponent in < p? pSp> > we obtain another subgroup in the formk k< pl 1 .-p~*below > . As noted above, this subgroup will be directlyp? -p> > . Since there are j choices of exponents to increaseby 1, there will be j distinct subgroups directly below < p;l p^ -p> > .Finally, suppose that there are m exponents such that ni = ki . We can add1 only to those j - m exponents which are not ni. Therefore, there arej - m subgroups directly below < pfl p$ - p.1 > .LEMMA 2. In the Hasse subgroup diagram of Cnj the subgroupP? P;. ... Pj< pf~p$.- - pj > has:0 subgroups directly above if ki = 0 for i = 1 to j,j subgroups directly above if ki # 0 for i = 1 to j,j - m subgroups directly above if m is the number of times ki =Ofori= 1 toj.The proof is similar to that of Lemma 1.The rank of a subgroup is its level or height in the Hasse subgroupdiagram. We define the rank of the identity subgroup to be zero. Therankof a subgroup is then the number of lattice points passed following acontinually ascending chain from the identity to the subgroup.Figure 1 shows the Hassesubgroup diagram for theRankgroup Since is4two lattice points above the 3identity subgroup, has ,/' >-,, y /'rank 2. Figures 2 and 3 show ' CQQ. Figure 2 shows the' \,,' >, ,/'Hasse subgroup diagram and Figure 3 shows a two- '-, \ 'dimensional representation to (4'show the rank structure moreclearly.Figure 1LEMMA 3. The rank of< pk>p$ - p>jl > in the Hasse subgroup diagram of C nl n2 ... IS(nl + n2 - + nj) - (kl ++ - + kj).PlP2Proof. Suppose < pFp$ ..- p>jl > is a subgroup of CpBlp;2 ...py.Using the fact that {e} can bewritten as ,we create a chain bysubtracting 1 from theexponent of pl . We continueuntil the exponent of pi iskl. We have now movednl - k, lattice points(subgroups) up the diagram.Continuing this for eachp, ,the chain is now (nl + n2 + - Figure 2P^
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Page 10 and 11: A WEIGHTED AM-GM-HM INEQUALITYDan K
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Page 25: Rachele DernbowskiSUNY, Stony Brook
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