If a subgroup $H$ contains $A_n$ and at least one odd permutation, then $H=S_n$ If we have $H \subseteq S_n$ which contains $A_n$ and at least one odd permutation, then  how can we show that $H=S_n$? 
I've looked at some examples and it's clear that it is true, I'm just not sure where to start on proving it. Any direction would be much appreciated, thanks. 
 A: Well, what is the order of $\langle H\rangle?$ The order of $\langle H\rangle$ divides $n!$, the order of $S_n$. Now, $\lvert A_n\rvert=\frac{\lvert S_n\rvert}{2}$, but we know that $\lvert H\rvert\ge \lvert A_n\rvert+1$, so that $\langle H\rangle$ has order at least $\lvert A_n\rvert+1$. Then, in order to divide $\lvert S_n\rvert$, it must be the case that $\lvert \langle H\rangle\rvert=n!$, so that $\langle H\rangle=S_n$.
A: In fact, a stronger statement is true.

If $H$ is a subgroup of $S_n$ that contains at least one odd permutation, then exactly half of the elements of $H$ are odd permutations.

To see this, consider $\operatorname{sgn}:S_n\to\{\pm1\}$ which maps each permutation to its sign. It is well-known that this function is a homomorphism. Now consider $\ker\operatorname{sgn}$.
A: Let $\sigma\in S_n-A_n$ and let $\tau\in H$ be an odd permutation.
Then $\sigma\circ\tau=\xi\in A_n\subseteq H$ because the product of two odd permutations is even, thus $\sigma=\xi\circ\tau^{-1}\in H$ because product of two elements in $H$, being $H$ a subgroup of $S_n$.
A: If $K\subseteq H$ are subgroups of the finite group $G$, then you should be able to prove that
$$
[G:K]=[G:H][H:K]
$$
where $[G:H]$ means the index of $H$ in $G$. Hint: use Lagrange’s theorem.
Now use $G=S_n$, $K=A_n$ and $[H:K]>1$.
