Exercise 68.2

Proof of $(a)$. For $1\ne g\in G_1,1\ne h\in G_2$, $\mathit{gh}\ne \mathit{hg}$ since otherwise we would have distinct reduced word representations of the same element of $G$. □

Proof of $(b)$. If $x\in G$ has even length, then without loss of generality, $x$ starts with an element in $G_1$ and ends with one in $G_2$, and so we cannot reduce $x^n$ to the identity, and $x$ has infinite order. If $x\in G$ has odd length, then without loss of generality, $x=\mathit{gh}{g}^{\prime }$ for $g,g^{\prime }\in G_1$, $h\in G$. Then, $g^{-1}\mathit{xg}=h{g}^{\prime }g$ has shorter length, for $g^{\prime }g$ reduces to one element in $G_1$. □

Proof of $(c)$. Suppose $x\in G$ has finite order. Then, by $(b)$ it must have odd length $2k+1$. We proceed by induction on $k$. For $k=0$, we see the length is $1$, and so $x\in G_i$ for some $i$, and has finite order in $G_i$. Now suppose $k>0$. Since $x$ has odd length, $x=g^{-1}\mathit{yg}$ for $g,y\in G$, $y$ of shorter length. $y$ has odd length for if not, $y$ has infinite order by $(b)$, and so $x$ also has infinite order since $x^n=g^{-1}{y}^{n}g$, which is a contradiction. Since $y$ is of finite order by the fact $g{x}^n{g}^{-1}=y^n$, $y$ is either equal to an element of $G_i$ with finite order or conjugate to one by inductive hypothesis. If the latter is true, $x=g^{-1}\mathit{yg}=g^{-1}{h}^{-1}\mathit{zhg}={(\mathit{hg})}^{-1}\mathit{zhg}$ for $h\in G$, and $z\in G_i$ having finite order for some $i$. $x$ is therefore conjugate to a finite order element of $G_i$. □