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Only the trivial bound $2|G|$, because the alternating groups $A_n$, $n\neq 6$ make this sharp. See e.g. the Wikipedia article.

No, it does not. As a simple example, let $G\cong S_3$, $\Omega = G/C_2 \sqcup G/C_2 \sqcup G/C_3$. You can easily check that the corresponding permutation character is isomorphic to the regular chara …

Note that $\text{Inn}(X)$ is isomorphic to $X/Z(X)$. So your requirement is equivalent to $X$ being simple modulo the centre. For example $SL_n(\mathbb F_{p^m})$ satisfies this, since the centre is gi …

I think that the commutator subgroup of $SL(2,\mathbb{Z})$ is an index 2 subgroup of $\Gamma(2)$, rather than $\Gamma(2)$ itself. Indeed, $\Gamma(2)$ has index 6 in $SL(2,\mathbb{Z})$, but you can wri …

In an answer to an earlier question, I showed how to prove that the square root counting function $r_2: S_n\rightarrow \mathbb{N},\;g\mapsto \#\{h\in S_n|h^2=g\}$ assumes its maximum at the identity, …

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 …

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 …

The answer is no. You can easily have a situation where $f$ is the trivial map, while $g$ makes $H$ act through an inner automorphism of $N$, so that in both cases $N\rtimes H\cong N\times H$. For con …

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 local-global principle you are citing comes from the fact that for any open subgroup $H\leq G$, $H^n(G,A)\stackrel{\text{Res}}{\longrightarrow}H^n(H,A)\stackrel{\text{Cor}}{\longrightarrow}H^n(G,A …

Yes, it is. If for any irreducible $\rho$, $\langle\text{Ind}_H^G(1),\rho\rangle$ is either 0 or dim $\rho$, then $H$ is the intersections of $\ker \rho$ for those $\rho$ for which this inner product …

To answer the question about natural examples/classification, Tim Dokchitser and I have completely classified all such sets in the following sense. If $\tilde{G}\leq G$, and $M$, $N$ are two $\tilde{ …

A quick google search produced this paper. It gives generators and relations for the Heisenberg group over rings of integers of quadratic fields and discusses its representations.

If I understood Joel David Hamkins's explanation correctly, then the problem of classifying finite ($p$-)groups is smooth for silly reasons. Attach to each group the following datum: randomly number t …

&cat[[h: h in Conjugates(G,H`subgroup) | S subset h]: H in Subgroups(G)]; will create a list of all subgroups of $G$ containing a given group $S$. The main caveat here is that the function Subgroups …