I'm reading "Contemporary Abstract Algebra," by Gallian.
This is Exercise 4.30.
Suppose $a$ and $b$ belong to a group, $a$ has odd order, and $aba^{-1}=b^{-1}$. Show that $b^2=e$.
My Attempt:
Let $\lvert a\rvert=2n+1$, $n\in\Bbb N\cup\{0\}$.
We have $$\begin{align}
b&=(b^{-1})^{-1} \\
&=(aba^{-1})^{-1} \\
&=ab^{-1}a^{-1},
\end{align}$$
so, by induction, $b=a(\underbrace{aba^{-1}}_{=b^{-1}})a^{-1}=\dots =a^{2n}ba^{-2n}$, so that
$$\begin{align}
b&=a^{2n}ba^{-2n} \\
&=a^{2n}(ab^{-1}a^{-1})a^{-2n} \\
&=a^{2n+1}b^{-1}a^{-(2n+1)} \\
&=b^{-1},
\end{align}$$
so $b=b^{-1}$, i.e., $b^2=e$.$\square$
Is the above proof okay? I think it is.
However, I'd be interested to see how properties of Baumslag-Solitar groups could be used here, since $aba^{-1}=b^{-1}$ brings them to mind.
Best Answer
Your proof is fine. Here is an alternative one using a bit more group theory.
Let $G$ denote the subgroup generated by $a$ and $b$. Let $H \cong \mathbb{Z}_n$ be the subgroup of $G$ generated by $b$. Since $aba^{-1}=b^{-1}$, it follows that $H$ is normal $G$. The quotient $G/H$ must be isomorphic to $\mathbb{Z}_m$ for some odd $m$, since the image of $a$ under the map $G \rightarrow G/H$ maps to a generator of $G/H$ and $a$ has odd order. The conjugation action of $a$ on $H$ hence gives rise to a homomorphism $$ \mathbb{Z}_m \rightarrow \mathrm{Aut}(H): 1 \mapsto \varphi=( b \mapsto aba^{-1}=b^{-1}) $$
Note that $\varphi$ is either trivial or of order two. Since $m\varphi =0$ for $m$ odd, it follows that $\varphi$ is trivial. Hence $b=b^{-1}$, i.e. $b^2=e$.