# Countable subgroups of compact groups

What is known about countable subgroups of compact groups? More precisely, what countable groups can be embedded into compact groups (I mean just an injective homomorphism, I don't consider any topology on the countable group)? In particular, can one embed S_\infty^{fin} (the group of permutations with finite support) into a compact group? Any simple examples of a countable group that can't be embedded into a compact group?

-

## 3 Answers

Your questions are related to Bohr compactification, a left adjoint to the inclusion of compact (= compact Hausdorff) groups into all topological groups. A discrete group G can be embedded into a compact group iff the natural map from G to its Bohr compactification is an injection. Such groups are called "maximally almost periodic". Take a look at this paper for a more in-depth treatment. An example from that paper of a countable group which cannot be embedded into a compact group is SL(n, K) for n ≥ 2 and K an infinite countable field.

-

As a complement to Reid's answer: a finitely generated group is maximally almost periodic if and only if it is residually finite. Indeed, if a group is residually finite, it embeds into its profinite completion, which is compact. Conversely, if a finitely generated group $G$ embeds into a compact group $K$, then using first that homomorphisms $K\rightarrow U(n)$ separate points of $K$, second that finitely generated linear groups are residually finite (Mal'cev theorem), we conclude that $G$ is residually finite.

-

By the way, $A_\omega ^{fin}$ (group of even permutation of countable set) can't be embedded into compact group (hence $S_\omega^{fin}$ also can't).

Proof(by a contradiction):

It is well-known (proof uses theory of representations) that every compact group is isomorphic to a subgroup of cartesian product of unitary groups, so WLOG we may assume that $A_\omega^{fin}$ is embedded in $\prod U({n_i})$. Now let $f_i$ be a homomorphism $f:A_\omega^{fin} \rightarrow U({n_i})$ such that $\prod f_i$ is an embedding in $\prod U({n_i})$. Obviously Ker $f_i$ can't be trivial for all i, so for some i $Ker(f_i) \neq A_\omega^{fin}$ , but since $A_\omega^{fin}$ is simple $Ker (f_i) =(0)$ and $f_i$ is injective.

So, we've just proved that if simple group can be embedded in compact group it can be embedded in $U(n)$ for some n.

Now, it's easy to finish the proof (there are lot's of methods, actually). For example note that there are infinitely many pairwise commuting different elements in $A_\omega^{fin}$ such that $x^2 = e$ but there are only finitely many such elements in $U(n)$.

Hope this helps.

-
Thanks a lot. This is a very nice example. –  Konstantin Slutsky Apr 27 '11 at 11:31