Example of $a,~b\in G$ such that $ab\in H\leq G$ and $a^2b^2\notin H.$

Question

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.

Answer 1

Consider $S_3$.

Let $a=(1 2 3)$ and $b=(2 3)$. Then $ab=(1 2)$ and $a^2b^2=(1 3 2)$

Let $H=\{1, ab\}$. Then $ab\in H$ but $a^2b^2\not\in H$

Answer 2

Take $G$ to be the free group on $a,b$, whose elements are the reduced words in the alphabet $a,b,a^{-1},b^{-1}$.

Take $H$ to be the cyclic subgroup generated by $ab$.

The non-identity elements of $H$ are the reduced words words $(ab)^n$ for $n \ge 1$ and $(b^{-1}a^{-1})^n$ for $n \ge 1$.

Since the reduced word $a^2 b^2$ does not have that form, it follows that $a^2 b^2 \not\in H$.

Answer 3

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.