Indecomposable modules of a $p$-group in characteristic $p$ Let $k$ be a field of characteristic $p>0$ and let $P$ be a $p$-group. Then the group ring $kP$ has only one simple module, the trivial module $k$, and hence there is only one projective indecomposable module, which is the projective cover of $k$, namely $kP$. So to understand the representation theory of $kP$, we look at the non-projective indecomposable modules. 
On page 92 of his book Representations and Cohomology:I, Benson says

Thus every non-projective indecomposable $kP$-module has $\DeclareMathOperator{Soc}{Soc}$$\Soc(kP)$ in its kernel.

In "its kernel"? If $M$ is the non-projective indecomposable module  in question, does he mean the kernel of the map giving the action, $kP\rightarrow \DeclareMathOperator{End}{End}\End(M)$?
For more context, he says

Thus $kP$ has a unique minimal left ideal, isomorphic also isomorphic to the trivial module. It is equal to the last non-zero power of the radical, and is hence a two-sided ideal. We write $\Soc(kP)$ for this minimal ideal. Since the projective indecomposable $kP$-module is also injective, whenever it is a submodule of another $kP$-module, it is a summand. Thus every non-projective indecomposable $kP$-module has $\DeclareMathOperator{Soc}{Soc}$$\Soc(kP)$ in its kernel. So studying the representation theory of $kP$ is almost the same as studying the representation theory of $kP/\Soc(kP)$. 

I think what I mentioned above is what he meant, though it's not completely clear. If the action is faithful, then $kP$ is a summand of $\End(M)$, since $kP$ is injective. Since $M$ is indecomposable, $\End(M)$ is a local algebra (every element of $\End(M)$ is nilpotent or invertible). So then every element of $kP$ is either nilpotent or invertible, (which I'm guessing one can always find a counterexample for.)
So the action $kP\rightarrow \End(M)$ has nontrivial kernel. Now why is the minimal left ideal of $kP$ contained in the kernel? 
 A: I think the best interpretation of this statement is "$\operatorname{soc}(kP)$ acts trivially on every nonprojective indecomposable $M$."  
To see why this is true, consider the fact that any nonzero submodule of $kP$ must contain the socle: any nonzero submodule of $kP$ contains a simple submodule of $kP$, but there's only one simple submodule of $kP$, namely the socle.   Now if $m \in M$ and $\operatorname{soc}(kP)$ doesn't kill $m$, the map $kP \to M$ given by $1 \mapsto m$ is injective (its kernel is a submodule not containing the socle, so is zero), so since $kP$ is injective, $M$ has a projective summand --- contradiction.
The socle of $kP$ is in fact a two-sided ideal, so $\overline{kP} = kP/\operatorname{soc}(kP)$ is an algebra.  What the above tells you is that every $kP$-module with no projective summands can be made into a $\overline{kP}$-module, and conversely every $\overline{kP}$-module is naturally a $kP$-module with the socle acting as zero, i.e. with no projective summands.  This is the sense in which the rep theory of $kP$ is almost the same as that of $\overline{kP}$.
