If $\operatorname{Sym}_X \simeq \operatorname{Sym}_Y$ then there is bijection from $X \to Y$

Question

If $\operatorname{Sym}_X \simeq \operatorname{Sym}_Y$ then there is bijection from $X \to Y$ ? ,

I proved the other way around, i think i need to build $f$ from $\psi : \operatorname{Sym}_X \to\operatorname{Sym}_Y$.

Answer 1

We have to recover $|X|$ from $\operatorname{Sym}(X)$.

Note that $\operatorname{Sym}(X)$ is finite iff $X$ is finite, and in that case, the cardinality of $\operatorname{Sym}(X)$ uniquely determines that of $X$. So we may assume that $\operatorname{Sym}(X)$ is infinite (equivalently $X$ is infinite). Let $H\subseteq 2^{\operatorname{Sym}(X)}$ consist of all elementary Abelian 2-subgroups in $\operatorname{Sym}(X)$. These are isomorphic to additive groups of vector spaces over the 2-element field, and all of them contain the identity element and pairwise commuting involutions, i.e., possibly infinite products of transpositions with pairwise disjoint support. Let $\kappa$ be the maximum dimension occurring over all these vector spaces (more precisely, we should define it as the supremum of all such dimensions, but it will turn out that there is an actual maximum, as well).

We claim that $\kappa= |X|$. First of all, it is possible to partition $|X|$ into $|X|$ pairs: the transpositions corresponding to these pairs form the basis of a vector space in $H$. So $\kappa\geq |X|$. For the reverse direction, note that in a vector space $V\in H$, for any two $f,g\in V$ and $x\in X$ it is not possible that $x, f(x)$ and $g(x)$ are three different elements, otherwise $f$ and $g$ do not commute. So for all $x\in X$ either $x$ is a fixed point of the whole $V$, or there exists a unique $y\in X$ such that "half" of the elements of $V$ fix $x$ and the other "half" transposes $x$ and $y$. After dropping the fixed points, we have a unique $y$ for all $x$. But then the set of these transpositions $(x~y)$ certainly generates $V$. The cardinality of such a set of transpositions is clearly at most $|X|$, so $\dim(V)\leq |X|$ for all $V\in H$, i.e., $\kappa\leq |X|$.