If $H$ is a subgroup of $S_n$, is it a subgroup of $A_n$? If $H$ is a subgroup of $S_n$, how do I prove that $H$ is a subgroup of $A_n$ if $|H|$ is odd?
Attempt: By contrapositive: if $|H|$ is odd and $H$ is not a subgroup of $A_n$ then I must compare the numbers of odd & even permutations in $H$. How do I do this?
 A: If $H$ is not contained in $A_n$, then $\langle H, A_n\rangle = S_n$, because $[S_n:A_n]=2$.
Since $A_n$ is a normal subgroup, we have
$$
H/ (H \cap A_n) \cong \langle H, A_n\rangle / A_n = S_n /A_n
$$
and so $[H:H \cap A_n]=2$. This implies that $|H|$ is even.
Here is the same argument again, now using the language of @Beni's answer:
If $H$ is not contained in $A_n$, then $H$ contains an odd permutation and so  $f: H \to \{ -1, +1 \}$ is surjective. Therefore, $\ker f$ is a subgroup of $H$ having index $2$, and $|H|$ must be even. (Of course, $\ker f = H \cap A_n$, but that is not needed here.)
A: The signature morphism $f : S_n \to \{1,-1\}$ which associates to each permutation $\sigma$ the value $1$ if it is even or $-1$ if it is odd can help.
If $H$ is not a subgroup of $A_n$ then $f(H) = \{1,-1\}$. There exists a permutation $\sigma \in H$ which is odd. The function $g: H \to H,\ g(a) = \sigma a$ is a bijection and changes signature. Therefore $H$ contains as many even permutations as odd permutations. Hence $|H|$ is even. 
Thus if $|H|$ is odd then necessarily $H$ is a subgroup of $A_n$.
