Exercise 1.3.13 ($\sigma$ has order $2$ in $S_n$ if and only if $\sigma$ is a product of commuting $2$-cycle)

Proof. Suppose that $\sigma \in S_n$ is a non void product of commuting transpositions:

Since the $2$-cycles $(a_1{a}_2),(a_3{a}_4),\cdots (,a_{2n-1}{a}_{2n})$ commute,

So $\sigma \ne e$ and ${\sigma }^2=e$, and we obtain $|\sigma |=2$.

Conversely, let $\sigma$ be a permutation of order $2$ in $S_n$, so that $\sigma \ne e$ and ${\sigma }^2=e$. Let $\sigma ={\tau }_1{\tau }_2\dots ,{\tau }_n$ be the cycle decomposition of $\sigma$. Then ${\tau }_1,{\tau }_2,\dots {\tau }_n$ commute. It remains to show that ${\tau }_1,{\tau }_2,\dots {\tau }_n$ are transpositions.

Write ${\tau }_1=(a_1{a}_2\dots a_m)$. Suppose for the sake of contradiction that $m>2$. Then $a_1\ne a_3$ and $\sigma (a_1)=\tau (a_1)=a_2$, so

Since ${\sigma }^2=e$, this is a contraction, which shows that $m=2$, so ${\tau }_1$ is a transposition, and similarly ${\tau }_2,\dots ,{\tau }_n$ are transpositions.

In conclusion, an element has order $2$ in $S_n$ if and only if its cycle decomposition is a product of commuting $2$-cycle. □