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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 $y\mapsto h^{-1}{y}^{-1}hy$.

A Frobenius group has a nontrivial proper subgroup $H$ satisfying $H\cap H^g=\{1\}$ for every $g\notin H$, with $H^g=g^{-1}Hg$. Such $H$ is a Frobenius complement. The Frobenius kernel theorem turns the identity together with elements outside all conjugates of $H$ into a normal subgroup $N$, giving $G=N⋊H$.

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 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.

A Frobenius group is a finite transitive group action on more than one point in which a nonidentity element fixes at most one point and a point stabilizer is nontrivial. Equivalently it has a nontrivial proper subgroup $H$ with $H\cap H^g=1$ for every $g\notin H$. The Frobenius complement is such a point stabilizer $H$.

The Frobenius kernel is the identity together with all fixed-point-free elements in this action, equivalentlyThe Frobenius kernel theorem asserts that this set is a normal subgroup, acts regularly on the points, and gives $G=N⋊H$. In particular $C_N(h)=1$ for $1\ne h\in H$.

Brauer's permutation lemma states that an automorphism of a finite group fixes equally many conjugacy classes and irreducible characters. More generally a group of automorphisms has equal permutation characters on those two sets. Indeed, the character table is invertible and intertwines the two permutation matrices: simultaneous permutation of rows and columns preserves character evaluation. Similar matrices have equal traces, and a permutation-matrix trace counts its fixed points. The fixed objects on the group side are classes, not individual elements.

Conversely, the same induction-norm formula shows that $I_G(\phi )=N$ for every nonprincipal $\phi \in \mathrm{Irr}(N)$. Conjugation by each $g\notin N$ fixes only $1_N$, so Brauer's permutation lemma says that it fixes only the identity conjugacy class of $N$. A nonidentity element centralizing $g$ would provide another fixed class. Therefore $C_N(g)=1$ for every $g\notin N$.

We first prove that $|N|$ and $[G:N]$ are coprime. If a prime $p$ divided both, a Sylow p-subgroup $S$ of $G$ would have $1<S\cap N<S$. The nontrivial normal subgroup $S\cap N$ meets $Z(S)$ nontrivially: the class equation for the conjugation action of the p-group $S$ on $S\cap N$ makes the number of fixed elements divisible by $p$, and one of them is the identity. Choose $1\ne z\in N\cap Z(S)$ and $s\in S\setminus N$. Then $z\in C_N(s)$, contradicting the previous conclusion.

Thus $N$ is a normal subgroup that is also a Hall subgroup. The Schur-Zassenhaus theorem supplies a group complement $H$, meaning $G=NH$ and $N\cap H=1$. The theorem's existence assertion requires only normality and coprimality of order and index. Here $H$ is nontrivial and proper by the assumptions on $N$.

For $n\in N\setminus \{1\}$, suppose $1\ne h\in H\cap nH{n}^{-1}$. Write $h=n{h}^{\prime }{n}^{-1}$ with $h^{\prime }\in H$. Projection to $G/N$ gives $hN=h^{\prime }N$, hence $h=h^{\prime }$ because $H\to G/N$ is injective. This makes $n$ centralize $h\notin N$, which is impossible. Hence $H\cap nH{n}^{-1}=1$. Writing an arbitrary element outside $H$ as $n{h}_0$ reduces its conjugate intersection to this case. It follows that $H$ is a Frobenius complement and $G$ is a Frobenius group.

Moreover, for each $1\ne h\in H$, the map $v\mapsto vh{v}^{-1}$ from $N$ into the coset $Nh$ is injective, since $C_N(h)=1$, and hence bijective. Thus every element outside $N$ belongs to a conjugate of $H$, while no nonidentity element of $N$ does. The Frobenius kernel is exactly $N$. This completes both directions of the Frobenius kernel criterion by irreducible induction.

Use the defining property of the Frobenius complement: $H\cap H^x=\{1\}$ whenever $x\notin H$. The induced character formula, extended linearly to any class function, isFor $h\in H\setminus \{1\}$, a summand requires $h\in H\cap xH{x}^{-1}$. The Frobenius intersection property forces $x\in H$. All $|H|$ remaining terms equal $\theta (h)$ since $\theta$ is a class function on $H$, so ${\theta }^G(h)=\theta (h)$. At the identity,Together these giveThe zero identity value is essential: at nonidentity elements the intersection argument already gives the equality, whereas induction multiplies the identity value by the subgroup index.

First show $C_N(h)=\{1\}$. If $z\in N$ commutes with $h$ and $z\ne 1$, then $z\notin H$ because $N\cap H=\{1\}$. But $z^{-1}hz=h$, so the nonidentity element $h$ lies in $H\cap H^z$, contradicting the Frobenius complement property. Thus conjugation by $h$ has no nonidentity fixed element in the Frobenius kernel.

Normality of $N$ makeswell-defined. If $f_h(a)=f_h(b)$, cancellation gives $a^{-1}ha=b^{-1}hb$, and hence $b{a}^{-1}$ commutes with $h$. Since $b{a}^{-1}\in N$ and $C_N(h)=\{1\}$, we obtain $a=b$. The map is injective, and an injection of a finite set into itself is surjective. ConsequentlyThis is the commutator bijection from a fixed-point-free automorphism, applied to $\alpha (y)=h^{-1}yh$.