Frobenius kernel implies fixed-point-free automorphism of prime order

Statement

Let $G$ be a finite group and $N$ be a Frobenius kernel in $G$: in particular, $N$ is a proper nontrivial complemented normal subgroup with the property that no nontrivial element of $N$ commutes with a nontrivial element outside $N$. Then, there exists a prime $p$ and a fixed point-free automorphism $\sigma$ of $N$ of order $p$.

Proof

Given: A finite group $G$, a Frobenius kernel $N$ in $G$.

To prove: There exists a prime $p$ and a fixed point-free automorphism $\sigma$ of $N$, where $\sigma$ has order $p$.

Proof: Let $H$ be a complement in $G$ to $N$ (so, $H$ is a Frobenius complement). Clearly $H$ contains elements of prime order, since it is nontrivial. Pick an element $g\in H$ of prime order. Let $\sigma$ be the automorphism of $N$ induced via conjugation by $g$.

If $a\in N$ satisfies $\sigma (a)=a$, then $g$ and $a$ commute. By our assumption, no nontrivial element of $N$ commutes with any nontrivial element outside $N$, so this forces that $\sigma$ has no non-identity fixed points.