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

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.

• 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 $2$.) – 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.

• Failure of normality does not imply that for all $u \in G$, $\nu \in H$ we have $u\nu u^{-1} \not\in H$. – Lee Mosher Jan 9 at 23:04
• 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 ». – Mindlack Jan 9 at 23:09
• +1, to me this is just best possible. – Andreas Caranti Jan 10 at 11:23