Let $H = \{a^2 : a \in S_6\}$. Prove that $H \neq A_6$.

Question

Let $H = \{a^2 : a \in S_6\}$. Prove that $H \neq A_6$.

So far every $a^2$ in $H$ that I could think of was an even permutation in $S_6$, and therefore also in $A_6$. I'm not sure how to go about this proof. Any ideas?

Answer 1

The question is not whether $a^2\in A_6$ for every $a\in S_6$. The question is if the sets $H$ and $A_6$ are equal. In order for them to be the same set, not only does every $a^2$ have to be in $A_6$, but also every element of $A_6$ must be in $H$, i.e. every even permutation is expressible as a square.

If you're familiar with cycle types ("disjoint cycle representation" or other names), you can pick one permutation of each cycle type and square it, record the cycle types of the resulting squares, then check to see if there's any cycle type that cannot be the square of a permutation.