Subgroup of $S_n$ that has no subgroups of index 2

Question

Let $n\geq5$, and $G$ a subgroup of $S_n$ s.t. $G$ has no subgroups of index 2 ($G$ is also simple). Is this enough to say that $G$ lays in $A_n$ (as $A_n$ is simple and has no subgroup of index 2)?

Is there any use of the 2nd isomorphism theorem to prove this? i'm working on a group of order 2016 and for n=8 ($o(G) \leq o(A_{8})$)

Answer 1

If $G$ is simple non commutative, $[G,G]=G\subset [S_n,S_n]=A_n$ done.

Answer 2

We do not need whether $A_n$ is simple or not; we need to know that its index in $S_n$ is $2$, which is true for all $n\geq 1$. You may ignore also the hypothesis that $G$ is simple.

Suppose $G$ is not in $A_n$ (but it is in $S_n$ by htpothesis). Then $GA_n=S_n$ and hence $G\cap A_n$ is subgroup of index $2$ in $G$ (use the formula for $|HK|=|H||K|/|H\cap K|$ or any other way).

This contradicts another hypothesis. Hence $G$ must be in $A_n$.