Does $A_n\leq G\leq S_n$ imply either $G=S_n$ or $G=A_n$?

Question

Is it true that if $A_n\leq G\leq S_n$ implies $G=A_n$ or $S_n$?

Here is my argument:

Since $|G|=m$ divides $n!$ we have $mk=n!$ for some $k$. But $|A_n|=n!/2\leq n!/k$, hence $k\leq 2$, hence $m=n!/k$ is either the full order of $S_n$ or that of $A_n$.

Answer 1

Promoting @ThomasAndrews' comment into an answer . . .

Yes, you're correct.

In general, if $G_1\le G_2$ and $\lvert G_2\rvert=p\lvert G_1\rvert$ for some prime $p$ for finite groups $G_1, G_2$, then for any $G$ such that $G_1\le G\le G_2$, either $G=G_1$ or $G=G_2$.