From: snowe@rain.org (Nick Halloway) Newsgroups: sci.math Subject: Determining finite groups Date: 15 Feb 96 02:32:22 GMT Does the number of elements of various orders determine a finite group up to isomorphism? For example, C_4 has 1 element order 1, 1 element order 2 and 2 elements order 4; C_2 x C_2 has 1 element order 1 and 3 elements order 2. This information determines finite abelian groups; is it enough to specify any finite group? If not, do the sizes of conjugacy classes, together with the orders of their elements, determine a finite group up to isomorphism? e.g. for C_2 x C_2 you would be given information: 1 conjugacy class of size 1, elements of order 1 3 conjugacy classes of size 1, elements of order 2. Can you transform a given finite group by redefining operations on the elements in some way so that the number of elements of different orders remains the same, but the group is not isomorphic to the original group? One thought I had was, given a group (G,*), define a group (G,&) by g&h = h*g. But then the two groups are isomorphic since g --> g^(-1) is an isomorphism. ============================================================================== From: rusin@vesuvius.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: Determining finite groups Date: 15 Feb 1996 18:40:43 GMT In article , Nick Halloway wrote: >Does the number of elements of various orders determine a finite group up to >isomorphism? For example, C_4 has 1 element order 1, 1 element order >2 and 2 elements order 4; C_2 x C_2 has 1 element order 1 and 3 >elements order 2. > >This information determines finite abelian groups; is it enough to >specify any finite group? No. For example, for any odd prime p there is a non-abelian group of order p^3 all of whose non-identity elements have order p. >If not, do the sizes of conjugacy classes, together with the orders >of their elements, determine a finite group up to isomorphism? e.g. >for C_2 x C_2 you would be given information: > >1 conjugacy class of size 1, elements of order 1 >3 conjugacy classes of size 1, elements of order 2. Still no. There's probably a pretty small example, but I can come pretty close to an existence proof. You might enjoy trying to see if you can actually make it work. Consider a group G of order p^n (p a prime). The conjugacy class of an element x is the orbit of x under the action of G on itself by conjugation, and so is in a natural way in one-to-one correspondence with G/Stab(x) = G/Centralizer(x). In particular, this is of order p^k for some k < n. Also the order of x is p^m for some m <= n. You intend to store the numbers a_{k,m}(G) = number of classes of cardinality p^k of elements of order p^m. Thus you have at most n^2 integers, all at most equal to p^n. There are thus only (p^n)^(n^2) = p^(n^3) different invariants you are storing for G. Here you need to strengthen the bounds just a little bit. For example, a_{k,m} is in fact less than p^(n-k), so there's an easy factor of (1/2) which can be inserted in front of the n^3. Let's assume you can further reduce it by say a factor of (1/8). (Realistically, O(n^3) looks pretty sloppy anyway -- at the very least I would imagine one could prove the number of possible sets of invariants a_{k,m} has a logarithm which is only o(n^3).) Well, the number N(p,n) of groups of order p^n is huge. As I recall, log_p N(p,n) is of the form (2/27)n^3 + smaller terms. In particular, for n sufficiently large, this is greater than the number of different sets of invariants listed above, and so by the pigeonhole principle there are two non-isomorphic groups with the same collection of invariants. More generally, the rule of thumb is, no matter what list of things to measure you prescribe in advance, you can find two finite 2-groups for which these measurements come out equal, but the groups are not isomorphic. I wouldn't dignify this with the title "conjecture", but it's certainly been my experience. >Can you transform a given finite group by redefining operations on >the elements in some way so that the number of elements of different >orders remains the same, but the group is not isomorphic to the >original group? Well, I don't see how this is different from your first question exactly, but if I understand the spirit of your question, I would suggest you look up group extensions, and the use of cocycles. This would let you create new groups which share a normal subgroup N and quotient group G/N with the original group, and even the action of the latter on the former, but which are just a little different from G. dave