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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 $g\in G\backslash H$ …

$\DeclareMathOperator\M{M}\DeclareMathOperator\Gal{Gal}\DeclareMathOperator\End{End}$As has been mentioned in the comments, the question for algebraically closed fields of characteristic $0$ is equiva …

Today in my research, I had to use fairly explicitly the rather tautological property of finite cyclic groups that every normal subgroup is characteristic, i.e. fixed by all automorphisms. This got me …

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" in general. There may be an elementary way of seeing this, but I will frame this in representation theoretic terms and will describe a general construction. The question is equivale …

The answer is "no", since for every sufficiently large prime $p$ there are simple non-trivial $\mathbb{F}_p[{\rm PSL}_2(\mathbb{F}_{2^f})]$-modules. You can take $N$ to be any such module and form the …

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 …

The answer is "no", in general, since there may be local obstructions. Suppose, for example, that $k$ and $n$ are odd prime powers, and let $L/\mathbb{Q}$ be the unique intermediate quadratic in $F$. …

A partial answer to Question 2: the following is a theorem of Burnside (see e.g. Isaacs, Theorem 3.8). Theorem. Let $\chi$ be an irreducible character, let $K$ be a conjugacy class of $G$, and let $g\ …

$\DeclareMathOperator{\SL}{SL}\DeclareMathOperator{\GL}{GL}$To determine the real representations of a finite group, it suffices to determine the complex irreducible representations and their Schur in …

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 …

The following is wrong, see comment section: The Schur index over $K$ is, among other things, the degree of a minimal field extension of $K$ over which the underlying representation can be realised o …

The answer to the first question is negative. The group ${\rm SL}_2(\mathbb{F}_3)$ has a $2$-Sylow subgroup isomorphic to $Q_8$, which is not determined by its character table, but ${\rm SL}_2(\mathbb …

$\DeclareMathOperator{\GL}{GL} \DeclareMathOperator{\SL}{SL}$ The maximum of the function counting square roots is attained at $x_0=1$ and this statement generalises quite well. Let $s(\chi)$ denote …

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