How is $S_n$ not simple for $n \geq 3$?

Question

I know the only normal subgroups of $S_n, n \geq 3$, are itself and the trivial subgroup. But why is this the case?

Answer 1

This is not true. $A_n$ is normal in $S_n$ for all $n \geq 3$.

Answer 2

To expand on Johanna's answer, one way to look at it is to look at the set:

$$\left\{(x_1,\dots,x_n)\in\mathbb R^n\mid \sum x_i=1\text{ and } \forall i\, x_i\geq 0\right\}$$

This is a simplex in $n-1$ dimensions, and $S_n$ acts on it in an obvious way - permuting the $x_i$. The action in some cases is orientation preserving, and in other cases, orientation inverting.

We can send orientation preserving maps to $+1$ and orientation-inverting maps to $-1$ and we get a group homomorphism, whose kernel is $A_n$. So $A_n$ represents the orientation-preserving maps, and is a normal subgroup.