From: mareg@mimosa.csv.warwick.ac.uk () Subject: Re: automorphisms of abelian p-groups Date: Fri, 19 Oct 2001 21:01:35 +0000 (UTC) Newsgroups: sci.math Summary: Computing the orders of Aut(P) for certain families of p-groups P In article <70b3c8fa.0110191148.5db1776@posting.google.com>, finelar@hotmail.com (larry fine) writes: >Suppose we have an abelian finite p-group: > > >(Z_p^n1)x(Z_p^n2)x..x(Z_p^n_i) > >and n1 >The book claims there's a chain of characterisic subgroups: > >1=C0 < C1 ... < Cn = G such that |C(n+1):C(n)| = p For a p-group G and i>=0, define O_i = O_i(G) = { g in G | g^{p^i} = 1 } and P_i = P_i(G) = { g^{p^i} | g in G }. Then O_i(G) and P_i(G) are characteristic subgroups of G, and you can construct your series as suitable intersections of these subgroups. Example: Z_p x Z_{p^2} x Z_{p^4}. Choose C0 = 1, C1 = P_3, C2 = P_2, C3 = P_1^O_1, C4 = P_1, C5 = O_1, C6 = O_2, C7 = G. >and this implies that |Aut(G)| = (p-1)*p^r for some integer r. This is not correct. Since p-1 divides |Aut(Z_{p^j})|, it is clear that |Aut(G)| is divisible by (p-1)^i. In fact, we have |Aut(G)| = (p-1)^i*p^r for some r. You can prove that any automorphism that centralizes G/P_1 has order a power of p. Also any automorphism that induces the identity on each factor C_{j+1}/C_j has order a power of p. Putting those facts together, you get |Aut(G)| = (p-1)^i*p^r. More details on request! Derek Holt. ============================================================================== From: mareg@mimosa.csv.warwick.ac.uk () Subject: Re: automorphisms of abelian p-groups Date: Tue, 23 Oct 2001 17:20:13 +0000 (UTC) Newsgroups: sci.math In article <70b3c8fa.0110221943.cb16545@posting.google.com>, finelar@hotmail.com (larry fine) writes: >Thanks for the hints, Derek! > >I just have a couple questions. > >1. I considered those groups O_i & P_i but although you can theoretically > figure out the chain, I wasn't able to prove that it can always be done. OK, let m (= your n_i) be such that p^m is the exponent of G, and then take the series 1 = P_m < P_{m-1} < ... < P_1 < P_0 = G. This series has elementary abelian layers P_i/P_{i+1}, and because all of the invariant factors of G are distinct, you can choose generators of each such layer which are the images of elements in G having distinct orders. So if we refine the series by intersecting with the O_j, then we get layers of order p. To be precise, we refine the layer P_{i+1} < P_i to P_{i+1} = P_{i+1}(O_0 ^ P_i) < P_{i+1}(O_1 ^ P_i) < P_{i+1}(O_2 ^ P_i) < ... < P_{i+1}(O_m ^ P_i) = P_i. The layers in this series all have order 1 or p, so we just remove any repeated terms to get a series with all factors of order p. >2. Can you please elaborate a bit on the second part of the proof? Specifically, > |Aut(G)| is obivously (p-1)^i * p^r * n for some n & r, but how do I show > that n is 1? And what do you mean by "induces identity". And isn't > P_1 not necessarily one of the chain elements, so where does it come in? > All of the subgroups in the series above are characteristic, so any automorphism x of G will fix all of these subgroups. It will then induce an automorphism on any of the factors C_j/C_i (using your notation below) with j > i. The set of automorphisms that induce the identity automorphism on C_j/C_i in this way forms a subgroup of Aut(G). I have now chosen the series to ensure that P_1 is in it. P_1 is intersting because it is the Frattini subgroup of G. There is a theorem (see for example Theorem 12.2.2 of M. Hall, "The Theory of Groups") which says that the subgroup of Aut(G) which induces the identity on P/Frattini(P) has order a power of p in any p-group. So to prove that |Aut(G)| = (p-1)^i * p^r for some r, it is enough to prove that the induced group of automorphisms of P/P_1 has order (p-1)^i *p^r for some r. Let the part of the series between P_1 and P be P_1 = Q_i < Q_{i-1} < ... Q_1 < Q_0 = G. Each factor Q_{i-1}/Q_i has order p, and so |Aut(Q_{i-1}/Q_i)| = p-1. Hence |Aut(G)| has order xy, where x divides (p-1)^i, and y is the order of the subgroup of G that induces the identity automorphism on each of the factors Q_{i-1}/Q_i. You can show that y is a power of p, and since we already