Alternating Group is Simple except on 4 Letters

Theorem
Let $n$ be an integer such that $n \ne 4$.

Then the $n$th alternating group $A_n$ is simple.

Proof
Recall that a group is simple if its normal subgroups are itself and the trivial subgroup.

Let $n < 4$.

$A_2$ is the trivial group and hence simple.

$A_3$ is the cyclic group of order $3$, hence a prime group.

By Prime Group is Simple, $A_3$ is simple.

We note that $A_4$ is a special case.

From Normality of Subgroups of Alternating Group on 4 Letters, $A_4$ has a proper normal subgroup $K_4$, and so is not simple.

It remains to investigate the case where $n \ge 5$.

Let $N$ be a non-trivial normal subgroup of $A_n$.

It is to be shown that $N = A_n$.

The first step is to show that $A_n$ contains a $3$-cycle.

Let $e$ denote the identity element of $A_n$.

Let $\alpha \ne e$ be an element of $A_n$ which fixes as many elements of $\N_n$ as possible.

From Existence and Uniqueness of Cycle Decomposition, we can express $\alpha$ in the form:
 * $\alpha = \alpha_1 \alpha_2 \dotsm \alpha_s$

where each of the $\alpha_i$ are disjoint cycles.

Let it be assumed that $\alpha_1, \ldots, \alpha_s$ are arranged in order of increasing length.