Countable subgroups of compact groups - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T08:14:42Z http://mathoverflow.net/feeds/question/3420 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/3420/countable-subgroups-of-compact-groups Countable subgroups of compact groups Konstantin Slutsky 2009-10-30T07:23:42Z 2011-04-27T14:36:59Z <p>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?</p> http://mathoverflow.net/questions/3420/countable-subgroups-of-compact-groups/3421#3421 Answer by Reid Barton for Countable subgroups of compact groups Reid Barton 2009-10-30T07:46:05Z 2009-10-30T07:46:05Z <p>Your questions are related to <a href="http://en.wikipedia.org/wiki/Bohr_compactification" rel="nofollow">Bohr compactification</a>, 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 <a href="http://www.math.wisc.edu/~kunen/gpbohr.ps" rel="nofollow">this paper</a> 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 &ge; 2 and K an infinite countable field.</p> http://mathoverflow.net/questions/3420/countable-subgroups-of-compact-groups/62857#62857 Answer by Ostap Chervak for Countable subgroups of compact groups Ostap Chervak 2011-04-24T20:11:49Z 2011-04-27T14:36:59Z <p>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).</p> <p>Proof(by a contradiction):</p> <p>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.</p> <p>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.</p> <p>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 $U(n)$.</p> <p>Hope this helps.</p> http://mathoverflow.net/questions/3420/countable-subgroups-of-compact-groups/62861#62861 Answer by Alain Valette for Countable subgroups of compact groups Alain Valette 2011-04-24T20:48:33Z 2011-04-24T20:48:33Z <p>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. </p>