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$\DeclareMathOperator\Aut{Aut}\DeclareMathOperator\Inn{Inn}$Let $S_\infty$ be the group of all permutations of a countable infinite set (enriched with the product Polish topology). Does there exist a (continuous) group embedding $\phi: S_\infty\to S_\infty$ such that $\phi(G)$ is conjugate to $\phi(H)$ for every pair of (closed) isomorphic groups $G, H\leq S_\infty$?

Note that two subgroups $A, B$ of a group $G$ are conjugate in $G$ if there is $g\in G$ for which $A=gBg^{-1}$.

Additional note: It would be stil of interest for me, if there is an embedding $\phi: S_\infty \to T$ for some Polish group $T$, such that $\phi(G)$ is conjugate to $\phi(H)$ in $T$ for every pair of (closed) isomorphic groups $G, H\leq S_\infty$?

My first guess was to consider $T$ to be the semidirect product of $S_\infty$ with $\Aut(S_\infty)=\Inn(S_\infty)=S_\infty$.

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    $\begingroup$ It seems clear to me from your body, but not from your subject, so I want to make sure: you want a single embedding that makes all pairs of isomorphic subgroups (of the original $S_\infty$) conjugate? $\endgroup$
    – LSpice
    Feb 22, 2022 at 18:28
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    $\begingroup$ Yes - a single embedding which works for all pairs of subgroups. $\endgroup$ Feb 22, 2022 at 19:55
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    $\begingroup$ The question in the title, asking only about some larger group and not requiring continuity, should be answerable (affirmatively) by doing a lot of HNN extensions. $\endgroup$ Feb 22, 2022 at 19:57
  • $\begingroup$ @AndreasBlass I think the natural embedding of $S_\infty$ into the group of permutations of $S_\infty$ works for the version not requiring continuity - the only thing to check is that isomorphic subgroups have the same index. I'm not sure if one can put a natural topology to make this a continuous embedding. $\endgroup$
    – Will Sawin
    Feb 22, 2022 at 21:17
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    $\begingroup$ Isomorphic subgroups don't actually have the same index (the subgroup fixing a point is isomorphic to the whole group). However, embedding them in the permutations of the union of infinitely many copies of $S_\infty$ will do the trick without topology. $\endgroup$
    – Will Sawin
    Feb 28, 2022 at 12:59

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No, there's no such endomorphism. More generally:

Proposition. for every infinite set $X$, there exists no set $Y$ and nontrivial continuous homomorphism $f:S_X\to S_Y$ for which the following assertion holds: for any two isomorphic discrete infinite cyclic subgroups of $S_X$, the images $f(S_X)$ and $f(S_Y)$ are conjugate.

Note: every group endomorphism of $S_\omega$ is continuous (automatic continuity, here essentially due to Dixon-Neumann-Thomas: see Prop 4.3 and Thm 4.4 in Rosendal's survey). [The notation $S_\infty$, which implicitly assumes that every infinite set is countable (Cantor disproved this) is inconvenient.]

Let $c$ be any nontrivial permutation of $X$ consisting only of infinite cycles (at least one) and fixed points. Let $c'$ be any permutation of $X$ with both at least one infinite cycle and finite cycles of all sizes.

Since both $c$ and $c'$ have an infinite cycle, the cyclic subgroups $\langle c\rangle$ and $\langle c'\rangle$ are infinite and discrete (hence closed).

Proposition. For every set $Y$ and homomorphism $f:S_X\to S_Y$, the subgroups $\langle f(c)\rangle$ and $\langle f(c')\rangle$ are not conjugate.

More precisely, $f(c)$ has no finite cycle of size $\ge 2$, while (unless $f$ is the trivial homomorphism) $f(c')$ has finite cycles of unbounded size.

(This immediately implies the first proposition, in a strong sense since the embedding property fails for a given pair of isomorphic closed subgroups.)

Lemma Let $H$ be an open subgroup of $S_X$. Then there exists a finite subset $F$ of $X$ such that $H$ is trapped between the pointwise and global stabilizer of $F$. $\Box$

[This is classical. The main case is when $H$ is transitive, in which case the conclusion is $H=S_X$. The general case easily follows.]

Through $f$, we can view $Y$ as an $S_X$-set. For both statements it enough to assume that $Y$ is a transitive $S_X$-set (through $f$). Hence $Y=S_X/H$ for some open subgroup $H$.

For the statement about $c$ (no nonsingleton finite cycle), we can use the lemma and find an open finite index subgroup $H'$ of $H$, such that $H'$ is the pointwise stabilizer of a finite subset. We can identify $S_X/H'$ with the set of pairwise distinct $n$-tuples in $X^n$ for the product action, for some $n\ge 0$. Indeed we see that $c$ has no nonsingleton finite cycle on $\omega^n$. Since the equivariant surjection $S_X/H'\to S_X/H$ is finite-to-one, we deduce that $c$ has also no nonsingleton finite cycle on $S_X/H$.

For the statement about $c'$ (existence of unbounded finite cycles), we can use the lemma again, now defining $H'$ as finite index overgroup of $H$, such that $H'$ is the global stabilizer of a finite subset $F$. If $F$ is empty, we have $H=H'=S_X$ and the statement is void. Assume now $F$ nonempty. We can identify $S_X/H'$ with the set of $n$-elements subsets of $X$ for some $n\ge 1$. Since $c'$ admits a $kn$-cycle for every $k\ge 2$, we see that $c'$ has a $k$-cycle for its action on $S_X/H'$. Hence $c'$ has a $kk'$-cycle for some $k'\ge 1$ on $S_X/H$.

Using automatic continuity, let me mention, out of curiosity:

Corollary. For $X=Y=\omega$ and $c,c'$ as above, the HNN extension of $S_\omega$ over the isomorphism $\langle c\rangle \to \langle c'\rangle$ mapping $c$ to $c'$, has no faithful action on any countable set.

(Indeed, finding groups of at most continuum cardinal with no faithful action on any countable set has been an open problem long ago, and while various examples are now known, such a construction seems to be a new one. Note that no group topology is involved in the above statement.)


This argument doesn't address general Polish groups as targets. However it suggests that it already makes sense to attempt to make discrete subgroups conjugate, even infinite cyclic ones.

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  • $\begingroup$ PS Uspenskii proved that the group $G$ of self-homeomorphisms of the Hilbert cube $[0,1]^\omega$ is "universal" in the sense that it is Polish and every Polish group is isomorphic to a closed subgroup therein. So the generalized question reduces to a question about continuous homomorphisms $S_\omega\to G$. But I don't know if much can be said about such homomorphisms in general. $\endgroup$
    – YCor
    Mar 1, 2022 at 11:28
  • $\begingroup$ This is nice. For the lemma, a quick proof is to take a minimal set $F$ such that the pointwise stabilizer of $F$ is contained in $H$. Since $H$ is open, $F$ is finite. A short argument shows that $F$ is unique, so $H$ must stabilize $F$ setwise. $\endgroup$ Mar 1, 2022 at 18:56
  • $\begingroup$ Thank you for this nice answer. I am stil interested in the general Polish group as a target (hoping that there is a positive answer). $\endgroup$ Mar 6, 2022 at 16:19

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