Determine integers n for which H ={$\alpha \in A_n | ~\alpha^2 = \mathcal{E}\} $ is a subgroup of $A_n$, where $\mathcal{E}$ is the identity .
Attempt : If $\alpha_1, \alpha_2$ are even permutation $\in H$ such that :
${\alpha_1}^2 = \mathcal{E}$ and ${\alpha_2}^2 = \mathcal{E}$
H is a subgroup $=>$ $\alpha_1 {\alpha_2}^{-1} \in H$
$=> [\alpha_1 {\alpha_2}^{-1}]^2 = \mathcal{E}$
$=> \alpha_1 {\alpha_2}^{-1}\alpha_1 {\alpha_2}^{-1}= \mathcal{E}$
$\alpha_1, \alpha_2$ are even permutations $=>{\alpha_2}^{-1}$ is also an even permutation
Hence overall, $=> \alpha_1 {\alpha_2}^{-1}\alpha_1 {\alpha_2}^{-1}$ is an even permutation which is consistent with the requirements of an identity permutation
Now, order of $ \alpha_1 {\alpha_2}^{-1} = 2$ where $\alpha_1, \alpha_2$ $\in A_n$
How do i move on from here?
Thank you
EDIT: order of $ \alpha_1 {\alpha_2}^{-1} = 2$ where $\alpha_1, \alpha_2$ $\in A_n$
But, since ${\alpha_2}^2=\mathcal{E} =>\alpha_2= {\alpha_2}^{-1}$
=> order of $ \alpha_1 \alpha_2 = 2$
where both $\alpha_1, \alpha_2$ $\in A_n$
Now, i must find $n$ such that $A_n$ contains elements of order $2$.
What can i do next?