Let $S_n$ denote the symmetric group on $\{1,\ldots,n\}$. Let $M$ be the subgroup $\{\sigma \in S_n \mid \sigma(1) = 1\}$. Show that $M$ is a maximal subgroup of $S_n$.
Here is what I've come up with so far:
Suppose we have a subgroup $H$ of $S_n$ such that $M \subseteq H \subseteq S_n$ . We must show that $H = M$ or $H = S_n$. Thus, it suffices to show that if $H$ is a subgroup of $S_n$ that contains $M$ together with at least one element of $S_n$ that is not in $M$, then $H$ must be all of $S_n$.
To this point, suppose $H$ contains $M$ and a permutation $\beta$ such that $\beta(1) \neq 1$.
Is there a reason why $H$ must then be all of $S_n$ ? I'm hoping there's a clever way to see why that must be true, and I need some help hashing this out.
Thanks!