How to prove that an even permutation of $A_n$ is a square of another permutation from $S_n$?

Question

I am trying to go through a proof which contains a statement that an even permutation from $A_n$ is a square of another permutation from $S_n$. My basic ideas are like this:

Suppose an even permutation is $y$. As the signature of an even permutation is $+1$, we are able to write it as a product of $t_{1}t_{2}t_{3}t_{4}...t_{2r-1}t_{2r}$ with each $t$ standing for a two cycle. Then, consider $t_{1}t_{2}$, it can be written as $(ij)(kl)$ if $t_{1}$ and $t_{2}$ are distinct, or as $(ij)(jk)$ if they are not distinct. However, $(ij)(kl) = (ikjl)^2$ and $(ij)(jk) = (ikj)^2$, which means each of the $t_1t_2$, $t_3t_4$,...,$t_{2r-1}t_{2r}$ is a square of another permutation of $S_n$, say $x_i$.

However, how does it imply that $y$ is a square of another permutation? If $y = x_1^2...x_r^2$, does it necessarily mean that $y$ is a square of something? Is $x_i$ commutative in this case? I need some hints, thanks.

Answer 1

Hint. A permutation is the square of another permutation if and only if, its cycle decomposition has an even number of cycles of length $m$ for every even number $m$.

Answer 2

Some more hints:

Begin by studying the cycle structures of squares $\sigma:=\pi\circ\pi$ of arbitrary permutations. For this it is sufficient to find out what happens to a cycle of odd length under squaring, and what happens to a cycle of even length under squaring.

Answer 3

That's not remotely true: the density of squares in $S_n$ goes to zero as $n\to +\infty$ (like $\frac{C}{\sqrt{n}}$, actually), while your claim would imply that the $\liminf$ of such density is at least $\frac{1}{2}$. Indeed, if we consider $$\sigma = (1\,2)(3\,4\,5\,6) \in A_6 $$ this is not the square of anything in $S_6$. A square in $S_6$ has the property that the cycles with a fixed even length appear with an even multiplicity.