# Is my solution correct? (If $S_\Omega \cong S_\Delta$ then $|\Omega|=|\Delta|$)

I've tried to solve the following exercise from Dummit & Foote's Abstract Algebra text (p. 151, Exercise 8). Here $\Omega$ is an infinite set, $D \leq S_\Omega$ is the subgroup of permutations which move only a finite number of points, and $A \leq D$ is the subgroup of permutations which act as even permutations on the set of points they move:

Under the notation of the preceding two exercises, prove that $|D|=|A|=|\Omega|$. Deduce that $$\text{if } S_\Omega \cong S_\Delta \text{ then } |\Omega|=|\Delta|.$$ [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 $|\Omega|$.]

My attempt:

First, since $A \leq D$, we have $|A| \leq |D|$. Fix two elements $x_1,x_2 \in \Omega$, and define $f:\Omega \setminus \{x_1,x_2\} \to A$ by $f(x)=(x_1 \; x_2 \;x)$. Since $f$ is injective we find $|\Omega \setminus \{x_1,x_2\}|=|\Omega| \leq |A|$. It remains to prove $|\Omega| \leq |D|$.

Let $p(n)$ be the partition number of $n$, and for all $n \geq 2$ let $D_n \subseteq D$ be the subset of all permutations $\sigma$ which move exactly $n$ points. Also let $\mathfrak{C}_n$ be a disjoint union of $p(n)$ copies of $\Omega^n=\Omega \times \Omega \times \cdots \Omega$. We define a mapping $\coprod \mathfrak{C}_n \to D \setminus\{1\}$ according to the following rule: For each $n \geq 2$ we name the possible partitions of $n$ by $\pi_1(n),\pi_2(n),\pi_{p(n)}(n)$. If $\mathbf{x}$ in the domain comes from $\mathfrak{C}_n$, and within it from the $k$th copy of $\Omega^n$, we rearrange the parenthesis around the elements of $\mathbf{x}$ so that they create a permutation having $\pi_k(n)$ as its cycle type in the obvious way. It can be seen that this map is a surjection from a set of cardinality $|\Omega|$ to a set of cardinality $|D|$.

We thus have $|D|=|A|=|\Omega|$.

For the last part, suppose $\varphi:S_\Omega \to S_\Delta$ is an isomorphism. We define $A=\{1 \neq H \trianglelefteq S_\Omega\},B=\{1 \neq K \trianglelefteq S_\Delta\}$, and claim that $\varphi:A \to B$ is an injection:

Since $\varphi^{-1}$ is also an isomorphism, for any $H \in A$, we have that the pre-image $(\varphi^{-1})^{-1}(H)=\varphi(H)$ is a normal subgroup of $B$. It is also obviously nontrivial. If $\varphi(H_1)=\varphi(H_2)$ we can apply $\varphi^{-1}$ to get $H_1=H_2$ to get infectivity, and any subgroup $K \in B$ can be written as $K=\varphi(\varphi^{-1}(K))$, where $\varphi^{-1}(K)$ is a nontrivial normal subgroup of $S_\Omega$.

According to preceding exercise (in which we have shown that the "infinite alternating group" is the unique minimal nontrivial normal subgroup of $S_\Omega$) we have that

\begin{equation} A_\Omega=\bigcap_{1 \neq H \trianglelefteq S_\Omega} H \end{equation}

Applying $\varphi$ to this gives

\begin{equation} \varphi[A_\Omega]=\bigcap_{1 \neq H \trianglelefteq S_\Omega} \varphi(H)=\bigcap_{1 \neq \varphi(H) \trianglelefteq S_\Delta} \varphi(H)=A_\Delta. \end{equation}

Restricting $\varphi$ to $A_\Omega$, this shows $|A_\Omega|=|A_\Delta|$, and thus $|\Omega|=|\Delta|$ as required.

Is my solution correct? If not, please help me fix it.

Thanks!