# Expressing $\sum_{g\in [G/H]}ge_Hg^{-1}\in Z(\mathbb{C}[G])$ in terms of primitive central idempotents?

Suppose $$G$$ is a finite group, and $$H$$ a subgroup. For an irreducible character $$\chi$$ of $$G$$, there is a central idempotent in the group algebra $$\mathbb{C}[G]$$: $$e_\chi=\frac{\chi(1)}{|G|}\sum_{g\in G}\chi(g^{-1})g.$$

I write $$e_H:=e_{1_H}=\frac{1}{|H|}\sum_{h\in H}h$$ for the idempotent in $$\mathbb{C}[H]$$ corresponding to the trivial character on $$H$$. If $$[G/H]$$ denotes a complete set of left coset representatives of $$H$$ in $$G$$, then by construction the element $$\sum_{g\in [G/H]}ge_Hg^{-1}$$ is central in $$\mathbb{C}[G]$$. Is there any way to somewhat explicitly extract which characters have their corresponding central idempotents in this linear combination, and/or their multiplicity? Just looking at the Mackey formula and since $$ge_Hg^{-1}=e_{gHg^{-1}}$$, my guess is it may be characters that are constituents of $$\operatorname{Ind}^G_H(1_H)$$, or something similar, but I am unsure.

• Yes, your central element is $\operatorname{Ind}_H^G\left(1_H\right)$ when regarded as a character of $G$. Indeed, look at the formula (4.1.3) in Darij Grinberg, Victor Reiner, Hopf Algebras in Combinatorics, arXiv:1409.8356v5 and set $U = 1_H$ (so that $\chi_U$ is the function that is constantly $1$). The right hand side of the formula differs from your sum in that it sums over all $g \in G$ as opposed to only over left coset representatives; but to balance this out, it is being divided by $\left|H\right|$. – darij grinberg Apr 12 at 5:35
• @darijgrinberg Thank you, Darij, I'll take a look. This book looks like a valuable resource to boot. – Denise Gi Apr 12 at 6:53