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Suppose $q$ is a prime $(\neq 2)$ and $G$ a finite group, for example the cyclic group $C_p$. Is there a way to determine all the $\textbf{indecomposable}$ $\mathbb{F}_{q^n}[G]$ modules for some $n\in \mathbb{N}$?.
I just found the irreducible modules for $n=1$. I would appreciate any help or hints where to find this in literature.
If $G$ has trivial Sylow $p$-subgroups, then the group ring is semi-simple and the indecomposables are exactly the irreducibles.
If $G$ has cyclic Sylow $p$-subgroups, then there is an algorithm to determine all indecomposable modules. You basically want to look at “Brauer tree algebras” (for instance Janusz's papers) to get the formula.
If $G$ has dihedral, quasi dihedral, or (generalized) quaternion Sylow 2-subgroups and $p=2$, then there is a reasonable description of the indecomposable modules, but they tend to be a bit infinite, and I've found them a bit hard to work with concretely.
If $G$ has non-cyclic, non-dihedral Sylow 2-subgroups, then there is (provably, I believe) no reasonable way to describe the indecomposables, though I've seen papers that provide some sort of description.
A module is called “absolutely indecomposable” if it remains indecomposable even when written over a larger field. For indecomposable modules over finite fields, it is very easy to tell what field a module is “really” over (these are called splitting fields, and are uniquely identified by a single positive integer more or less easily calculated).
For instance, the $F_3[C_4]$ module $x \mapsto \begin{bmatrix}0&-1\\1&0\end{bmatrix}$ is indecomposable (and irreducible), but the same $F_9[C_4]$ module decomposes. This is basically because $F_3$ was missing some eigenvalues, $±i$.
At any rate, given an indecomposable module $M$ over a finite field $F$, there is a field extension $K$, such that $K\otimes_F M$ is a direct sum of absolutely indecomposable modules, and every direct summand is of the form $M_0^\sigma$ where $M_0$ is absolutely indecomposable and $\sigma \in \operatorname{Gal}(K/F)$. One can simply take $K$ to be a field containing the $|G|$th roots of unity.
Back to the example: $x \mapsto \begin{bmatrix}0&-1\\1&0\end{bmatrix}$ decomposes as $x \mapsto i$ and $x\mapsto to -i$, but $i^\sigma = i^q = i^3 = -i$ so the Frobenius element of the Galois group swaps the two summands of the indecomposable, but not absolutely indecomposable, module.
A great example to have in mind when thinking about indecomposable (but not irreducible) representations is the representations of $\mathbb{Z}$ over the complex numbers. Here representations are just invertible matrices, two reps are equivalent if the matrices are conjugate, and direct sum decomposition means that you can conjugate so that it's block diagonal. If you think about it a little, the irreducible representations are 1-dimensional eigenspaces (so there's one for each non-zero complex number), and the indecomposables are the Jordan blocks.
The story for representations of cyclic groups over an algebraically closed field of finite characteristic is quite similar. Now you're only looking at matrices whose pth power is 1, but again the indecomposables are certain nice Jordan blocks. I won't spell it all out so as not to spoil your fun (per your comment).
