TITLE: Example of $a,~b\in G$ such that $ab\in H\leq G$ and $a^2b^2\notin H.$
QUESTION [5 upvotes]: Let $G$ be a group and $H$ be a subgroup of $G$. Let also $a,~b\in G$ such that $ab\in H$.
True or false?   $a^2b^2\in H.$

Attempt. I believe the answer is no (i have proved that the statement is true for normal subgroups, but it seems that there is no need to hold for arbitrary subgroups). I was looking for a counterexample in a non abelian group  of small order, such as $S_3$, or $S_4$, but i couldn't find a suitable combination of $H\leq S_n$, $\sigma$ and $\tau\in S_n$ such that $\sigma \tau \in H$ and $\sigma^2 \tau^2 \notin H.$
Thanks in advance for the help.

REPLY [2 votes]: Let $u \in G$, $v \in H$.
Take $a=u$, $b=u^{-1}v$. Then $ab \in H$. 
Moreover, $a^2b^2=uvu^{-1}v$, thus $a^2b^2 \in H \Leftrightarrow uvu^{-1}v \in H \Leftrightarrow uvu^{-1} \in H$. 
Thus if $H$ is not normal, the property does not hold.