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Let $G$ be a closed subgroup of $S_\infty$ and let $f:G\rightarrow S_\infty$ be a continuous surjective homomorphism. Under which conditions $f$ has a right split, i.e. there exists some $g:S_\infty\rightarrow G$ such that $g$ is a continuous homomorphism and $fg=id_{S_\infty}$?

Added: Alternatively, can you find a closed subgroup $G$ of $S_\infty$ such that for all $f:G\rightarrow S_\infty$, $f$ does not right split?

According to Wikipedia, $f:G\rightarrow S_\infty$ right splits iff $G$ is the semi-direct product of $ker(f)$ and $S_\infty$. So, the question is equivalent to finding some closed subgroup $G$ of $S_\infty$ such that $G$ is not the semi-direct product of any of its normal subgroups and $S_\infty$.

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  • $\begingroup$ This sounds a bit general to me to expect any useful answer. Plenty of closed subgroups of $S_\infty$ admit such homomorphism $f$. $\endgroup$
    – YCor
    May 11, 2016 at 22:42
  • $\begingroup$ @YCor. This is not my area, so I am not sure. Looking around on the web, I found for instance that if $ker(f)$ is a complete group, i.e. $ker(f)$ has no outer automorphism and is centerless, then $f$ splits. It is true that $S_\infty$ is centerless and has no outer automorphisms. Does this imply the same is true for $ker(f)$? $\endgroup$
    – Ioannis Souldatos
    May 11, 2016 at 22:53
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    $\begingroup$ No, it's not true. The result you mention is true in full generality. An example of a closed subgroup of $\mathrm{Sym}(\mathbf{Z})$ is the subgroup of odd permutations (= satisfying $f(-n)=-f(n)$ for all $n$. This naturally maps to $S_\infty$ with kernel isomorphic to the abelian group $(\mathbf{Z}/2\mathbf{Z})^\mathbf{N}$, which has itself as a center and has plenty of (outer) automorphisms. Of course in this example, the extension is split. $\endgroup$
    – YCor
    May 11, 2016 at 23:04
  • $\begingroup$ Then do you think there is a counterexample? I.e. some $f$ as above that does not split? $\endgroup$
    – Ioannis Souldatos
    May 11, 2016 at 23:14
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    $\begingroup$ Here's a counterexample (non-split $f$). Let $G=S_\infty$ act on $\mathbf{N}=\{0,1,\dots\}$ in the standard way. Then the amalgam $H=G_0\ast_{G_0\cap G_1} G_1$ is a Polish group with canonical a non-split homomorphism onto $G$ (induced by the standard inclusion $G_0\to G$, $G_1\to G$. Then $H$ has a basis of neighborhoods of identity neighborhoods consisting of open subgroups of countable index (the family $(G_0\cap G_1\cap\dots\cap G_n)_n$, and this implies that it's isomorphic to a closed subgroup of $S_\infty$ (see Proposition 4.1 here: math.univ-lyon1.fr/~melleray/Rosendal.pdf). $\endgroup$
    – YCor
    May 12, 2016 at 14:29

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