Exercise 4.14

Proof. Suppose $|a|=m,|b|=n,m\wedge n=1$.

$\text{∙}$ If ${(\mathit{ab})}^k=e$, then $a^k=b^{-k}$, so $a^{\mathit{kn}}=b^{-\mathit{kn}}={(b^n)}^{-k}=e$, thus $m\mid \mathit{kn}$, with $m\wedge n=1$, therefore $m\mid k$.

Similarly, $b^{\mathit{km}}=a^{-\mathit{km}}={(a^m)}^{-k}=e$, thus $n\mid \mathit{km},n\wedge m=1$ : $n\mid k$.

As $n\mid k,m\mid k,n\wedge m=1$, we conclude $\mathit{nm}\mid k$.

$\text{∙}$ Conversely, if $\mathit{nm}\mid k,k=\mathit{qnm},q\in \mathbb{Z}$, so ${(\mathit{ab})}^k=a^k{b}^k={(a^m)}^{\mathit{qn}}{(b^n)}^{\mathit{qm}}=e$.

This proves $|\mathit{ab}|=\mathit{nm}$. □