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Let $G$ be a finite group. In an an earlier question, Fedor asked whether the square root counting function $r_2:G\rightarrow \mathbb{N}$, which assigns to $g\in G$ the number of elements of $G$ that …

The question is in the title, but here is some background/reminders: A subgroup $H\neq\{1\}$ of a finite group $G$ is called a Frobenius complement if $H\cap H^g = \{1\}$ for all $g\in G\backslash H$ …

This is the second follow-up to this question on square roots of elements in symmetric groups and is concerned with generalisations to $n$-th roots. Let $G$ be a finite group and let $r_n(g)$ be the n …

There is no complete classification, but some structural results are known. To give you something to search for: such groups are called $\mathbb{Q}$-groups. There is a whole book devoted to their stru …

The answer is no. Counterexamples include the Weyl groups of types E6, E7, and E8. For a proof that the representations of these Weyl groups are indeed all realisable over $\mathbb{Q}$ (equivalently o …

Let $p$ be an odd prime and let $F/\mathbb{Q}$ be a Galois extension with Galois group $D_{2p}$, let $K$ be the intermediate quadratic extension of $\mathbb{Q}$, and $L$ an intermediate degree $p$ ext …

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 is an answer to the second question: I ran an experiment with $S_6$ (which was the best guess due to the famous "oddness" of 6). There are two subgroups in $S_6$ isomorphic to this involution cen …

I think I finally have a correct answer for arbitrary $p$. As F. Ladisch notes, $G=C_{p^3}$ has only finitely many indecomposable modular representations. For the following argument, I will not only n …

What is known about the Sylow 2-subgroups of $\rm{PGL}_n(\mathbb{F}_q)$, where $q$ is any prime power? For example, according to Theorem 7.9 in Isaacs's Character Theory, these Sylow 2-subgroups canno …

Here is a direct way to obtain Denis Serre's formula: Just note that $x_i^k\frac{\partial^k}{\partial x_i^k}$ multiplies a monomial in the determinant by $\frac{r!}{(r-k)!}$ where $r$ is the power of …

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 …

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 …

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 …