Isomorphic Sylow p-subgroups of FSym$(\mathbb{N})$ Let FSym$(\mathbb{N})$ denote the finitary symmetric group on the natural numbers. Are all Sylow p-subgroups of FSym$(\mathbb{N})$ isomorphic (maybe to the iterated wreath product $C_p\wr C_p\wr\dots$ of cyclic $p$-group $C_p$)?
It is well-known that the Sylow $p$-subgroups of $S_{p^k}$, for the positive integer $r$, are all isomorphic to $C_p\wr C_p\wr\dots\wr C_p$ (with $k$ copies of $C_p$) [see for example Proposition 19.10 of the book “A Course in Group Theory” by J.H. Humphreys].
 A: $\DeclareMathOperator\FSym{FSym}\DeclareMathOperator\Sym{Sym}\newcommand\N{\mathbf{N}}$The answer is no, as already mentioned in the comments: a transitive Sylow subgroup, or more generally any subgroup without finite orbit, has a trivial center (clear, since for a nontrivial element, its centralizer preserves its support), while any Sylow with a finite orbit of cardinal $\ge p$ has a nontrivial center (and they indeed exist in $\FSym(X)$ for any infinite $X$).
The main question seems to classify $p$-Sylow subgroups. Let us provide a full reduction to the transitive case.
Proposition 1. Let $P$ be a $p$-Sylow, $(X_i)$ its orbit decomposition, $P_i$ the image of $P$ in $\FSym(X_i)$. Then $P=\bigoplus_i P_i$, and all finite $X_i$ have a $p$-power cardinal.
Indeed, $P\subset\bigoplus_i P_i$ (as elements have finite support), and by maximality, $P=\bigoplus_i P_i$. The second assertion is clear from the finite case.$\Box$
For the converse:
Proposition 2. *Let $X$ be a set with a partition $(X_i)$ such that each finite $X_i$ has a $p$-power cardinal. Let $P_i$ be a transitive $p$-Sylow in $\FSym(X_i)$. Then $P=\bigoplus P_i$ is a $p$-Sylow in $\FSym(X_i)$ iff for each $n<\infty$, the number $n_i$ of $i$ such that $|X_i|=p^n$ is $<p$.
The condition is clearly necessary, using the finite case. Conversely, suppose it holds. Choose $f\notin \bigoplus P_i$ such that $\langle P,f\rangle$ is a $p$-subgroup.  Let $J$ be the set of $i$ such that $f(X_i)\neq X_i$, and $X'=\bigcup_{i\in J}X_i$. Note that $J$ is not empty, and that $X'$ is $\langle P,f\rangle$-invariant.
a) Suppose that $X'$ is finite. The assumptions $n_i<p$ implies that $\prod_{i\in J}P_i$ is a $p$-Sylow in the finite group $\Sym(X')$, so we get a contradiction.
b) So $X'$ is infinite; since $J$ is finite (as the support of $f$ is finite), there is $i\in J$ with $X_i$ infinite. Let $W$ be the (finite) support of $f$ and $W_i=W\cap X_i$. By a classical lemma of B.H. Neumann, there exists $s\in P_i$ such that $s(W_i)\cap W_i=\emptyset$. Then for some $k\ge 1$, some cycle of $f$ visits a point in $X'-X_i$, then $k$ points in $X_i$, and then goes back to $X'-X_i$. By direct verification, the commutator $f'=[s,f]=sfs^{-1}f^{-1}$ contains a $(2k+1)$-cycle whose support meets $X_i$ in a subset of cardinal $2k$. Apply the same with $f'$: find $s'\in P_i$ so that $f''=[s',f']$ contains a $(4k+1)$-cycle. Since $f'$ and $f''$ are both in a $p$-Sylow, we deduce that both $2k+1$ and $4k+1$ are $p$-powers. Since $1<(4k+1)/(2k+1)<2\le p$, we deduce a contradiction. $\Box$
This entirely reduces to understanding transitive Sylow subgroups.
And indeed it's known that all transitive $p$-Sylow are conjugate when $X$ is infinite countable. This is due to Ivanyuta (Sylow p-subgroups of the countable symmetric group. (Russian) Ukrain. Mat. Ž 15 1963 240–249), see also here.
(To complete the picture, one should prove that for $X$ uncountable there's no transitive $p$-Sylow.) So a conjugacy class of $p$-Sylow is determined by some cardinal $n_\omega$ (the number of infinite orbits), and for each $i$, the number $0\le n_i<p$ of orbits of cardinal $p^i$.
Actually it is claimed in both papers that $p$-Sylow subgroups of $\FSym(X)$ are isomorphic iff they're conjugate: this is slightly false: they are isomorphic iff they have the same $n_i$ for each $0<i\le\omega$ (but $n_0$ can vary).
