# Is $Alt_\omega$ a dense subgroup of a non-discrete locally compact topological group?

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.

• 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? – Andreas Thom Feb 23 '18 at 13:52
• @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 $S_\omega^{fin}$). – Taras Banakh Feb 23 '18 at 14:00
• 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! – Taras Banakh Feb 23 '18 at 14:20
• @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 – YCor Feb 27 '18 at 22:40
• @YCor Thank you for the info. I made the corresponding changes in the text of OP. – Taras Banakh 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).

• Anton, thank you for the answer! Truly speaking, I expected that it will be opposite. – Taras Banakh Feb 27 '18 at 19:30
• 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. – YCor Feb 27 '18 at 22:37
• 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).. – Anton Klyachko Feb 28 '18 at 13:02
• Taras, you are welcome. The symmetric group and finitary symmetric group are quite different. – Anton Klyachko Feb 28 '18 at 13:03