Characterisation of the squares of the symmetric group

Solution 1:

Observe that the square of a $k$-cycle is again a $k$-cycle when $k$ is odd, and is the product of two $k/2$ cycles when $k$ is even.

It follows that an element $\sigma\in S_n$ is a perfect square if and only if, for every even value of $k$, the cycle structure for $\sigma$ has an even number of $k$-cycles.

So $S_n^2 = A_n$ for $n\leq 5$, since elements of $A_n$ have cycle structure $(*\;*\;*)$, $(*\;*\;*\;*\;*)$, or $(*\;*)(*\;*)$, all of which are perfect squares.

For $n \geq 6$, there are even permutations of the form $(*\;*)({*}\;{*}\;{*}\;{*})$, which cannot be perfect squares. Thus $S_n^2$ is properly contained in $A_n$. However, $S_n^2$ still contains all 3-cycles, and the $3$-cycles generate $A_n$, so $S_n^2$ is not a subgroup of $A_n$ for $n\geq 6$.