Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 4 / 1

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.

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.

Let $V$ afford the irreducible character $\chi$. The positive character inner product $⟨{\chi }_H,1_H{⟩}_H$ is the multiplicity of the trivial representation in $V$ restricted to $H$, so there is a nonzero vector $v$ fixed by every element of $H$. Since $N\subseteq \ker \chi$, every element of $N$ acts as the identity on all of $V$. By $G=NH$, every element of $G$ therefore fixes $v$.

The line $\mathbb{C}v$ is a nonzero invariant subspace of the irreducible representation $V$, so it must equal $V$. The action on this line is trivial, provingThe hypothesis on the kernel of a character is what makes an $H$-fixed vector into a $G$-fixed vector.

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

Because $N$ is an abelian group, every $\phi \in \mathrm{Irr}(N)$ is a linear character, hence a homomorphism to ${\mathbb{C}}^{\times }$. Its inertia group of a character contains $N$, whose inner conjugations act trivially on $N$. Write an arbitrary $g\in G$ as $g=nh$ with $n\in N$ and $h\in H$, using $G=NH$. Conjugation by $n$ does not affect $\phi$, so $g$ stabilizes $\phi$ if and only if $h$ does.

Suppose $h\ne 1$ stabilizes $\phi$. Then $\phi (h^{-1}yh)=\phi (y)$ for every $y\in N$, and its multiplicativity givesPart (iv) makes $y\mapsto [h,y]$ surjective onto $N$, so $\phi (x)=1$ for every $x\in N$. This would be the trivial character, contrary to the hypothesis. Therefore no nonidentity $h\in H$ stabilizes $\phi$, andThe argument uses both abelianness, to make $\phi$ multiplicative, and the fixed-point-free commutator bijection.