Frobenius kernel ($N=(G\setminus {\bigcup }_g{H}^g)\cup \{1\}$)

The Frobenius kernel consists of the identity and the elements lying in no conjugate of a Frobenius complement. The Frobenius kernel theorem proves that this conjugacy-invariant set is a normal subgroup, rather than assuming closure as part of its definition.

For $1<N◃G$ proper, the displayed condition is equivalent to $N$ being a Frobenius kernel. For a normal subgroup the induction norm is $|I_G(\phi ):N|$. Trivial inertia for each nonprincipal character and Brauer's permutation lemma imply that every element outside $N$ has trivial centralizer in $N$. This forces $N$ to be a normal Hall subgroup; a complement is then fixed-point-free on $N$.

Let $N$ be a finite group and $\alpha$ an automorphism whose only fixed element is the identity. If $\alpha (a{)}^{-1}a=\alpha (b{)}^{-1}b$, then $\alpha (b{a}^{-1})=b{a}^{-1}$, so $a=b$. The displayed map is therefore a bijection. For conjugation by a nonidentity element of a Frobenius complement, it gives the commutator map $y\mapsto h^{-1}{y}^{-1}hy$.

For a Frobenius complement $H$, a class function $\theta$ with $\theta (1)=0$ satisfies $({\theta }^G{)}_H=\theta$. For each nontrivial $\eta \in \mathrm{Irr}(H)$ of degree $d$, the virtual character $(\eta -d{1}_H{)}^G+d{1}_G$ has norm one and positive degree by Frobenius reciprocity. It is therefore irreducible, restricts to $\eta$ and is identically $d$ on the candidate kernel $N$. The intersection of the kernels of these characters is exactly $N$: outside $N$, separation follows from the character of the regular representation of $H$. This proves normality; $N\cap H=1$ and $|N|=[G:H]$ then give the semidirect product.