Exercise 3.5.13 (Every element of order $2$ in $A_n$ is the square of an element of order $4$ in $S_n$)

Proof. By Exercise 1.3.13, an element $\sigma$ of order $2$ in $A_n$ is a product of commuting (disjoint) transpositions. Since $\sigma$ is even, the number of transpositions is even:

Moreover

is the square of a $4$-cycle, so

Since the transpositions are disjoint, these $4$-cycles are disjoint, therefore

is the square of the element $\alpha =(a_1{a}_3{a}_2{a}_4)\cdots (a_{2n+1}{a}_{2n+3}{a}_{2n+2}{a}_{2n+4})\in S_n$. Since $\alpha$ is a product of disjoint $4$-cycles, ${\alpha }^4=()$. But ${\alpha }^2\ne ()$, because ${\alpha }^2(a_1)=a_2\ne a_1$. So $\alpha$ has order $4$.

Every element of order $2$ in $A_n$ is the square of an element of order $4$ in $S_n$. □