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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 $N$.

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.

Suppose first that $N$ is the Frobenius kernel. An element $g\notin N$ fixes a point in the Frobenius action. If $n\in C_N(g)$, then $n$ maps that point to another point fixed by $g$. Since $g$ has only one fixed point and $N$ acts regularly, $n=1$. ThusNo nonidentity conjugacy class of $N$ can be fixed by such a $g$. If the class of $n\ne 1$ were fixed, $gn{g}^{-1}=vn{v}^{-1}$ for some $v\in N$, so $v^{-1}g\notin N$ would centralize $n$, a contradiction. Conjugation by $g$ consequently fixes just the identity class. By Brauer's permutation lemma it fixes just one irreducible character, necessarily $1_N$.

For every nonprincipal $\phi \in \mathrm{Irr}(N)$, its inertia group of a character is therefore $I_G(\phi )=N$. Since $N$ is normal, the restriction of the induced character is the sum of its conjugates, counted over $G/N$:Frobenius reciprocity and character orthogonality giveAn actual character of norm one is irreducible. Every nonprincipal irreducible character of $N$ therefore induces irreducibly to $G$.

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.

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 $\alpha (y)=h^{-1}yh$.