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There are quite a lot of results in the literature about the number of commutators needed to express an element of the derived group, and the character table is useful here. These are generally rather …

One relationship is that the number of $p$-singular elements ( that is, elements whose order is divisible by $p$) is divisible by the number of Sylow $p$-subgroups of $G$. This is a consequence of a t …

Well, it depends what you are prepared to accept as a reasonable answer. A finite group is completely reducible if and only if the intersection of all its maximal normal subgroups is trivial. This is …

Let's suppose that $p >3$ (otherwise, the groups is solvable in any case). I also work with $G = {\rm SL}(2,\mathbb{Z}/p^n \mathbb{Z})$, but there is an obvious correspondence between what happens for …

Just a remark : if $A$ and $B$ are non trivial finite $p$-groups, each acting faithfully as transitive permutation groups, then the action of $ A \wr B $ is never primitive as a permutation action. Fo …

It is still the case that the $\mathbb{F}_{q}$-valued trace functions of the (say) $\ell$ non-isomorphic simple $\mathbb{F}_{q}$-modules $V_{1},V_{2}, \ldots V_{\ell}$, are linearly independent, where …

If you are speaking of a finite dimensional irreducible $G$-module (for a possibly infinite group $G$), then there is little difference from the standard argument for finite groups when dealing with …

Let $d = \phi(n)$ and assume $n>2$.Then the primitive $n$-th roots of unity occur in $d/2$ complex conjugate pairs, and the GM-AM inequality applied to the (positive) contributions from each pair give …

David Speyer has answered the question, but let me add some background. The general result is that if $R$ is a principal ideal domain with field of fractions $K$, and $G$ is a finite group, then every …

If $|G|= \prod_{i=1}^{r} p_{i}^{n_{i}}$ where the $p_{i}$ are distinct primes and each $n_{i}$ is a positive integer, then any minimal generating set of $G$ has at most $\sum_{i=1}^{r}n_{i}$ elements …

I think that the assumption that $A_{m} \cap A_{n} = 1$ may get in the way. You may need to drop it for inductive reasons. For example, you can take a group $G$ such that $G = A_{m}A_{n}$ for subgroup …

Certainly not in general : For an extreme case, consider the case that $n = q-1,$ where $q$ is a power of $2$. Let $H$ be a subgroup of order $(q-1)^{3}$ of ${\rm GL}(V),$ generated by a scalar matrix …

The result is true for Hall subgroups in solvable groups, but not in general. I don't have the references to hand, but it's a Theorem of Brauer, or maybe E.C. Dade, or maybe Suzuki, that if $G$ has a …

There's an old book by H.F. Blichfeldt called "Finite Collineation Groups" (University of Chicago Press, 1917 I believe) which deals with the classification of finite linear groups in low dimensions, …

Not really an answer, but this is already difficult in the complex case, when $C$ is a central subgroup. For example, if $C = Z(G),$ and we induce a faithful irreducible $C$-module to $G,$ the number …