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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.

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  • $\begingroup$ Since $S_n^2$ is clearly invariant under conjugation, it is a group exactly when it is equal to $A_n$, provided $n\geq5$. $\endgroup$ – Mariano Suárez-Álvarez Jul 24 '15 at 8:38
  • $\begingroup$ Or more simply: It's not too difficult to see that $A_n$ is generated by $3$-cycles, and any $3$-cycle is a square. I think this is a good exercise for the learner, so I'll refrain from giving an answer, but also note that $(1\,2\,3\,4\,5\,6) \in S_6$ is not a square. $\endgroup$ – John Brevik Jul 24 '15 at 8:42
  • $\begingroup$ But $(1 2 3 4 5 6)\notin A_6$. So what do you want to tell us with that fact? $\endgroup$ – principal-ideal-domain Jul 24 '15 at 8:45
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    $\begingroup$ Sorry -- hasty post on my part. I meant $(1\,2\,3\,4)\,(5\,6)$. $\endgroup$ – John Brevik Jul 24 '15 at 9:05
  • $\begingroup$ More generally and element of $A_n$ containing an even length cycle of length more than $n/2$ cannot be a square, and they exist for all $n \ge 6$. $\endgroup$ – Derek Holt Jul 24 '15 at 9:25
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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 $S_n$ for $n\geq 6$.

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