Let S(A) be the set of all bijective functions from A ---> A. Similarly we define S(B). Then both S(A) and S(B) are groups under function composition.
If S(A) is isomorphic to S(B) , then is it true that |A| = |B| ?
For finite sets this follows immediately. I ve been stuck to proving ( or disapproving) that this result holds.
My strategy was constructing an injective function from A to B, and then by following symmetry such a function will also exist from B to A as well. Then using F. Bernstein's Theorem, we could conclude that both sets are of equal cardinality. However, I was not able to do so. Any help would be appreciated.
Here's an easy self-contained proof. In a group $G$, let $\Xi(G)$ be the set of elements $f$ in $G$ of order 2, such that for every conjugate $g$ of $f$, $fg$ has finite order. The fact that for $X$ infinite, the isomorphism class of $S(X)$ determines $|X|$ immediately follows from:
Proposition. Let $X$ be a set and $f$ and element of order $2$ in $S(X)$. Then $f\in\Xi(S(X))$ if and only if $f$ has finite support. In particular, for $X$ infinite $\Xi(S(X))$ has cardinal $|X|$.
Proof: if $f$ has support of finite cardinal $n$, clearly $fg$ has support of cardinal $\le 2n$ and hence has finite order. Otherwise, there exists an injective sequence $(x_n)_{n\in\mathbf{Z}}$ such that $f(x_{2n})=x_{2n+1}$ for all $n\in\mathbf{Z}$. Let $s$ be the permutation such that $s(x_n)=x_{n+1}$ for all $n\in\mathbf{Z}$ (and say identity elsewhere), and $g=sfs^{-1}$. Then $g(x_{2n+1})=x_{2n+2}$ for all $n\ge 0$. Hence $gf(x_{2n})=x_{2n+2}$ for all $n\ge 0$. So $gf$ has infinite order. $\Box$