Why is $A_n$ maximal in $S_n$?

Question

I am struggling to see why exactly $A_n$ is maximal in $S_n$.

A subgroup $M$ of a group $G$ is called a maximal subgroup if $M\neq G$ and the only subgroups of $G$ which contain $M$ are $M$ and $G$.

Here is what I know...

The index $|S_n:A_n|=2$ because $$ |S_n:A_n|=\frac{|S_n|}{|A_n|}=\frac{n!}{n!/2} = 2. $$

Since $|S_n:A_n|=2$, we know that $A_n\unlhd S_n$. I feel like the reason $A_n$ is maximal stems from the fact that the index is 2. Can someone please help explain to me why $A_n$ is maximal? Thanks.

Answer 1

You’re making it harder than it really is. You know that the order of a subgroup must divide the order of the group. If $A_n<H<S_n$, then $2=[S_n:A_n]=[S_n:H][H:A_n]$, so either $[S_n:H]=1$ and $H=S_n$, or $[H:A_n]=1$, and $H=A_n$.

Answer 2

Suppose there is a subgroup $A_n \subsetneq H \subsetneq S_n$. Then we'd have $|A_n| < |H|$ and hence $|S_n|/|A_n| > |S_n|/|H|$. But since $|S_n|/ |A_n| = 2$, and since the order of any subgroup must divide the order of the group itself, we are forced to conclude that $|S_n|/|H| = 1$, or in other words $|S_n| = |H|$. This contradicts the fact that $H$ is a proper subgroup of $S_n$.

Answer 3

Because if $H$ is a subgroup of $G$ you have that $order(H) |order(G)$; the order of $A_n$ is $\frac {order(S_n)} 2$ so it is maximal.

Answer 4

Suppose there were another subgroup $M < N < G$ with $N \neq M$. Let $x \in N \setminus M$. Now $N \supseteq M \cup (x \cdot M) = M \cup (G \setminus M) = G$, therefore $N = G$.