Let $G$ be an algebraic group with closed normal subgroup $N$. Suppose that $N$ and $G/N$ are both unipotent. Does it imply that $G$ is also unipotent?
By definition, G is unipotent if and only if every nonzero representation has nonzero fixed vectors. Consider a representation V of G. As N is unipotent, $V^N$ is nonempty. Because N is normal, $V^N$ is stable under G, hence under G/N, and hence has nonzero fixed vectors (because G/N is unipotent). Therefore G is unipotent. 


$G$
is connected though it's not an issue in characteristic 0. Anyway, the answer is definitely yes. If you already have the full BorelChevalley structure theory of (connected) linear algebraic groups in hand, it's an easy consequence: modulo the unipotent radical of$G$
, you get a reductive group (which can't be unipotent if nontrivial). Naturally you might prefer shortcuts. – Jim Humphreys Feb 15 '13 at 1:06