# “Infinite alternating group” is unique minimal normal subgroup of “Symmetric group of infinite set”

Let $\Omega$ be an infinite set.
Let $D$ be the subgroup of $S_\Omega$ consisting of permutations which move only a finite number of elements of $\Omega$ and let $A$ be the set of all elements $\sigma \in D$ such that $\sigma$ acts as an even permutation on the (finite) set of points it moves.
Prove that if $H\lhd S_{\Omega}$ and $H\neq1$ then $A\leq H$.

This is the exercise $4.6.7$ page $151$ from Abstract Algebra, Dummit and Foote.

From its previous exercise, $A$ is an infinite simple group.
Note that $H\cap A\lhd A$.
So we have either $H\cap A=1$ or $A$.
If $H\cap A=A$, then we are done.
Assume now that $H\cap A=1$.
If $H$ is finite, then $H$ cannot contain any even permutation. Hence $H=\{1,\alpha\}\cong \Bbb{Z}_2$.
This implies that $H\leq Z(S_{\Omega})=1$ which is a contradiction.
Now the place where I face problem is when $H$ is infinite.

If $H \cap A = 1$, then $[H,A] = 1$ so $H \le C := C_{S_\Omega}(A)$.
Now let $a \in \Omega$ and let $A_a$ be the stabilizer of $a$ in $A$. Since $A_a$ contains $(b,c,d)$ for any $b,c,d, \in \Omega \setminus \{a\}$, $a$ is the unique fixed point of $A_a$.
But $C$ centralizes $A_a$ and so it must permute its fixed point set. So $C$ fixes $a$, and this is true for all $a \in \Omega$, so $C=1$ and hence $H=1$.