Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 2 / 2 / b

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 $G$.

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.