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

Question

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., $b^2=e$.$\square$

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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.

Answer 1

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. $b^2=e$.

Answer 2

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

$$\begin{align} x^{2m+1}=b &\implies (x^{2m+1})^2=b^2 \\ &\implies (x^2)^{2m+1}=b^2 \\ &\implies (a^2)^{2m+1}=b^2 = (a^{2m+1})^2\\ & \implies e=b^2. \end{align}$$