Representation theory problem 6.2 Isaacs.

Question

Let $ N \unlhd G $ and suppose $G/N $ is abelian. Let $C$ be the group of linear characters of $G/N $ so that $C$ acts on $Irr(G)$ by multiplication. Let $\vartheta \in Irr(N)$. Show that

$\vartheta^G=f \sum \chi_i $ where $f$ is an integer and $\chi_i \in Irr(G)$ contistute an orbit under $C$.

Try: I tried using Gallagher's theorem but couldn't use it somehow that would give me that $\beta\chi$ is a constituent, where $\chi \in Irr(G) $ such that $\chi_N =\vartheta$ and $\beta \in Ιrr(G/N) $. But First I don't know if such $\chi$ exists. Second I also know that those $\beta\chi$ for $\beta$ running in $G/N$ would give me indeed irreducible characters so they would be ok for the $\chi_i$'s the problem asks for. Thats what I have thought so far.

COROLLARY (Gallagher) Let $N \unlhd G$ and let $\chi \in Irr(G)$ be such that $\chi_N = θ \in Irr(N)$ . Then the characters $\beta\chi$ for $\beta \in Irr(G/N)$ are irreducible, distinct for distinct $\beta$ and are all of the irreducible constituents of $\vartheta^G$.

Answer 1

If $\phi$ is a class function on $N$ and $\psi$ is a class function on $G$, there is the following well-known formula $(\phi \psi_N)^G=\phi^G\psi$. Then if $\chi$ is an irreducible character of $G$ over $\theta$, one has that $(\chi_N 1_{N})^G=\rho_{G/N} \chi$. Since $\theta$ is a constituent of $\chi_N$, $\theta^G$ is a constituent of $\rho_{G/N}\chi$. From this follows that the irreducible constituents of $\theta$ are of the form $\lambda \chi$, with $\lambda\in \hat{G/N}$. Then they have the same multiplicity in $\theta^G$ by Frobenius reciprocity (their restriction is always $\chi_N$). It is also clear from above that they are an orbit for the multiplicative action of $\hat{G/N}$.

Answer 2

Let $N \unlhd G$ with $G/N$ abelian. Then we can identify $Irr(G/N)=\{\lambda \in Irr(G): N \subseteq ker(\lambda) \}$. Note that $\lambda(1)=1$, if $\lambda \in Irr(G/N)$. In addition (see for example Problem (5.3) in M.I. Isaacs, Character Theory of Finite Groups), if $\chi \in Irr(G)$, then $(\chi_N)^G=(\chi_N \cdot 1_N)^G=\chi \cdot (1_N)^G$. And, in general, for a linear $\lambda \in Irr(G)$, the product character $\lambda\chi \in Irr(G)$.
Now, $(1_N)^G=\sum\limits_{\lambda \in Irr(G/N)}\lambda$, the regular character of $G/N$. Hence $$(\chi_N)^G=\sum\limits_{\lambda \in Irr(G/N)} \lambda\chi \tag{1}$$

Let $\vartheta \in Irr(N)$ and let $\chi \in Irr(G)$ lying above $\vartheta$, say (applying Clifford) $$\chi_N=e\sum\limits_{i=1}^{t} \vartheta_i,\tag{2}$$ where the $\vartheta_i$'s are the distinct $G$-conjugates of $\vartheta$ and $t=|G:I_G(\vartheta)|$, the index of its inertia subgroup. Since $N$ is a normal subgroup of $G$, we have $\vartheta^G=\vartheta_i^G$ for each $i$, hence from $(1)$ and $(2)$ it follows $$et\vartheta^G=\sum\limits_{\lambda \in Irr(G/N)} \lambda\chi \tag{3}$$ Now let us calculate the multiplicity of each of $\lambda\chi$'s appearing as an irreducible constituent in the right hand-side of $(3)$ using Frobenius Reciprocity. Since $N \subseteq ker(\lambda)$, we have $\lambda_N=1_N$, hence $$[(\chi_N)^G,\lambda\chi]=[\chi_N,(\lambda\chi)_N]=[\chi_N,(\lambda_N)(\chi_N)]=[\chi_N,\chi_N]=e^2t$$ where the last equality follows from $(2)$. So we can rewrite $(3)$ as $$et\vartheta^G=e^2t\sum\chi_i \tag{4}$$ where the $\chi_i$ now constitute an orbit under the action of $Irr(G/N)$ on $\chi$. From $(4)$ it follows that $$\vartheta^G=e\sum\chi_i \tag{5}$$ If $s$ is the size of the orbit above, then evaluating the character degrees at both sides in $(2)$ and $(5)$ yields $\vartheta(1)|G:N|=es\chi(1)=es\cdot et\vartheta(1)=e^2st\vartheta(1)$, so we find $|I_G(\vartheta):N|=e^2s$. With the notation above, we get the following consequence (see also, R.L. Roth, A dual view of the Clifford Theory of Characters of Finite Groups, Can. J. Math., Vol. XXIII, No. 5, 1971, pp. 857-865, Theorem(2.1))

Corollary (R.L. Roth, 1971) Let $N \unlhd G$ with $G/N$ abelian and $\chi \in Irr(G)$. Then the action of $Irr(G/N)$ fixes $\chi$ if and only if $\vartheta^G=e\chi$.
Proof If every $\lambda \in Irr(G/N)$ fixes $\chi$, then from $(5)$ we get $\vartheta^G=e\chi$. Conversely, since $N$ is normal, $\vartheta^G$ vanishes outside $N$, whence $\lambda\chi=\chi$ outside $N$. And of course the restriction $(\lambda\chi)_N=\chi_N$, since $N \subseteq ker(\lambda)$. So $\lambda\chi$ and $\chi$ agree on the whole $G$. $\square$