A finite group $G$ is called comaximal if for any non-trival irreducible representations $V$ and $W$ of $G$, it exists $n \in \mathbb{N} \ $ such that $(V^{\otimes n},W)\ge 1$.
A finite group $G$ is called simple if for any subgroup $H \subset G$: ($\forall g \in G, \ gHg^{-1} \subset H$) $\Rightarrow H \in \{1,G\}$.
Question: Is comaximal equivalent to simple?
This is true for complex representations, where it becomes a statement about characters. If $G$ is simple, and $\chi$ is any non-trivial irreducible character of $G$, a theorem of Burnside/Blichfeldt, for any other irreducible character $\mu$ of $G$, we have $\langle \chi^{n}, \mu \rangle \neq 0$ for some positive integer $n.$ On the other hand, if we have that property of tensor powers, then every non-trivial irreducible character of $G$ must be faithful. Hence $G$ can have no proper non-trivial normal subgroup $N$ (otherwise, we could take $\chi$ a non-trivial irreducible character of $G/N$, and then every irreducible character of $G$ would have $N$ in its kernel).
If we work in characteristic $p >0$, we have to assume that $O_{p}(G) = 1$. Then a slightly more complicated version of this argument shows that the tensor condition is equivalent to $G/O_{p}(G)$ being simple.