Frobenius's theorem on Frobenius kernels

Remarks

Counting is easy. Let $n = \abs{\Omega}$; the $n$ point stabilisers are the conjugates of $H$, and any two of them meet only in $e$. So their union has $n(\abs{H} - 1) + 1$ elements, and $K$ has $\abs{G} - n\abs{H} + n = n$ elements. The hard part is that $K$ is closed under multiplication. The standard proof uses characters. Each irreducible character of $H$ is extended to a character of $G$, and $K$ is the intersection of the kernels of these characters.

Example: for the maps $x \mapsto ax + b$ with $a \ne 0$ acting on $\F_q$, the kernel $K$ is the group of translations $x \mapsto x + b$, and the complement $H$ is the group of maps $x \mapsto ax$. For $S_3$ acting on $3$ points, $K = A_3$.

John Thompson later proved that the Frobenius kernel is always nilpotent.

Going deeper

For a student

The theorem is a famous example of the power of character theory: its statement is purely about permutations, but the standard proof needs characters. It gives the structure $G = K \rtimes H$ with $\abs{K} = \abs{\Omega}$ and $\abs{H}$ dividing $\abs{K} - 1$, since $H$ acts on $K \setminus \set{e}$ by conjugation with all orbits of size $\abs{H}$. The Frobenius kernel and complement are central in the study of finite groups, for example in the proof of the Feit–Thompson theorem.