[Math] If a subgroup of a symmetric group has an odd permutation, then it has a subgroup of index 2.

Question

I want to show that for $n\geq 2$ and $H\leq S_n$ if $H$ contains an odd permutation then it necessarily has a subgroup of index 2. I am not sure how to start, if it has an odd permutation then it might not necessarily contain a transposition, perhaps it is possible to consider the group generated by odd permutation, but it can have many possible orders.

Answer 1

Consider the restriction of $\operatorname{sign}$ to $H$. Since $H$ has an odd permutation, this restriction is still surjective. Hence, $\ker( \operatorname{sign}\mid H)$ has index $2$ in $H$.