Suppose $a, b\in G$ such that $\lvert a\rvert$ is odd and $aba^{-1}=b^{-1}$. Show that $b^2=e$. 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.

We have
$$\begin{align}
aba^{-1}=b^{-1}&\implies ab=b^{-1}a=x\\
&\implies (ab) \cdot (b^{-1}a)=x^2 \\
&\implies a^2=x^2.
\end{align}$$
Now:
$$\begin{align} x^{2m+1}&= (x^2)^m \cdot x\\ &= (a^2)^m \cdot (ab)\\ &= a^{2m+1}\cdot b\\ &=b.\end{align}$$
and
$$\begin{align} x^{2m+1}=b &\implies (x^{2m+1})^2=b^2 \\ &\implies (x^2)^{2m+1}=b^2 \\ &\implies (a^2)^{2m+1}=b^2 = (a^{2m+1})^2\\ & \implies e=b^2. \end{align}$$
