Showing that $A_{\infty}$ is a simple group.

Question

Let $S=\{1,2,\dots\}=\mathbb{N}$, and let $G$ be the subgroup of Sym($S$) such that $$G=\{\sigma\in\text{Sym}(S):\sigma(j)=j\text{ for all but finitely many }j\}.$$

I'd like to show that $A_{\infty},$ the subgroup of $G$ generated by all 3-cycles $(a,b,c)$, is a simple group.

I honestly have no idea how to even begin with this problem, and would really appreciate any help or suggestions. Thank you.

Answer 1

Let $B$ be a normal subgroup of $A_{\infty}$, $B\cap A_n$ is normal, this implies that $B\cap A_n = A_n$ or is trivial for $n>4$ since $A_n$ is simple.

If there exists $n_0>4$ such that $B\cap A_{n_0}= A_{n_0}$, for every $n>n_0$, $B\cap A_n= A_n$ since $A_n$ is simple and $B\cap A_n$ is normal and not trivial. This implies that $B=A_{\infty}$

Answer 2

1. Show that finite alternating groups $A_n$ are simple for $n\geq5$.
2. Show that simplicity is closed under taking direct unions.

The second one is easy. For the first one, you can do an induction on $n$ (where you might want to use the fact that $A_n$ contains $n$ subgroups isomorphic to $A_{n-1}$, namely the point stabilizers).