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

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 …

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 …

$\DeclareMathOperator{\GL}{GL} \DeclareMathOperator{\PGL}{PGL} \DeclareMathOperator{\mcd}{mcd} \newcommand{\C}{\mathbb{C}}$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 …

As Francesco says, the smallest example is $D_{10}$. It is a normal subgroup of the Frobenius group of order $20$ and that extension is not split, as can be seen from looking at the $2$-Sylows. There …

See Gorenstein, Finite Groups, Chapter 5, Theorem 4.10. Such groups are as follows: if $p$ is odd, then $G$ is cyclic; if $p=2$, then $G$ is either cyclic, or generalised quaternion of order $2^l$ f …

This is not a complete answer, but it's too long for a comment. Here is what I would try (for $G=C_p\rtimes C_2$ for any $p$): your $M$ sits in the exact sequence $$ 0\rightarrow M\rightarrow \mathbb{ …