For a partition $\lambda$ let $S^{\lambda}$ be the corresponding Schur functor. Is it true that for every $\lambda$ there exists an irreducible representation $V$ of a finite nonabelian group $G$ such that $S^{\lambda}(V)$ is still irreducible?
This is not obvious to me even for the symmetric and exterior powers (although maybe I'm not thinking hard enough), so any partial results would be appreciated.
Since the Guralnick and Tiep paper is very long, I thought I would summarize my understanding of it. Note that I only learned about this result from moonface last night, so I am hardly an expert. This answer is community wiki, in case anyone can improve on my summary.
We want to establish the following result: Let $G$ be a finite noncommutative subgroup of $GL(V)$, with $V$ a $\mathbb{C}$ vector space. Let $k \geq 6$. Then $\mathrm{Sym}^k(V)$ is reducible. (If $G$ is commutative and $V$ is one dimensional then, of course, $\mathrm{Sym}^k(V)$ is always one dimensional and hence irreducible.)
Our proof is by induction on $|G|$. Our base case will be when $G$ is a central extension of a simple group.
Choose a nontrivial normal subgroup $H$ of $G$ such that $G/H$ is simple. If we can't do this then $G$ is simple and we are in the base case. Let $V \cong \bigoplus U_i$ be the decomposition of $V$ into $H$-isotypic components. If there is more than one summand, then $\bigoplus \mathrm{Sym}^k(U_i)$ is a nontrivial $G$-subrep of $\mathrm{Sym}^k(V)$. So we may assume that $V \cong W \otimes X$, where $W$ is an $H$-irrep and $H$ acts trivially on $X$. Then one can show (proof of lemma 2.5) then $G$ is contained in $GL(W) \times GL(X)$. (This is nontrivial but not deep; you could give it as a problem in a graduate-level representation theory course.)
Then $\mathrm{Sym}^k(W) \otimes \mathrm{Sym}^k(X)$ is a subrepresentation of $\mathrm{Sym}^k(V)$. This subrep is proper unless $\dim X=1$ or $\dim W=1$. If $\dim X=1$, then $V$ is an irrep of $H$ and $\mathrm{Sym}^k(V)$ is irreducible as an $H$-rep, so we are done by induction.
If $W$ is one dimensional, then $H$ acts on $V$ by scalars, so $H$ is central. So we are in our base case: a central extension of a simple group. This case is done by group cohomology and the classification of simple groups. See section 4 for groups of Lie type, section 6 for alternating groups and section 7 for sporadic groups.
By averaging, over $\mathbb R$ every representation of a finite group can be given a positive definite inner product so that the action is orthogonal. And my memory is that Schur functurs of representations of $\operatorname{O}(n)$ are not usually irreducible. For example, $\operatorname{Sym}^2(\mathbb R^n)$ has a one-dimensional summand over $\operatorname{O}(n)$, given by the trace w.r.t. the inner product on $\mathbb R^n$. This suggests to me that over any finite group, the essential image of $\operatorname{Sym}^2$ contains very few irreps. But maybe I am in error.
$f$can be computed explicitly or if the boundedness is somehow shown to be true or false? The question is intriguing, but also puzzling. Classical Schur-Weyl theory gives good algorithmic knowledge of the constituents of tensor powers of $V$` for general linear groups over$\mathbb{C}$; but neither this nor the finite subgroups are known in closed form as$n$grows (eventually all finite groups appear). $\endgroup$