A question about Lang's proof of simplicity of $A_n$

Question

Theorem 5.5. If $n\geq 5$ then the alternating group $A_n$ is simple.

Proof. Let $N$ be a non-trivial normal subgroup of $A_n$. We prove that $N$ contains some $3$-cycle, whence the theorem follows by (b). Let $\sigma\in N$, $\sigma\ne id$, be an element which has the maximal number of fixed points...

I don't understand why the highlighted part holds: why should such an element exist? Can someone describe the set which $\sigma$ supposedly belongs to in set-builder notation?

Answer 1

For $s \in N$, let $\mathrm{fp}: N \rightarrow \Bbb{Z}_{\geq 0}$, taking $s$ to its number of fixed points. Since $n$ is finite, $\mathrm{fp}(s)$ is finite for each $s \in N$. Since $A_n$ is finite, $N$ is finite, so $M = \max \{\mathrm{fp}(s) : s \in N \smallsetminus \{\mathrm{id}\}\}$ exists. Take $\sigma \in \{s \in N : \mathrm{fp}(s) = M\}$.

Answer 2

Well, the number of fixed points is at most $n-2$.
So an element of $\{0,1,\dots,n-2\}$. So just take an element of $N$ with the maximal number of fixed points.