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

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.

• Just take $H\cap A_n$. By your hypothesis, $H\cap A_n\neq H$. So what could $[H:H\cap A_n]$ be? – Prahlad Vaidyanathan Nov 18 '13 at 12:03

1 Answer

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 $H$.

• This is essentially a fancy way to express the comment by Prahlad Vaidyanathan. – lhf Nov 18 '13 at 12:21
• Why does $\ker(\mathrm{sign}|_H)$ have index 2 in $H$? – Jimmy R Nov 18 '13 at 14:58
• @JimmyR because $H/\ker \cong \{\pm 1\}$. – lhf Nov 18 '13 at 15:00