1. Nov 27, 2007

### MathsManiac

Hi,

I recall being told in an algebra course in college that there exist groups with matching order tables and that are nonetheless not isomorphic. That is, if you list out the orders of all the elements in one group and all the orders of the elements in the other, the lists are "the same", and yet you can't match them up to yield an isomorphism. I vaguely recall being told that the smallest example of a pair of non-isomorphic groups like this consists of two groups of order 60. I thought also that there was a name for such groups. (i.e. an adjective that fills the gap in the sentence: "The fact that two groups are ***** is a necessary but not sufficient condition for them to be isomorphic," with the sneaky pair of groups I'm looking for being the smallest counter-example to the sufficiency.

Are my recollections correct? If so, can anyone give me the details of this pair of groups (and the elusive adjective, the non-recollection of which is preventing me from finding the info I want on the web, I suspect)?

2. Nov 29, 2007

### Chris Hillman

Just to clarify: are you looking for two nonisomorphic groups of order n, each one element of order one (duh!), $n_2$ elements of order two, $n_3$ elements of order three, and so on?

3. Nov 29, 2007

### mathwonk

cycles?

4. Nov 29, 2007

### Chris Hillman

Just noticed this bit:

MathsManiac, is your real question: "what is an example of a property which is neccessary but not sufficient for two finite groups to be isomorphic?" If so, there are much simpler examples of such properties than how many elements have each possible order!

Mathwonk, you doubtless know how to solve what I thought was the problem in my Post #2, but I intended to try to get the OP to figure it out with some hints!

Unless MathsManiac's real question concerns permutation groups and the answer is "Polya cycle index".

Last edited: Nov 29, 2007
5. Nov 30, 2007

### MathsManiac

Yes.

6. Nov 30, 2007

### MathsManiac

No. Sorry if this muddied the water. It was just also irritating me that I couldn't remember what the property I was describing was called.

I'm in a situation where other people need to be convinced that establishing that the shared property I described (and which you clarified correctly in your first reply) is not sufficient to establish isomorphism.

For example, I ask somebody to prove that two groups are isomorphic. They check the order of each element in each group. The orders match (in the sense described). They say "Therefore, the groups are isomorphic" (without actually giving the isomorphism). Now, even if these particular groups happen to be isomorphic, I wish to point out that the reasoning is flawed, because this in principle is not sufficient to establish isomorphism. I would like to be able to point to an actual counterexample to prove my point.