The way Frobenius groups act on other groups is an important part of the eventual proof by J.G. Thompson that Frobenius kernels are nilpotent. For this purpose, it is enough to prove that a finite group which admits a fixed-point free automorphism $\alpha$ of prime order $p$ is nilpotent, which Thompson did in his famous thesis.

Before Thompson's thesis it was known that a finite solvable group $G$ which admits a fixed-point free automorphism of prime order $p$ is nilpotent. This was proved by G. Higman and J. Witt (independently I believe) in the 1950s. I will return to this in a moment.

Once the solvable case is dealt with, precursors to Thompson's theorems on normal $p$-complements were used by Thompson to reduce the general case to the case that $G$ is solvable.

Here is a quick outline, making use of the more modern version of these normal $p$-complement theorems (and standard properties of coprime automorphisms): let $G$ be a finite group which admits a fixed-point-free automorphism $\alpha$ of prime order $p$ such that $G$ is not nilpotent, and $G$ has minimal order subject to this. Then $|G| \equiv 1$ (mod $p$).

Now we may assume that $G$ is not solvable. Hence we may choose an odd prime divisor $q$ of $|G|$. Then $G$ has an $\alpha$-invariant Sylow $q$-subgroup $Q$, and $N_{G}(ZJ(Q))$ is also $\alpha$-invariant. There are two cases: if $ZJ(Q) \lhd G$, then $\alpha$ induces a fixed-point free automorphism on $G/ZJ(Q)$, so that $G/ZJ(Q)$ is nilpotent, and $G$ is solvable, contrary to assumption.

Hence $N_{G}(ZJ(Q))$ is a proper subgroup of $G$, so is nilpotent by the minimality of $G$. Then Thompson's normal $q$-complement theorem tells us that $G$ has a normal $q$-complement too. Since $q$ was an arbitrary odd prime divisor of $|G|$, we quickly see that $G$ has a normal Sylow $2$-subgroup $U$ say, and that $G/U$ is nilpotent by minimality of $G$. Then $G$ is again solvable, contrary to assumption.

So this reduces the question (having used very deep theorems of Thompson) to proving the result in the case that $G$ is solvable.

The proof in the solvable case really reduces to representation theory.
It reduces quickly to the case that $Q = F(G)$ is a minimal normal subgroup of $G$, so is an elementary Abelian $q$-group for some prime $q$. Furthermore, $G/M$ is an elementary Abelian $r$-group for some prime $r$. Let $R$ be an $\alpha$-invariant Sylow $r$-subgroup of $G$. Consider the semi-direct product $H = G\langle \alpha \rangle = Q(R\langle \alpha \rangle )$.

Then $R\langle \alpha \rangle$ is itself a Frobenius group of order $p|R|$, and is acting faithfully as a group of automorphisms of $Q$. It is possible to prove by representation-theoretic methods (essentially Clifford's Theorem) that in such an action, $\alpha$ can't act without non-trivial fixed-points on $Q$, contrary to the assumption that it does.

This latter method of proof has been adapted and generalized over the years by many people (such as T. Berger, E.Dade, A. Turull, for example) to use representation-theoretic methods to restrict the structure of groups $G$ which admit a group of automorphisms $A$ with restrictions on the structure of $C_{G}(A)$.