# Maximal subgroup of $S_n$

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!

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 $$S_n$$.
• This proof assumes that $r\neq k$. I think that it should be mentioned that if $r=k$, then $\beta$ is already of the form $(1,k)\tau$, with $\tau\in M$. Commented May 11, 2021 at 9:37
• @Adren: There is no need to separate cases. If $r=k$, then $(r,k)$ is the identity. So there is no such assumption. Commented May 11, 2021 at 10:22
• I agree. It's only a question of notation : some people consider that the symbol $(i,j)$ denotes a transposition (with the implicit assumption that $i\neq j$). Commented May 11, 2021 at 10:28