Suppose $H$ is a subgroup of $S_n$ but $H$ is not a subgroup of $A_n$. Prove that |$H$ ∩ $A_n$| = $\frac{1}{2}|H|$? i know that|$S_n$|=$n!$ and |$A_n$|= $\frac{n!}{2}$ but i'm not sure where to go from there. Thanks.
 A: Following @Jyrki Lahtonen's hint in the comments, let's define $\operatorname{sgn}\colon S_n\to \{1,-1\}^\times$ by $\operatorname{sgn}(\sigma):=1$ if $\sigma$ in even, and $\operatorname{sgn}(\sigma):=-1$ if $\sigma$ in odd. Since $\operatorname{sgn}(\sigma\tau)=1$ if and only if $\operatorname{sgn}(\sigma)=\operatorname{sgn}(\tau)$, then $\operatorname{sgn}(\sigma\tau)=\operatorname{sgn}(\sigma)\operatorname{sgn}(\tau)$. Therefore, $\operatorname{sgn}$ is a group homomorphism with kernel $\{\sigma\in S_n\mid \operatorname{sgn}(\sigma)=1\}=A_n$. Now, if $H\le S_n$ but $H\nsubseteq A_n$, then $\operatorname{sgn}(H)=\{1,-1\}^\times$ (because $\iota_{S_n}\in H$ and some odd permutation $\tau\in H$), and thence $\operatorname{sgn}_{|H}$ is a surjective homomorphism on $\{1,-1\}^\times$ with kernel $A_n\cap H$. By the First Homomorphism Theorem, $H/A_n\cap H\cong \{1,-1\}^\times$, whence $|H|/|A_n\cap H|=2$ and finally $|A_n\cap H|=|H|/2$.
A: $|H:H\cap A_n|=|HA_n:A_n|$  divides $|S_n:A_n|=2$ and is not $1$. So this index is 2 and one can use Lagrange.
