From: lfr942f@cnas.smsu.edu (Reid Leslie F) Newsgroups: sci.math Subject: Re: Determining finite groups Date: 6 Mar 1996 16:57:50 GMT Nick Halloway (snowe@rain.org) wrote: : I posted a question earlier: can a finite group be determined, given : information about the number of conjugacy classes with a given size and : order of elements. [snip] : Any thoughts? I think I have an explicit counterexample to your original question. Let p be a prime greater than 3. Let H=C_p x C_p x C_p x C_p. We have Aut(H)=GL_4(Z_p). We will construct two semidirect products of C_p and H, one where the generator of C_p is sent to the matrix 1 1 0 0 and another where is is sent to 1 1 0 0 0 1 1 0 0 1 0 0 0 0 1 0 0 0 1 1 0 0 0 1 0 0 0 1. If we denote the first group by G_1 and the second by G_2, then explicit presentations are G_1= and G_2=. Now all of the nonidentity elements of G_1 and G_2 have order p (this is where we need p>3; it is not the case in G_1 if p=2 or 3) and both groups have p^3-p^2 conjugacy classes of order p^2, p^3-p conjugacy classes of order p, and p^2 conjugacy classes of order 1. The groups are not isomorphic since modding out by their centers gives nonisomorphic groups: G_1/Z(G_1)=G_1/= G_2/Z(G_2)=G_2/=, the former being nonabelian, while the latter is abelian. I think the key was getting two different Jordan canonical forms having the same number of 1's on the superdiagonal. Let me know if this does the trick, Les Reid Southwest Missouri State University ============================================================================== From: snowe@rain.org (Nick Halloway) Newsgroups: sci.math Subject: Re: Determining finite groups Date: 7 Mar 96 20:54:42 GMT Posted for Avinoam Mann The questin was: suppose G is a finite group, and for each order of an element of G we know how many conjugacy classes of elements of this order G has, and also the sizes of these classes. Is G then determined (up to isomorphism)? The answer is "no". E.C.Dade, answering a question of R.Brauer, has constructed, for each prime p at least 5, two non-isomorphic groups of exponent p, order p^7, and the same character table. Having the same char. table is much stronger than having the same number of classes of each size. Dave Rusin suggested an asymptotic approach as follows. Let f(n,p) be the number of groups of order p^n, and let g(n,p) be log_pf(n,p). Then it is known that g(n,p) is asymptotic (as n goes to infinity) to (2/27)n^3. Now suppose that G, of order p^n, has a(j,k) classes of size p^j of elements of order p^k. THen both j and k are between 0 and n, so there are (n+1)^2 pairs (j,k). For each pair, the number a(j,k) is at most p^(n-j). The total no. of possible a(j,k) is then about p^((1/2)n^3). Now if the constant 1/2 here could somehow be improved to be less than 2/27, we'll obtain that there are many p-groups, for large n, with the same a(j,k). One possibility to improve the estimate is to note that since x belongs to the centraliser of x, the size p^j is at most p^(n-k), so j + k is at most n. This replaces the 1/2 by 1/3, which is not yet good enough. But, if we look at G.Higman's paper in which he obtains the lower bound (2/27)n^3 for g(n,p), we see that he uses only groups of exponent p or p^2. Thus we may assume that k is 1 or 2, and this gives 2n pairs (j,k), and a quadratic bound for log_p(no. possible a(j,k)), so certainly for large n we have many groups with the same a(j,k). References: Dade's paper is the first in J.Alg. (v.1, p.1).(1964). Higman's paper is at Proc. London Math. Soc., late 50's. C.C.Sims' paper, giving the upper bound for g(n,p), is at a later volume of the same Proc. Avinoam Mann