Let $S_\omega$ be the group of bijections of the countable ordinal $\omega:=\{0,1,2,\dots\}$ and $Alt_\omega$ be the subgroup of $S_\omega$ consisting of even permutations of $\omega$ (i.e., the compositions of even number of transpositions). By Onofri-Schreier-Ulam Theorem, $Alt_\omega$ is the smallest non-trivial normal subgroup in the permutation group $S_\omega$.

It is known that

$\bullet$ the group $Alt_\omega$ is not isomorphic to a subgroup of a compact topological group;

$\bullet$ the group $S_\omega$ is not isomorphic to a dense subgroup of a non-discrete locally compact topological groups.

These two results motivate the following

Problem 1. Is the group $Alt_\omega$ isomorphic to a dense subgroup of a non-discrete locally compact topological group?

Since the group $Alt_\omega$ is simple, this problem can be reformulated in the following form.

Problem 2. Let $h:Alt_\omega\to G$ be a homomorphism of $Alt_\omega$ to a locally compact topological group $G$. Is the image $h(Alt_\omega)$ discerete?

Remark. By the answer of @YCor to this question, for every homomorphism $h:S_\omega\to G$ to a locally compact topological group $G$ the image $h(S_\omega)$ is discrete.

  • $\begingroup$ What is the definition of $Alt_\omega$, there is a subgroup of even permutations of finite support, is that what you mean? Why don't you ask it for $S^{fin}_\omega$ first? $\endgroup$ Feb 23 '18 at 13:52
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    $\begingroup$ @AndreasThom You are right: $Alt_\omega$ is the subgroup of even permutations (with automatically finite support). I wrote the definition in the third line of the question: the composition of even number of pemutations. Why $Alt_\omega$? Because it is a smallest normal subgroup of $S_\omega$ and the answer for $Alt_\omega$ will imply the answer for all larger subgroups of $S_\omega$, including $S_\omega^{fin}$. I think that the questions for $Alt_\omega$ and $S_\omega^{fin}$ could be equivalent (as $Alt_\omega$ has index 2 in $S_\omega^{fin}$). $\endgroup$ Feb 23 '18 at 14:00
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    $\begingroup$ I cannot understand what is wrong with this question that it has got two downvotes. A question as many other questions. At least it has a natural motivation. What is wrong? Istead of demonstrating the power in downvoting, better suggest a solution! $\endgroup$ Feb 23 '18 at 14:20
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    $\begingroup$ @TarasBanakh the result that $Alt_\omega$ is the minimal nontrivial normal subgroup in $Sym(\omega)$ was first proved by Onofri in 1929, reproved 4 years later by Schreier-Ulam, and owes nothing to Baer (who generalized to uncountable cardinals). math.stackexchange.com/a/2645097/35400 $\endgroup$
    – YCor
    Feb 27 '18 at 22:40
  • $\begingroup$ @YCor Thank you for the info. I made the corresponding changes in the text of OP. $\endgroup$ Feb 27 '18 at 23:01

Corollary 1.5 of this Vissarion Belyaev's paper says that

A group containing an infinite inert residually finite subgroup embeds into a non-discrete locally compact group.

Here, a subgroup $H$ of a group $G$ is called inert if $H\cap g^{-1}Hg$ has a finite index in $H$ for any $g\in G$.

So, in the alternating group $Alt_\omega$, for instance, the (abelian) subgroup generated by the 3-cycles $(1\;2\;3),(4\;5\;6),(7\;8\;9),\dots$ is inert. infinite, and residually finite .

Thus, the answer to Question 1 is yes (and to Question 2 is no).

  • $\begingroup$ Anton, thank you for the answer! Truly speaking, I expected that it will be opposite. $\endgroup$ Feb 27 '18 at 19:30
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    $\begingroup$ The emphasized statement is (essentially) originally due to Schlichting, 1980. At least Schlichting's construction applies here. On the other hand, I think the first explicit dense embeddings of the alternating group into a locally compact group are in Willis' J. Algebra 2007 paper. $\endgroup$
    – YCor
    Feb 27 '18 at 22:37
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    $\begingroup$ Yves, it seems that Belyaev was unaware of Schlichting's work. But this is not just "a construction". This is actually a criterion: a group $G$ is isomorphic to a dense subgroup of a totally disconnected non-discrete locally compact group if and only if $G$ contains an infinite inert residually finite subgroup (I am citing Belyaev's paper).. $\endgroup$ Feb 28 '18 at 13:02
  • $\begingroup$ Taras, you are welcome. The symmetric group and finitary symmetric group are quite different. $\endgroup$ Feb 28 '18 at 13:03

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