This fact that finite alternating groups are simple can be extended to a result about an infinite group. Let $G$ be the subgroup of the group of permutations on a countably infinite set $M$ (which we may take to be the set of natural numbers for concreteness) which is generated by cycles of length $3$ . Note that any since every element of this group is a product of a finite number of cycles, the permutations of $G$ are such that only a finite number of elements of our set are not mapped to themselves by a given permutation.

We will now show that $G$ is simple. Suppose that $\pi$ is an element of $G$ other than the identity. Let $m$ be the set of all $x$ such that $\pi (x) \neq x$ . By our previous comment, $m$ is finite. Consider the restriction $\pi_m$ of $\pi$ to $m$ . By the theorem of the parent entry, the subgroup of $A_m$ generated by the conjugates of $\pi_m$ is the whole of $A_m$ . In particular, this means that there exists a cycle of order $3$ in $A_m$ which can be expressed as a product of $\pi_m$ and its conjugates. Hence the subgroup of $G$ generated by conjugates of $\pi$ contains a cycle of length three as well. However, every cycle of order $3$ is conjugate to every other cycle of order $3$ so, in fact, the subgroup of $G$ generated by the conjugates of $\pi$ is the whole of $G$ . Hence, the only normal subgroups of $G$ are the group consisting of solely the identity element and the whole of $G$ , so $G$ is a simple group.