Every element of order 2 in Alt(n) is the square of an element of order 4 in Sym(n)

Prove that every element of order 2 in $A_n$ is the square of an element of order 4 in $S_n$.

We know that every element of order 2 in $S_n$ (hence also $A_n$) is a product of commuting 2-cycles by a previous theorem. Write $\sigma \in A_n$ of order 2 as $\sigma = (a_{1,1}\ a_{1,2})(b_{1,1}\ b_{1,2})(a_{2,1}\ a_{2,2})(b_{2,1}\ b_{2,2}) \cdots (a_{2k,1}\ a_{2k,2})(b_{2k,1}\ b_{2k,2})$. Note that the number of 2-cycles in the decomposition of $\sigma$ is even since $\sigma \in A_n$.

Note that $(a_{i,1}\ b_{i,1}\ a_{i,2}\ b_{i,2})^2 = (a_{i,1}\ a_{i,2})(b_{i,1}\ b_{i,2})$. Thus $\sigma = (a_{1,1}\ b_{1,1}\ a_{1,2}\ b_{1,2})^2 (a_{2,1}\ b_{2,1}\ a_{2,2}\ b_{2,2})^2 \cdots (a_{2k,1}\ b_{2k,1}\ a_{2k,2}\ b_{2k,2})^2$ $= \left( (a_{1,1}\ b_{1,1}\ a_{1,2}\ b_{1,2})(a_{2,1}\ b_{2,1}\ a_{2,2}\ b_{2,2}) \cdots (a_{2k,1}\ b_{2k,1}\ a_{2k,2}\ b_{2k,2}) \right)^2$ using a previous theorem, and the element $(a_{1,1}\ b_{1,1}\ a_{1,2}\ b_{1,2})(a_{2,1}\ b_{2,1}\ a_{2,2}\ b_{2,2}) \cdots (a_{2k,1}\ b_{2k,1}\ a_{2k,2}\ b_{2k,2})$ has order 4 in $S_n$ by §1.3 #15.