Endomorphism ring of finite-dimensional representation If $G$ is any group and $V$ is a finite-dimensional representation of $G$, then we can form the endomorphism ring $E = End_G(V)$. Suppose that $V$ is indecomposable, i.e. not a direct sum of subrepresentations. 
Quals problem:(i) show that the nilpotent elements of $E$ form a two-sided ideal $I$ of $E$. (ii) Show that this ideal is nilpotent, i.e. there exists $n$ such that $I^n = 0$. 
For (i) The part that I'm blanking on is why a sum of nilpotents would be nilpotent. A useful side fact here is that any element of $E$ is a scalar plus a nilpotent, so I was thinking perhaps it's just a linear algebra fact that a sum of nilpotents can't be unipotent. 
 A: This is essentially Fitting's lemma: for any endomorphism $f$ of $V$ we have
$$V=\mathrm{ker}(f^n) \oplus \mathrm{im}(f^n)$$ where $n$ is the dimension of $V$ (key point of proof of this lemma: the sequence of subspaces $\mathrm{ker}(f) \subseteq \mathrm{ker}(f^2) \subseteq \cdots$ stabilizes after at most $n$ steps for dimension reasons). 
Since $f$ commutes with $G$ the summands here are $G$-submodules, and by indecomposability one or the other is equal to $V$: so $f$ is either nilpotent or invertible. Evidently if $f$ is nilpotent then $fg$ and $gf$ are not invertible for any endomorphism $g$; these are therefore nilpotent. If $f$ and $g$ are nilpotents with $f+g$ invertible, then for some endomorphism $h$ 
$$hf+hg=h(f+g)=1 \quad \implies \quad hg=1-hf.$$ But $hf$ being nilpotent means $1-hf$ is invertible, contradicting $hg$ nilpotent. So $f+g$ is not invertible and is therefore nilpotent.
(We have proved that the endomorphism ring of an indecomposable module is a local ring; the converse is also true).
