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

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 $G$.