Simplicity of the alternating group

Statement

Let $n \geq 5$. Then the alternating group $A_n$ is simple: its only normal subgroups are $\{e\}$ and $A_n$.

Remarks

A usual proof shows that a nontrivial normal subgroup of $A_n$ contains a $3$-cycle, then all $3$-cycles, and then uses that the 3-cycles generate $A_n$. The group $A_4$ is not simple, because the Klein four-group $\{e, (1\ 2)(3\ 4), (1\ 3)(2\ 4), (1\ 4)(2\ 3)\}$ is normal in $A_4$. The group $A_3$ is simple because it has prime order $3$. For $n \geq 5$, the only normal subgroups of $S_n$ are $\{e\}$, $A_n$ and $S_n$.

Explanations

L1 human

A simple group cannot be broken into a normal subgroup and a quotient. From five symbols on, the even permutations form such an unbreakable group. This fact is a key step in the proof that there is no general formula in radicals for polynomials of degree $5$ or more.