invariant subspaces of general linear groups for finite fields Let $K$ be a finite field, let $n\ge 1$ be an integer and let $G=\mathrm{GL}(n,K)$ be acting linearly on a finite dimensional $K$-vector space $V$. Although $G$ is a reductive group, it is not linearly reductive, so one cannot decompose $V$ as a direct sum of irreducible $G$-modules. Is there a simple way to describe all invariant $K$-vectors subspaces $W\subset V$?
 A: The short answer is no.
If $L$ is an infinite field containing $K$ then one may restrict a polynomial representation of $\mathrm{GL}(n,L)$, defined with polynomial coefficients in $K$, to a representation of $G$. Provided $K$ is sufficiently large, the composition factors do not change. As for representations of $G$, polynomial representations of $\mathrm{GL}(n,L)$ are in general not completely reducible. Finding their composition factors is a long-standing open problem in modular representation theory. It is well-known to include the decomposition number problem for the symmetric group as a special case (and in fact is equivalent to it).
Even the composition factors of easily defined representations such as $\mathrm{Sym}^r E$ where $E$ is the natural $n$-dimensional representation of $\mathrm{GL}(n,K)$ are unknown in general.
As a small example, taking $G = \mathrm{GL}(2,\mathbb{F}_8)$, the $G$-module $\mathrm{Sym}^4 E$ is $5$-dimensional and indecomposable with proper subspaces of dimensions $2$ and $3$. The corresponding composition series has irreducibles labelled by the highest weights $(4)$, $(2,2)$ (the determinant) and $(3,1)$, from  bottom to top. The behaviour rapidly gets even more intricate.
