Indecomposble Direct Summand of Nilpotent Endomorphism of Quiver Representation

Question

I was reading the proof of Gabriel's Theorem (Theorem 8.12) from Ralf Schiffler's 'Quiver Representations'. The proof begins by showing that for an indecomposable module $M$, the vector space $Hom(M,M) \cong k$ (where $k$ is the underlying field) by inducting on the dimension of $M$. In the induction step, he considers an arbitrary indecomposable quiver representation $M$, assuming the result to be true for all indecomposable representations of dimension less than that of $M$. Assuming for the sake of contradiction that $End(M) \not\cong k$, the indecomposability of $M$ implies that every endomorphism of $M$ is of the form $\lambda id_M + g$ for some nilpotent endomorphism $g$ and some $\lambda \in k$, so our assumption implies that there is some $f := \lambda id_M + g \in End(M)$, and by passing to a higher power of $g$ if necessary we may assume (without loss of generality) that the least exponent $m \in \mathbb{N}$ for which $g^m$ is identically zero is $m=2$. Moreover among all nonzero endomorphisms of $M$ whose square is zero we may choose the one with $dim$ $im$ $g$ minimal. Then $g^2 = 0 \neq g$ so that $im$ $g \subset ker$ $g$, and this is where he says, "and hence there exists an indecomposable summand $L$ of $ker$ $g$ such that $im$ $g \cap L$ is nonzero". Unfortunately it is not obvious to me why the last statement (in quotes) follows immediately from previous arguments. [It seems to me that the minimality in the dimension of $im$ $g$ (besides the fact that $g \neq 0 = g^2$) will be probably used somewhere, but I don't exactly see where.]