Maximal subgroup of $S_n$

Question

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!

Answer 1

It is enough to show that $H$ contains all transpositions of the form $(1,j)$, $j=2,\ldots,n$.

Let $k\neq 1$ be such that $\beta(k)=1$, and let $\beta(1)=r\neq 1$. Multiplying $\beta$ on the left by $(r,k)\in M$ we get a permutation that sends $1$ to $k$ and $k$ to $1$. Thus, $(r,k)\beta = (1,k)\tau$ for some $\tau$ that is disjoint from $\{1,k\}$, so $\tau\in M$. Therefore, $(1,k)\in H$.

Now conjugating with $(k,j)\in M$, $2\leq j\leq n$, we get $(1,j)\in H$, proving that $H$ contains all transpositions of the form $(1,j)$, and hence contains $S_n$.