know that |Aut(G)| is divisible by (p-1)^i, this proves that |Aut(G)| is equal to (p-1)^i times a power of p. Derek Holt. [quote of previous message deleted --djr] ============================================================================== From: mareg@mimosa.csv.warwick.ac.uk () Subject: Re: automorphisms of abelian p-groups Date: Wed, 24 Oct 2001 08:35:27 +0000 (UTC) Newsgroups: sci.math In article , genkisaito@hotmail.com (Nobuo Saito) writes: > >If G is of type (p, p, ...,p) i.e. an elementary abelian p-group, >then Aut(G) is isomorphic to the GL(n, Z/pZ), i.e. the general linear >group >over a finite field Z/pZ. >In this case, I think, the structure of Aut(G) is rather well known: >1) Each element of Aut(G) can be explicitly written(as a regular >matrix) > and multiplication of two elements can be explicitly written. >2) |Aut(G)| can be explicitly written by p and n. >3) The composition factors of Aut(G) can be explicitly determined. >4) Irreducible representations over the complex number field can be >explicitly >determined(right? I'm not sure.). >etc. > >Note:In the above statements, at least p is a _general_ prime, >not a given particular prime like 5. >I'm talking about like GL(3, Z/pZ), GL(n, Z/pZ) >where n is a general positive integer and p is a general prime number. > >If G is of type (p^n, p^n,...,p^n), then Aut(G) is isomorphic >to the general linear group over a finite ring Z/p^nZ. >So in this case also, I think, the structure of Aut(G) is rather well >known. >(I may be wrong. For example, can irreducible representations >be explicitly determined?) > >The question is: >What about Aut(G), where G is an abelian p-group of type (p^n1, >p^n2,...)? >Here, p is a general prime, but (n1, n2,..) can be given particular >integers >like (1, 2, 3). Avinoam Mann has pointed out that this problem is discussed in a book on Group Theory by Speiser, and that the order of |Aut(G)| is determined. I haven't looked this up, but I think one can say the following. Suupose G has di direct summands of type p^ni, for i = 1,...,r, where the ni are all distinct. Then Aut(G) has a normal subgroup P of order a power of p, and G/P is the direct product of the groups GL(di,p). The chief factors of G within P can be regarded as modules for G/P, over the field of order p, and the module structure of each factor is of the form Hom(Vi,Vj), for some (not necessarily distinct) i,j with 1 <= i,j <= r, where Vi is the natural module for GL(di,p). I can believe that with a bit of thought one could work out exactly how many chief factors there are of each possible type (this will be a function of the ni), and hence the order of P, but I am too lazy to try and do that now, and it is presumably to be found in Speiser's book. >P.S. >What's CFSG? The Classification of Finite Simple Groups, but you do not need that for this particular problem. Derek Holt. ============================================================================== From: mareg@mimosa.csv.warwick.ac.uk () Subject: Re: automorphisms of abelian p-groups Date: Sun, 28 Oct 2001 16:12:32 +0000 (UTC) Newsgroups: sci.math In article <1ebd23f6.0110261054.1dfae19d@posting.google.com>, rkbqk@yahoo.com (Ron J.) writes: >> >> Suupose G has di direct summands of type p^ni, for i = 1,...,r, where >> the ni are all distinct. >> >> Then Aut(G) has a normal subgroup P of order a power of p, and G/P is >> the direct product of the groups GL(di,p). >> > >which book is that out of? Or is there an easy way to prove that >last statement? This can be proved in the same way as in the special case when all of the invariant factors have distinct orders. You get a series of characteristic subgroup 1 < C_1 < C_2 < ... C_{r+1} < G, where C_1 is the Frattini subgroup, which is the subgroup generated by all p-th powers, and the other C_i are defiend as < C_1, O_j > where O_j is the subgroup of elements of order p^j for appropriate j. Then C_{i+1}/C_i is elementary of order p^di, and so the group of automorphisms induced on each C_{i+1}/C_i is GL(di,p), and the kernel of the map onto the direct product of the GL(di,p), induces the identity automorphism on each C_{i+1}/C_i, and can be shown to be a p-group. Derek Holt.