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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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 $n\geq 6$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.