Given than $A_n$ is simple for $n\ge 5$, prove $A_n$ is the only non-trivial proper normal subgroup of $S_n$.
My work:
Assume it is false, then there exists some non-trivial proper normal subgroup $K$ of $S_n$, where $K\neq A_n$. We know $A_n$ is a normal subgroup of $S_n$. Let $T=K\cap A_n$, hence $T$ is a normal subgroup of $S_n$, which implies $T$ is a normal subgroup of $A_n$, namely,
$$\{e\}\mathrel{\unlhd}T\mathrel{\unlhd}A_n\mathrel{\unlhd}S_n$$
So $T=\{e\}$ or $T=A_n$, otherwise $A_n$ is not simple.
(1) If $T=A_n$, it implies $A_n\subseteq K$. Since the index $[S_n: A_n]=2$, $A_n$ is the maximal proper subgroup of $S_n$, hence $K=A_n$, which contradicts with $K\neq A_n$.
(2) If $T=\{e\}$, it implies that all elements in $K$ are odd permutations (except the identity element $e$). But
$$\text{odd permutation}\cdot \text{odd permutation}=\text{even permutation}$$
so for any non-identity elements $a, b\in K$, we have $ab=e$, since $e$ is the only even permutation elements in $K$.
(i) If $|K|\ge 3$, then $K=\{e, a, b, \dots \}$, and $ab=e=aa\Longrightarrow a=b$, contradicts with $a\neq b$.
(ii) If $|K|=2$, then $K=\{e, a\}$, and $|a|=2$,
I am not sure about the following argument:
First we write $a$ as disjoint cycles, such as
$$a=(1,2,3)(6,9, 17, 8)...(., .,)$$
The number appears only once in the $(.,.)$. For example, it never happens like $(1,2)(1,3)$. In this case, $1$ appears twice, and we will simplify it as $(1,3,2)$. But this will lead to the order of $|a|\ge 3$, which contradicts with the fact that $|a|=2$.
Since the element $a$ has order $2$, we can simplify $a$ as the distinct disjoint transpositions, and each number appears only once, such as
$$a=(1,2)(i,j)...(p, q)$$
If $3$ doesn't appear in any of transposition $(.,.)$ of $a$, then we take an element $(1,2,3)$ from $S_n$ we have $(1,2)(1,2,3)\neq (1,2,3)(1,2)$
If $3$ appears in some transposition $(3,k)$ of $a$, we have $(1,2)(3,k)(1,2,3)\neq (1,2,3)(1,2)(3,k)$
Therefore, $K$ is not normal and we reach a contradiction.
