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