When is $H^m=\{ h^m \mid h\in H\}$ a subgroup $H$? Let $H$ be a group and $H^m=\{ h^m \mid h\in H\}$.
I know that this is  a subgroup of $H$ when $H$ is abelian.
But I want to know that what happens if $H$ is not abelian.
For which $n$, $H^n$ is a subgroup of $H$ and for which $n$ it is not?
I tried for the nonabelian group $S_3$ and found that $S_3^{2k}$ is a subgroup of $S_3$ for all $k\in \mathbb{Z}$ and $S_3^{2k+1}$ is not a subgroup of $S_3$ for all $k\in \mathbb{Z}$.
But I don't know whether it is true for arbitrary groups or not and I want to prove it if it is correct.
Any help would be appreciated.
 A: For any finite simple group of even order, the squares are not a subgroup.
To see this, note the following:
Lemma:  If the squares from a group $H$ are contained in a subgroup $G$ then $G$ is normal.
Pf:  Let $h,g$ be arbitrary elements of $H,G$ respectively.  We want to show that $hgh^{-1}\in G$.  But $$hgh^{-1}=h^2h^{-1}gh^{-1}gg^{-1}=h^2\left(h^{-1}g\right)^2g^{-1}\in G$$ and we are done.
Thus, $A_5$ in particular is a counterexample
Remark:  For completeness we should note that the squares can't be the entire group.  As the group has even order (by assumption) there is an element of order $2$, hence the squaring map is not injective.  As we are speaking of finite groups here, it follows that the squaring map is not surjective, and we are done.
A: It is certainly true that for every finite group  $H$ there exists $N$ such that whether or not $H^n$ is a subgroup depends only on $n\mod N$ and contain all multiples of $N$. For this $N$ you can take, for example, $|H|$. That is because if $m\equiv n\mod N$, then $H^m=H^n$, and $H^N=\{1\}$.
For infinite groups the set of $n$'s  for which $H^n$ is a subgroup may be empty (say, the free group of rank $\ge 2$).
