Frobenius group

Remarks

Equivalently, a finite group $G$ is a Frobenius group if it has a subgroup $H$ with $\set{e} \ne H \ne G$ and $H \cap gHg^{-1} = \set{e}$ for all $g \in G \setminus H$. Then $H$ is a point stabiliser for the action on the left cosets of $H$, and it is called a Frobenius complement. The set $K$ of elements that fix no point, together with $e$, is the Frobenius kernel; by Frobenius's theorem it is a normal subgroup.

Example: the group of maps $x \mapsto ax + b$ with $a \ne 0$, acting on a finite field $\F_q$ with $q \ge 3$. If $a \ne 1$, the equation $ax + b = x$ has exactly one solution, and if $a = 1$ and $b \ne 0$ it has none, so clause 1 holds. The map $x \mapsto ax$ with $a \ne 0, 1$ fixes $0$, so clause 2 holds. For $q = 3$ this is $S_3$ acting on $3$ points. For odd $n \ge 3$, the dihedral group of order $2n$ acting on the corners of a regular $n$-gon is also a Frobenius group.

Non-example: $S_4$ acting on $\set{1, 2, 3, 4}$. The transposition $(1\,2)$ fixes the two points $3$ and $4$, so clause 1 fails. The cyclic group generated by $(1\,2\,3)$ acting on $\set{1, 2, 3}$ satisfies clause 1 but fails clause 2.