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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?

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3 Answers 3

up vote 17 down vote accepted

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.

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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.

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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.

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

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