I'm reading "Contemporary Abstract Algebra," by Gallian.
This is inspired by Exercise 5.73 and Exercise 5.74 ibid. I have a preference for answers using only the tools available in the textbook so far; lemmas may be established, though, and are actually welcome.
Question: For which $n\in \Bbb N$ is $$H_n:=\{\alpha^2\mid \alpha\in S_n\}\cong A_n?\tag{$\mathcal{H}_n$}$$ Here $S_n$ is the symmetric group of order $n!$ and $A_n$ is the alternating group of order $\frac{n!}{2}$.
Thoughts.
The OEIS isn't helpful so far. This is probably because the only data I have is that $(\mathcal{H}_n)$ is true for $n=4$ and $n=5$ (by Exercise 5.73) but not $n=6$ (by Exercise 5.74).
This seems like a natural question to ask, so I doubt it's original here.
I think $(\mathcal{H}_1)$ and $(\mathcal{H}_2)$ are true by inspection but $(\mathcal{H}_3)$ is a case where I just have to shut up and calculate.
For $n\ge 7$, perhaps the lack of an isomorphism is inherited from the falsehood of $(\mathcal{H}_6)$.
Please help :)
Since $H_n\subseteq A_n$, if they have the same cardinality they must be equal. So, $H_n\cong A_n$ iff $H_n=A_n$. Now note that the square of an odd cycle is a cycle of the same length, and the square of an even cycle is a product of two disjoint cycles of half the length. In particular, even cycles in a square must come in pairs of the same length (since they can only come from the square of an even cycle). So, a product of two disjoint cycles of different even lengths is an element of $A_n$ which cannot be a square. Such a product exists as long as $n\geq 6$, so $H_n\neq A_n$ for any $n\geq 6$.