Injective Restriction of Brauer Characters

TL;DR

This paper characterizes conditions under which the restriction of Brauer characters from a finite group G to a subgroup H is injective, providing insights into the structure of Brauer characters and their restrictions.

Contribution

It offers a new characterization of when the restriction map of Brauer characters is injective for subgroups of finite groups.

Findings

Identifies specific conditions for injectivity of Brauer character restriction

Provides a theoretical framework for understanding Brauer character restrictions

Enhances understanding of subgroup structure via Brauer characters

Abstract

If G is a finite group and H is a subgroup of G, we characterize when restriction of Brauer characters from G to H is injective.

Equations

$$\mathbb{Z}[\mathrm{IBr}(G)]\to \mathbb{Z}[\mathrm{IBr}(H)]$$

Full text

Research supported by the Prometeo/Generalitat Valenciana, Proyecto MTM2016-76196-P and FEDER funds.

Let $G$ be a finite group, let ${\rm Char}(G)$ be the set of complex characters of $G$ and let $H$ be a subgroup of $G$. In [IN], M. Isaacs and this author have shown that restriction of characters ${\rm Char}(G)\rightarrow{\rm Char}(H)$ is never injective unless $H=G$.

If $p$ is a prime, we choose a maximal ideal of the ring of algebraic integers in ${\mathbb{C}}$ containing $p$, and we calculate Brauer characters of finite groups with respect to this ideal. Then it is also natural to ask when restriction from the Brauer characters ${\rm Br}(G)$ of $G$ to ${\rm Br}(H)$ is injective. (Or equivalently, when restriction from the ring of generalized Brauer characters ${\mathbb{Z}}[{\rm IBr}(G)]$ to ${\mathbb{Z}}[{\rm IBr}(H)]$ is injective.)

Essentially, the same proof in [IN] gives the following.

[TABLE]

is injective if and only if every $p$-regular element of $G$ is in some $G$-conjugate of $H$.

There are interesting non-trivial examples where the hypotheses of Theorem A are satisfied, even with $|{\rm IBr}(G)|=|{\rm IBr}(H)|$. For instance, if $G=A_{5}$ and $H=D_{10}$ with $p=3$; or if $G=A_{5}$ and $H=S_{3}$ for $p=5$.

The following is an immediate consequence.

[TABLE]

is an isomorphism, then $K\mapsto K\cap H$ is a bijection between the set of $p$-regular conjugacy classes of $G$ and of $H$.

The converse of Corollary B is not true. For instance, if $G=A_{5}$, $H=D_{10}$ and $p=3$, then $K\mapsto K\cap H$ is a bijection between the set of $p$-regular conjugacy classes of $G$ and of $H$. Hence, the restriction ${\rm Res}:{\mathbb{Z}}[{\rm IBr}(G)]\rightarrow{\mathbb{Z}}[{\rm IBr}(H)]$ is injective by Theorem A. However, the matrix of ${\rm Res}$ does not have determinant $\pm 1$, so it cannot be an isomorphism of free abelian groups.

There are some solvable examples too: if $p=2$, $G$ is the unique group of order $72$ with Fitting subgroup $C_{3}\times C_{3}$ and cyclic Sylow 2-subgroup, and $H$ is any subgroup of $G$ isomorphic to $S_{3}$.