Let $G$ be a finite group and $H \le G$ with $g \in G$ such that all elements of the coset $Hg$ are conjugate in $G$. Let $\chi$ be a $\mathbb C$-character of $G$ such that $[\chi_H, 1_H] = 0$. Show that $\chi(g) = 0$.

This is exercise 2.1 (b) from M. Isaacs, Character Theory of Finite Groups; it is given the hint to look at the trace of $\sum_{h \in H} \mathcal X(hg)$, where $\mathcal X$ is a $\mathbb C$-representation for $\chi$.

A $\mathbb C$-representation is a homomorphism $\mathcal X : G \to GL(n, \mathbb C)$ which leads naturally to a represententation of the algebra $\mathbb C[G]$. Further for two class functions $\varphi$ and $\vartheta$ on a group $G$, the inner product is $$ [\varphi, \vartheta] = \frac{1}{|G|} \sum_{g\in G} \varphi(g) \overline{\vartheta(g)}. $$ For a character $\chi$, the notation $\chi_H$ just means the restriction to $H$.

Do you have any ideas how to solve this problem?


Following we assume that the part (a) of the exercise 2.1 is correct.

Exercise 2.1 (a): Let $\mathcal{X}$ be an irreducible $F$-representation of $G$ over an arbitrary field. Show that $\sum_{g \in G}{\mathcal{X}(g)}=0$ unless $\mathcal{X}$ is the principal representation.

Proof of Exercise 2.1 (b):
Suppose that $\chi_1, ..., \chi_k \in Irr(H)$ are irreducible constituents of $\chi_H$ and $\mathcal{X}_1, ..., \mathcal{X}_k$ are corresponding irreducible representations of them. The proof consists of three stepts:

(1) We are going to show that $\sum_{h \in H}{\mathcal{X}(h)}=0$. Consider the diagonal block decomposition of matrices $\mathcal{X}(h)$ to their irreducible components, in which $\mathcal{X}_1(h), ..., \mathcal{X}_k(h)$ make the diagonal blocks of $\mathcal{X}(h)$. So it is sufficient to show that $\sum_{h \in H}{\mathcal{X}_i(h)}=0$ for each $1 \leq i \leq k$.
These latter equalities follow from part (a) of the excercise, since by $[\chi_H, 1_H]=0$ none of the irreducible representations $\mathcal{X}_i$ are the principal representation. So we have $\sum_{h \in H}{\mathcal{X}(h)}=0$.

(2) Since $\mathcal{X}$ is a representation of $G$, by (1) we have $\sum_{h \in H}{\mathcal{X}(hg)}=\sum_{h \in H}{\mathcal{X}(h)\mathcal{X}(g)}=\big(\sum_{h \in H}{\mathcal{X}(h)}\big)\mathcal{X}(g)=0$.
So $\sum_{h \in H}{\mathcal{X}(hg)}$ is a zero matrix.

(3) Remember that $\chi$ is a character of $G$ afforded by the representation $\mathcal{X}$. For each $h \in H$ we have $\chi(hg)=\chi(g)$, since all elements of the coset $Hg$ are conjugate in $G$. So
$\sum_{h \in H}{\chi(hg)}=\sum_{h \in H}{\chi(g)}=\mid H \mid \chi(g)$.

Now since $tr\big(\sum_{h \in H}{\mathcal{X}(hg)}\big)=\sum_{h \in H}{\chi(hg)}=|H| \chi(g)$ and by step (2) we know that$\sum_{h \in H}{\mathcal{X}(hg)}$ is a zero matrix, we must have $tr\big(\sum_{h \in H}{\mathcal{X}(hg)}\big) =|H| \chi(g)=0$ and so $\chi(g)=0$. Q.E.D


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