TITLE: Characterisation of the squares of the symmetric group
QUESTION [5 upvotes]: I found out that for $n\le 4$ we have $S_n^2=A_n$ with $G^2$ defined by
$$G^2:=\{g^2 \mid g\in G\}$$
for any group $G$. Surely we have $S_n^2\subseteq A_n$ for all $n\in\mathbb N$. Is there a characterisation of the squares of the symmetric group? When is $S_n^2$ a group and what does it look like in general?
Edit: So far we found out, that $S_n^2$ generates $A_n$ because each 3-cycle is a square (since we have $g=g^{-2}$) and $A_n$ is generated by the 3-cycles. So we simplified the original question to the question for which $n$ we have $S_n^2=A_n$ and how to characterise the elements which a missing in case we don't have the equality.

REPLY [4 votes]: 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 $n\geq 6$.