## The symmetric group on a set A is determined only by the cardinality of A for infinite sets A

Under the notation of the preceding two exercises (here and here), prove that $|D| = |A| = |\Omega|$. Deduce that if $S_\Omega \cong S_\Delta$ then $|\Omega| = |\Delta|$. [Hint: Use the fact that $D$ is generated by transpositions. You may assume that countable unions and finite direct products of sets of cardinality $|\Omega|$ also have cardinality $|\Omega|$.]

Since $D$ is generated by transpositions, every element of $D$ is a finite product of transpositions. Thus $|D| \leq |\bigcup_{n \in \mathbb{N}} \prod_{i=1}^n \Omega \times \Omega|$. Because countable unions and finite direct products of sets of cardinality $k$ have cardinality $k$, $|D| \leq |\Omega|$.

Now the set $\Omega$ may be well ordered. For every element $i \in \Omega$, there is a permutation $\sigma \in A$ such that $i$ is minimal in $M(\sigma)$ under this order; for example, choose any two elements $j$ and $k$ larger than $i$ and let $\sigma = (i\ j\ k)$. Thus $|A| \geq |\Omega|$.

Thus $|\Omega| \leq |A| \leq |D| \leq |\Omega|$, and we have $|D| = |A| = |\Omega|$.

Now let $\Omega$ and $\Delta$ be sets and suppose $\varphi : S_\Omega \rightarrow S_\Delta$ is an isomorphism. We now prove a lemma.

Lemma: If $\varphi : G \rightarrow H$ is a group isomorphism and $N \leq G$ is a $\leq$-minimal nontrivial normal subgroup, then $\varphi[N] \leq H$ is a $\leq$-minimal nontrivial normal subgroup. Proof: $\varphi[N]$ is a subgroup since $\varphi$ is a homomorphism and is normal since $\varphi$ is surjective. Since $|\varphi[N]| = |N|$, $\varphi[N]$ is nontrivial. Now suppose $K \leq H$ is a nontrivial normal subgroup. $\varphi^{-1} : H \rightarrow G$ is a group isomorphism, so that $\varphi^{-1}[K] \leq G$ is a nontrivial normal subgroup, hence $N \leq \varphi^{-1}[K]$. Thus $\varphi[N] \leq K$, so that $\varphi[N] \leq H$ is in fact a $\leq$-minimal nontrivial normal subgroup. $\square$

Now $A_\Omega \leq S_\Omega$ is a minimal nontrivial normal subgroup, so that $\varphi[A_\Omega] \leq S_\Delta$ is a minimal nontrivial normal subgroup. By this previous exercise, then, $\varphi[A_\Omega] = A_\Delta$. Thus $|A_\Omega| = |A_\Delta|$, and by the above argument, $|\Omega| = |\Delta|$.