Consider the group of all permutations on the set of positive integers, and its subgroup of permutations with finite support, i.e. those permutations of such that there is such that for all . Let’s call this group .

Because of Cayley’s Theorem it turns out that every finite group can be embedded into . I strongly believe that is the “smallest” such group, that is I think that the following statement is true:

**(S)** If is any group such that every finite group can be embedded into then contains a subgroup isomorphic to .

Statement **(S) **is one of the things that I think have a natural and easy proof — but I am unable to find it. Any help would be appreciated.

Things are less clear to me when we turn the question “upside down”: Consider the class of groups such that every for every finite group there is a surjective group homomorphism . Is there a group belonging to the class $latex \mathcal{C}$ such that for every there is a surjective group homomorphism ? I suspect that would be some quotient of the free group on …

The direct sum of all finite symmetric groups also has the property that all finite groups can be embedded into it, and all elements only have finitely many conjugates. In your Sym_{fin}(N) every element has infinitely many conjugates, so I don’t think that group can be a subgroup of mine.

For your other problem, I think the group must be the free group with countably infinite many generators itself.

Excellent point – thanks! It looks like the statement I was convinced of is false…

I was wrong with the second problem. A group that maps onto every finite group doesn’t need to be free at all. There is such a group that has no elements of infinite order, so does not map onto the free group. It is the direct sum of representatives of all isomorphism classes of finite groups. (That group also works for the first problem.)

I think in both cases there is no universal group.

But if there is, you are most likely right expecting it to be some quotient of the free group on N. That description fits to every countably generated group.

Many thanks for your thoughts. – When I was (mistakingly) thinking about the “smallest” part of my blog post, another quite natural question came up. Are there groups such that each embeds into the other one, but are non-isomorphic? Maybe your examples can be useful, but I’m not sure yet.

Take as G the direct sum of symmetric groups of degree 2, 3, 4… and as H the same without degree 2. There is an easy embedding G->H and an obvious embedding H->G. The transposition in the first component of G has no conjugates, while all nontrivial elements in H have conjugates, so they are not isomorphic.

Another, maybe more drastic example is that the free group of any countable rank embeds into the free group of rank two.

Nice example, thanks!

Statement (S) is false. Consider the the direct sum of Sym(n) for all n. First, this group clearly has an embedding of any finite group by Cayley’s theorem, but can enjoy no subgroup of the form Sym(N,fin), since the latter is generated by cyclic elements of order n. The image of any such cyclic subgroup must be contained in some finite direct sum. Let the maximal index of that partial sum be n’. Note that n<n'!, so that the image of the cyclic subgroup will commute with any cyclic subgroup of order n'!+1. These relations will not hold for Sym(N,fin), so Sym(N,fin) cannot embed into the direct sum of Sym(n).

I hope that helps.