I'm a noob in group theory, so please forgive me if I put some silly questions. I try to solve the following problem (6.11 from Isaac's book, Character theory of finite groups):
Let $A\unlhd G$ with $A$ Abelian. Let $\chi$ be an irreducible monomial character of $G$. Prove that $\chi$ is a relative monomial character with respect to $A$, i.e. there exists a subgroup $H$ with $A\subset H\subset G$ and an irreducible character $\psi$ of $H$ such that $\psi_A$ is irreducible and $\psi^G=\chi$.
Since $\chi$ is monomial, there exists a subgroup $K$ and a linear character $\lambda$ of $K$ such that $\chi=\lambda^G$. The hint given is that: Every constituent of $(\lambda^{AH})_A = (\lambda_{A\cap H})^A$ has multiplicity $1$. I don't know how to prove this hint and how to solve the problem using it. The linear character $\psi_A$ should be a constituent of $(\lambda^{AH})_A$? I don't grasp the connection between $\psi$ and $\lambda$. Ito's theorem (according to which $\chi(1)||G:A|$) plays a role in solving this problem?
Best Answer
Let $A \unlhd G$ and $\chi$ a monomial irreducible (complex) character say $\chi=\lambda^G$, with $\lambda$ linear character of a subgroup $K$ of $G$. Consider the subgroup $KA$. Observe that $\lambda^{KA}$ is irreducible. By using Problem (5.2) in Isaacs' book (all references are in Character Theory of Finite Groups), we have $(\lambda^{KA})_A=(\lambda_{K \cap A})^A$. Since $A$ is abelian, we can find a linear character $\mu$ of $A$, such that $\mu_{K \cap A}=\lambda_{K \cap A}$. Now apply (6.17) Corollary (Gallagher): we must have $$(\lambda^{KA})_A=(\lambda_{K \cap A})^A=\sum_{\nu \in Irr(A/K \cap A)} \mu \nu$$ Observe that all $\nu$ and hence $\mu\nu$ are linear, since $A$ is abelian. The corollary also says that all the $\mu\nu$ are distinct and are all of the irreducible constituents of $(\lambda_{K \cap A})^A$. Hence $(\lambda^{KA})_A$ is the sum of distinct conjugates of the $\mu$. Now apply Clifford's Theorem ((6.11) Theorem), there must be a linear character $\psi \in I_{KA}(\mu)$, the inertia group of $\mu$, with $[\psi_A,\mu] \neq 0$, such that $\psi^{KA}=\lambda^{KA}$. Hence $\chi=\lambda^G=(\lambda^{KA})^G=(\psi^{KA})^G=\psi^G$ by transitivity of induction (see Problem (5.1)). So $H=I_{KA}(\mu)$ is the requested subgroup.