0.9 Frobenius group

Def. A finite group $G$ is said to be a Frobenius group if $G$ contains a proper non-identity subgroup $H$ such that $H\cap g^{-1}Hg=\{1\}$ for all $g\in G\backslash H$.

Prop. The center of Frobenius group is trivial.

---

From wiki: Suppose a finite group $G$ contains a subgroup satisfying specific properties. Using that information, what can be said about the structure of $G$ itself? A classical and beautiful application of character theory is provided in elucidating the structure of Frobenius groups.

Namely, let $\{1\}<H<G$. Assume that $H\cap g^{-1}Hg=\{1\}$ whenever $g\in G\backslash H$. Then $H$ is a so-called Frobenius complement in $G$; the group $G$ is then a Frobenius group by definition. It was proved by G. Frobenius in 1901, see here, that the set

$$N=\{G\backslash (\bigcup _{x\in G}{x}^{-1}Hx)\}\bigcup \{1\}$$

is in fact a normal subgroup of $G$. Almost a century later, Frobenius’ proof that $N$ is a subgroup of $G$ is still the only existing proof; it uses character theory! The normal subgroup $N$ is called the Frobenius kernel of $G$.

It can be shown that $G=NH$, that $N\cap H=\{1\}$ and that the orders of $N$ and $H$ are relatively prime. Therefore, by the Schur–Zassenhaus theorem, all Frobenius complements in $G$ are conjugate to each other. Below, let $t$ be an element of a group and let $S$ be a subset of that group; let $C_S(t)$ denote the set $\{s\in S:s^{-1}ts=t\}$.

A finite Frobenius group $G$ with Frobenius complement $H$ and corresponding Frobenius kernel $N$ satisfies:

1. $C_G(n)\le N$ for all $1\ne n\in N$;

2. $C_H(n)=\{1\}$ for all $1\ne n\in N$;

3. $C_G(h)\le H$ for all $1\ne h\in H$;

4. every $x\in G\backslash N$ is conjugate to an element of $H$;

5. if $1\ne h\in H$, then $h$ is conjugate to every element of the coset $Nh$;

6. each non-principal complex irreducible character of $N$ induces irreducibly to $G$.

As a converse, assume that some finite group $G$ contains a normal subgroup $N$ and some subgroup $H$ satisfying $NH=G$ and $N\cap H=\{1\}$. Then the statements 1)–6) are all equivalent to each other, and if one of them is true, then $H$ is a Frobenius complement of $G$, turning $G$ into a Frobenius group with $N$ as corresponding Frobenius kernel.

Even more general, if some finite group $G$ with proper normal subgroup $N$ satisfies 1), then, applying one of the Sylow theorems, it is not hard (I don’t think so…) to see that all orders of $N$ and $G/N$ are relatively prime. Whence there exists a subgroup $H$ of $G$ satisfying $NH=G$ and $N\cap H=\{1\}$ (by the Schur–Zassenhaus theorem). Thus, again $G$ is a Frobenius group with Frobenius complement $H$ and Frobenius kernel $N$.