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Not a complete answer. Your definition of an extendible map says that $\beta$ is an endomorphism of the forgetful functor $U$ from the under category $G \downarrow \mathrm{FinGrp}$ to $\mathrm{Set}$ s …

The rings $\mathbb{Z}/n$ are the only examples. I assume that "primitive character" just means that it is faithful, or equivalently that it does not factor through a proper quotient; this is the meani …

For what it's worth, I had exactly the same question as you and worked out the proof that Noah sketches in some detail at The orthogonality relations for representations of finite groups, although I d …

Kurokawa's Zeta functions of categories contains the following definitions. Definition 1: In a category $C$ with a zero object, a simple object is an object $X$ such that, for every object $Y \in C$, …

Your setup is much more specific than necessary. All you need is two rings $R, S$ with commuting actions on an abelian group $M$ (which is therefore an $(R, S^{op})$-bimodule) such that $M$ is semisim …

As mentioned in the comments, conjecturally almost all finite groups are $2$-step nilpotent $2$-groups, so conjecturally the answers to 1) and 3) are that the limits both exist and both equal $1$; tha …

Yes. It's necessary and sufficient to show that every finite abelian group admits a nondegenerate quadratic form valued in a finite cyclic group. The following slightly stronger statement is true: eve …

A simple bound on the largest dimension of a complex irreducible representation (which is either equal to or half of the largest dimension of a real irreducible representation) is the following: we kn …

Here are some things you probably know. For a representation $W$ of $G$, let $\text{Inv}(W)$ denote the subspace of $G$-invariants. For an irreducible representation $V$ with character $\chi$, the F-S …

This is (part of) Theorem 6.13 in Serre's Finite Groups: An Introduction, which says the following. Say that an action of a group $H$ on another group $N$ is almost free if the action on $N \setminus …

The answer to questions 0 and 1 is yes. Here is a generalization. Claim: Let $\pi$ be a finitely generated group and $G$ be a finite group. Then $$\frac{|\text{Hom}(\pi \times \mathbb{Z}, G)|}{|G|}$$ …

This computation could in principle be done using Clifford theory. Clifford theory tells you how to describe the representation theory of a group $G$ given that it can be described as an extension $$ …

I'll work over $\mathbb{C}$ below for simplicity, although it can be replaced by an algebraically closed field of characteristic $0$. Assuming that "fusion subcategory generated by" means what I thi …

The fusion ring, as a ring with basis, contains the same information as the character table. So your question, phrased in language more familiar to finite group theorists, is: Is a finite simple g …

Here is a more conceptual approach to Clifford theory. Let me work with a slightly more general setup: namely, suppose we have a short exact sequence $$1 \to N \to G \to H \to 1$$ of finite groups, …