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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.

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.

We prove closure of $N$ through character kernels, rather than assuming that a conjugacy-invariant set is a subgroup. Let $\eta \in \mathrm{Irr}(H)\setminus \{1_H\}$, write $d=\eta (1)$, and define the generalised characterHere $1_H,1_G$ are the trivial characters. Since $(\eta -d{1}_H)(1)=0$, part (i) gives $({\Psi }_{\eta }{)}_H=\eta$ and ${\Psi }_{\eta }(1)=d$. Put $\vartheta =\eta -d{1}_H$. By Frobenius reciprocity and character orthogonality,ConsequentlyTo justify the resulting irreducibility, write any generalised character as an integer combination ${\sum }_{\chi }{a}_{\chi }\chi$ of irreducible characters. Its squared norm is $\sum a_{\chi }^2$, so norm one makes it $\chi$ or $-\chi$ for one irreducible $\chi$. Positive degree rules out the negative sign. Thus ${\Psi }_{\eta }$ is an irreducible character, an instance of the fact that positive-degree norm-one virtual characters are irreducible.

If $x\in N\setminus \{1\}$, no conjugate of $x$ lies in $H$, so the induction formula gives ${\vartheta }^G(x)=0$. At $1$ it is also zero. Hence ${\Psi }_{\eta }(x)=d$ for all $x\in N$. For a finite-group representation we may choose an invariant positive Hermitian form, making every representing matrix unitary. Its trace equals its dimension only when all its unit-modulus eigenvalues are $1$. Thus the kernel of a character is $\{g:{\Psi }_{\eta }(g)={\Psi }_{\eta }(1)\}$, and $N\subseteq \ker {\Psi }_{\eta }$.

Conversely, if $g\notin N$, then $g$ is conjugate to some $h\in H\setminus \{1\}$. The character of the regular representation of $H$ isIf every nontrivial $\eta$ satisfied $\eta (h)=\eta (1)$, the sum at $h$ would instead equal $\sum \eta (1{)}^2=|H|$, a contradiction. Choose an $\eta$ for which these values differ. Since ${\Psi }_{\eta }$ restricts to $\eta$, it has ${\Psi }_{\eta }(g)=\eta (h)\ne d$, so $g\notin \ker {\Psi }_{\eta }$. ThereforeThis proves the Frobenius kernel theorem's subgroup assertion. The definition also gives $N\cap H=\{1\}$. Using the permitted cardinality $|N|=[G:H]$, normality ensures $NH$ is a subgroup andThus $N$ is the Frobenius kernel, and $H$ supplies the complementary factor.