Search Results

Usually, "irreducible" means having no subrepresentations. In the integral context, there is no such thing, since for any $\mathbb{Z}_p[G]$-module $M$, $pM$ is a proper submodule. The right question i …

First, let me remark that the category of permutation representations in characteristic 0 is very far from being equivalent to the category of $\Gamma$-sets, since non-isomorphic $\Gamma$-sets can ind …

Your last statement is not true in general. Let $G=C_3$ and take your favourite finite field that does not contain the cube roots of unity, e.g. $K=\mathbb{F}_5$. Then the two non-trivial one-dimensio …

I will assume our algebra to have an identity. Question 1. How about: a representation is decomposable if and only if there exist two non-zero idempotent matrices $A_1$ and $A_2$ such that both com …

I will work in ${\rm GL}$ instead of ${\rm PGL}$. The corresponding question over ${\rm GL}_3(\mathbb{Z})$ is essentially$^1$ equivalent to asking how many faithful $\mathbb{Z}[G]$-modules, free of r …

This should have been a comment, but got too long. Even if we assume that everything is semi-simple and that the field is algebraically closed, as you seem to be doing, the scalars you are talking ab …

I think, the article by Vahid Dabbaghian-Abdoly, Journal of Symbolic Computation Volume 39, Issue 6, June 2005, Pages 671-688, entitled "An algorithm for constructing representations of finite groups" …

No classification of such groups is known. As you say, for every character to be a virtual permutation character, necessary conditions are that all irreducible characters are $\mathbb{Q}$-valued; eq …

As Keith says, such relations between permutation representations are often called Brauer relations, because Brauer was the first one to note that such isomorphisms of permutation representations give …

Yes, it is. If for any irreducible $\rho$, $\langle\text{Ind}_H^G(1),\rho\rangle$ is either 0 or dim $\rho$, then $H$ is the intersections of $\ker \rho$ for those $\rho$ for which this inner product …

To answer the question about natural examples/classification, Tim Dokchitser and I have completely classified all such sets in the following sense. If $\tilde{G}\leq G$, and $M$, $N$ are two $\tilde{ …

A quick google search produced this paper. It gives generators and relations for the Heisenberg group over rings of integers of quadratic fields and discusses its representations.

Let me try this. I think that the answer this time is positive. Step 1: We will first reduce to $p$ power order. Let $M$, $M'$ be matrices over $\mathbb{F}_p$ of order $p^na$, where $p\nmid a$. Then, …

$\DeclareMathOperator{\GL}{GL} \DeclareMathOperator{\PGL}{PGL} \DeclareMathOperator{\mcd}{mcd} \newcommand{\C}{\mathbb{C}}$Such groups do not exist. Indeed, suppose that $G$ has even order and satisfi …

In the infinite family $G_p=(\mathbb{Z}/p\mathbb{Z})\rtimes (\mathbb{Z}/p\mathbb{Z})^{\times}$, as $p$ runs over prime numbers, the bound is tight up to $O(1)$, and the representation in question is f …