Equivalence of two definitions of a Frobenius group? My usual definition of a Frobenius group is a semidirect product $G=K\rtimes H$, where $H\cap H^x=1$ for $x\in G\setminus H$, and $K=(G\setminus\bigcup_{g\in G}H^g)\cup\{1\}$. 
Apparently a Frobenius group can also be defined as $G=K\rtimes H$, where $C_K(h)=1$ for all nonidentity $h\in H$. 
Assuming this alternative definition, how is $G\setminus\bigcup_{g\in G}H^g\subseteq K$? This is the one part I can't verify. I want to express $g\in G\setminus\bigcup_{g\in G}H^g$ as a product $hk\in HK$, and I was hoping to show that $h$ is centralized by something nontrivial in $K$, to conclude $h=1$, so that $g=n\in N$.
 A: Call "Frobenius complement" a subgroup $H$ of a group $G$ with the property $H^g \cap H=1$ for all $g \in G\setminus H$. Then you can define as "Frobenius group" group $G$ that has a Frobenius complement $H$. Equivalently you can define a "Frobenius group" a group $G$ that acts transitively on a set $\Omega$ and such that every element $g \in G$ fixes at most one element of $\Omega$. We have the remarkble 

Theorem: Let $H$ a Frobenius complement of a Frobenius group $G$, then $G$ has a normal subgroup $N$ such that $G=N \rtimes H$. Moreover $$ N=\left(G \setminus \bigcup_{g \in G}H^g \right) \cup \{1\}.$$

The proof works with characters and you can find it in every book on group theory or character theory. This is one of the few results in group theory without character-free proof.
Returning to your question, much more is true:
Proposition Let $G=N \rtimes H$. Are equivalent


*

*$C_G(n) \leq N$ for all $1 \neq n \in N$.

*$C_G(h) \leq H$ for all $1 \neq h \in H$.

*$n^h\neq n$ for $1\neq n \in N$ and $1\neq h \in H$. 

*$H$ is a Frobenius complement. 

*Every $x \in G\setminus N$ is conjugate to an element of $H$.


This is more or less the content of the Problem 7.1 of "Character Theory of Finite Groups", Isaacs. You can find the  equivalence of the first four points in " Finite Group Theory", Isaacs (again!).
For the equivalence of the last two points, assume $H$ Frobenius complement. $H=H^g$ implies $g \in H$ and this means that $H=N_G(H)$, i.e. there are $[G:H]$ distinct conjugates of elements in $H$ in $G$, and two different such conjugates have trivial intersection. Then there are $[G:H](|H|-1)$ non-identity conjugates with elements in $H$. Then there are $|G|-[G:H](|H|-1)=[G:H]$ elements not conjugate to any element of $H$. But $N \cap H=1=N\cap H^g$ for all $ g \in G\setminus H$, then $N$ is constituted by elements not conjugated to any of $H$. So $|N| \le [G:H]=|N|$ and then $N$ is exactely the set of the elements that are not conjugated to any element of $H$. From here 5 follows.
Assume now $5$. Then if $H_1\dots H_t$ are the different conjugates of $H$ (this means $t=[G: N_G(H)]$ of course) we have that $$ G\setminus N \subseteq \bigcup_{i=1}^n H_i \setminus \{1\}.$$ If the $H_i$'s have trivial intersection, then 
$$ |G \setminus N| \le \sum_{i=1}^t |(H_i|-1)=t(|H|-1)$$
But $G \setminus N$ is the set of laterals of $N$ except $N$ itself and 
$$ |G\setminus N|=([G:N]-1)|N|=(|H|-1)[G:H]$$
Then $[G:H] \le [G:N_G(H)]$. But also the reverse inequality holds and equality must occur with the condition that all conjugates of $H$ trivially intersect $H$.
Note that the 1 $\implies$ 4 is always true and make $G$ splits over $N$. Indeed if $P \in Syl_p(G)$ and $p\mid |N|$ then $P\cap N$ is normal in $P$ and $P\cap N \cap Z(P)>1$ So $N$ absorbs all the $p$-Sylows for every prime $p$ that divides $|N|$. This means that $N$ is a Hall subgroup and has a complement $H$ by Schur-Zassenhaus. Let $g \in G \setminus H$, $g=nx$ with $1 \neq n \in N$ and $ x\in H$. If $ y \in H^g \cap H$, $y=h^n$ for $h \in H$. It is not difficult to prove that $[h,n]=1$, that means $H \ni h \in C_G(n) \le N$.
