TITLE: The set $\{g^2 | g \in G\}$ in a group $G$
QUESTION [16 upvotes]: Let $G$ be a group. Prove or disprove that $H =\{g^2 | g \in G\}$ is a subgroup of $G$.

I tried testing the permutations of $A_4$, however squaring each cycle yielded a cycle in $A_4$ so I'm lacking a counter-example (if there is one). In a nutshell I'm looking for a subgroup such that when you square the permutation cycle, it yields a cycle not in that subgroup.
Or, I could be way off base and figure out that there isn't a counter-example and I need to prove that indeed $H$ is a subgroup of $G$.

REPLY [2 votes]: You can see that if $H$ is a subgroup, then $H$ is also a normal subgroup. Thus a natural way to search for counterexamples is to look at groups with few normal subgroups.
Let $G$ be a finite, nonabelian simple group. Then if $H$ is a subgroup, it must equal $G$, because $g^2 = 1$ for every $g \in G$ implies that $G$ is abelian. If there is an element of order $2$ in $G$, then the map $g \mapsto g^2$ is not injective, thus not surjective, and thus $H$ cannot equal all of $G$.
By Cauchy's theorem, this shows that any finite nonabelian simple group of even order works. So for example, you could pick $G = A_n$ for any $n \geq 5$.
(And by Feit-Thompson, any finite nonabelian simple group works..)