If $G = HK$ is a Frobenius group with the complement $H$ being maximal, then $K$ is an elementary abelian $p$-group Let $G$ be some finite Frobenius group with Frobenius complement $H$ and kernel $K$. I want to show that if $H$ is a maximal subgroup in $G$, then $K$ is an elementary abelian $p$-group.
This is an exercise from the book Theory of Finite Groups by Kurzweil and Stellmacher, appearing after the fourth chapter.
My attempts: By maximality of $H$ the action on its coset space is primitive, and $K$ must act transitive. Also it must act regular as $K$ fixes no coset, so I can deduce that $K$ must be minimal normal in $G$. So, if I can establish that $K$ is abelian or that it has some abelian minimal normal subgroup itself I would be done. I have a proof if the order of $H$ is even, by looking how an involution of $H$ would act on $K$, but I cannot find an argument if $H$ contains no involution. So any ideas how to tackle this case?
 A: I was struggling to solve this, but after looking at the results in the book about Frobenius groups I realized that it can be done using  4.1.8 (a) from the book.
You have proved that $K$ must be a minimal normal subgroup of $G$, and if $K$ is abelian then you are done. Also if $K$ is a $p$-group for some prime $p$ then $Z(K) \lhd G$, so $K=Z(K)$ and we are done.
Now let $P \in {\rm Syl}_p(K)$ for some prime $p$ dividing $|K|$, and let $N = N_G(P)$. Then by the Frattini Argument $G = KN$, so $|N| = |N \cap K||G|/|K| = |N \cap K||H|$. Note also that, since $K \ne P$, we cannot have $N=G$ since that would contradict the minimality of $K$ as a normal subgroup of $G$.
We now apply 4.1.8 (a) in the book to the subgroup $N=U$. If $N \cap H^x = 1$ for all $x$, then $N \le K$, which is false because $|N| = |N \cap K||H|$ and $|H|>1$. So there exists $x \in G$ with $|N \cap H^x|>1$ and, since $1 \ne P \le N$, we do not have $N \le H^x$. So $N$ is a Frobenius group with kernel $N \cap K$ and complement $H_0$, where $H_0$ is conjugate to a subgroup of $H$.
So $|N| = |N \cap K||H_0|$ and hence $|H|=|H_0|$ so $H_0$ is equal to a conjugate of $H$, which contradicts the maximality of $H$ (and hence also of all of its conjugates) in $G$.
