Can every countable locally-finite group be embedded into $S_\infty$?

Question

Let's define $S_\infty$ as the group of all permutations of $\mathbb{N}$ with finite support.

It is not hard to see, that every finite group can be embedded in $S_\infty$. That is because any finite group $G$ can be embedded into $S_{|G|}$ by Cayley Theorem and $S_{n} \cong \{\sigma \in S_\infty|\sigma(m)=m \forall m > n\}$.

On the other hand, all subgroups of $S_\infty$ are locally-finite, because $S_\infty$ is locally-finite itself. Indeed, if we look at $\sigma_1, ... , \sigma_n \subset S_\infty$ with respective finite supports $S_1, ... , S_n$, then $\langle \sigma_1, ... , \sigma_n \rangle \leq S_{|S_1 \cup ... \cup S_n|}$.

I wonder whether every countable locally-finite group can be embedded into $S_\infty$. The problem is, that even if they are, that statement can not be proved using the construction from Cayley theorem, because permutations arising from action of infinite group upon itself by left multiplications do not generally have finite support.

Answer 1

I don't think that the Prüfer $p$-group $C_{p^\infty}$ for a prime $p$, which has presentation $$\left\langle x_i\ (i \in {\mathbb Z_{> 0}}) \mid x_1^p=1,\,x_i^p = x_{i-1}\ (i > 1)\right\rangle,$$ can be embedded into $S_{\infty}$.

Suppose, for a contradiction that there were such an embedding. Then, since $x_i$ has order $p^i$, its image must have at least one orbit of length $p^i$. But then the images of $x_j$ must move at least $p^i$ points for all $j \le i$.

Since this is true for all $i>0$, the images of the $x_i$ in an embedding cannot have finite support, contradiction.

Answer 2

Derek Holt's answer (+1) surprised me since this question's construction is similar (but simpler) than Philip Hall's construction.

In the question, we take the union of symmetric groups using a standard embedding: $S_n \hookrightarrow S_{n+1}$ that takes a permutation of $\{1,2,\ldots,n\}$ to the same permutation of $\{1,2,\ldots,n+1\}$ by defining $n+1$ to be a fixed point.

In Hall's construction, we take the union of symmetric groups using another standard embedding: $S_n \hookrightarrow S_{n!}$ by letting an element of the domain act as a permutation of the domain, the so-called regular representation.

Since both were unions of finite symmetric groups it seemed plausible that both would work (Hall's construction achieves a nicer property for finite groups -- isomorphic finite subgroups are conjugate -- but we didn't need that extra property). However, Derek Holt's answer points out a critical difference:

The number of moved points is unchanged by the embedding $S_n \hookrightarrow S_{n+1}$, but it grows quite a bit in $S_n \hookrightarrow S_{n!}$. However, the growth is very "regular". A $p$-cycle becomes $n!/p$ disjoint $p$-cycles -- this is plenty of room for the $p$-th power of an element of order $p^2$, so locally cyclic groups have no problem embedding now.