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.

  • 1
    $\begingroup$ Just out of curiosity, what pair $\sigma,\tau\in S_3$ did you try in your ques for an example with $\sigma^2\tau^2\notin\langle\sigma\tau\rangle$? (I assume they did not commute, and were not both of order $2$.) $\endgroup$ – bof Jan 9 at 23:15

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 $a^2b^2\not\in H$


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 $a^2 b^2 \not\in H$.


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.

  • $\begingroup$ Failure of normality does not imply that for all $u \in G$, $\nu \in H$ we have $u\nu u^{-1} \not\in H$. $\endgroup$ – Lee Mosher Jan 9 at 23:04
  • $\begingroup$ Yes. That may be not clear enough in my post: I proved that if the property held for every pair $(a,b)$, then the subgroup was normal. I think that in this case you answer « False ». $\endgroup$ – Mindlack Jan 9 at 23:09
  • $\begingroup$ +1, to me this is just best possible. $\endgroup$ – Andreas Caranti Jan 10 at 11:23

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