In the book "Endliche Gruppen I" by Huppert on p. 159 there is a theorem (Satz 3.2) attributed to Galois about (solvable) primitive permutation groups. Part (f) reads as follows: Let $G$ be a solvable primitive permutation group and let $N$ be a minimal normal subgroup. Then all complements of $N$ in $G$ are conjugate.
The proof of this part proceeds in the following steps: One assumes without loss of generality that $N$ is a proper subgroup of $G$ and chooses some $K \unlhd G$ containing $N$ such that $K/N$ is a minimal normal subgroup of $G/N$. The minimal normal subgroup $N$ is of order $p^m$ and $K/N$ is of order $q^k$ for some distinct primes $p$ and $q$.
If $V$ is a complement of $N$ in $G$, then one defines $Q = K \cap V$ and Huppert shows that $Q$ is a Sylow $q$-subgroup of $K$. Of course, all Sylow $q$-subgroups of $K$ are conjugate. Huppert then claims that it suffices to show that $V = N_G(Q)$.
I am not sure why this implies the claim about the complements being conjugate in $G$. Of course we have a map $$ \{ \text{complements of } N \} \to \{ \text{Sylow }q\text{-subgroups of }K \} $$ by mapping $V \mapsto K \cap V$. And this map is surjective (since Sylow subgroups are conjugate). We also know that the latter set has cardinality $[G:N_G(Q)] = [G:V] = |N|$. But can we estimate the number of complements of $N$ in $G$?
Ok, I found a solution. It works as follows. Suppose $V$ and $V'$ are complements of $N$ in $G$ and let $Q = K \cap V$, resp. $Q' = K \cap V'$. Since $Q$ and $Q'$ are Sylow $q$-subgroups of $K$ there is some $g \in G$ such that $(Q')^g = Q$. Note that $K$ is normal in $G$. Hence $Q = K \cap V = K \cap (V')^g$. And moreover, $Q \unlhd V$ and $Q \unlhd (V')^g$, again since $K \unlhd G$. Hence $(V')^g \leq N_G(Q) = V$ and so $(V')^g = V$ since they have the same order.