Is a quotient of a reductive group reductive?


**Edit** [Pete L. Clark]: As Minhyong Kim points out below, a more precise statement of the question is: 


Is the quotient of a reductive linear group by a Zariski-closed normal subgroup reductive?



Is a quotient of a reductive group reductive?
For a *linearly* reductive (i.e. all representations are completely redudible) group $G$, any quotient $G/H$ is also linearly reductive. Given a representation $V$ of $G/H$, you get an induced representation of $G$ (with the same underlying vector space). Then any decomposition of $V$ as a representation of $G$ is also a decomposition as a representation of $G/H$. Since $G$-invariant subspaces of $V$ are the same thing as $G/H$-invariant subspace of $V$, the notions of irreducible subrepresentations agree. So you get a complete decomposition of $V$ as a representation of $G/H$.


In characteristic zero, reductive is equivalent to linearly reductive, so you have your answer. In general, a group is reductive if its unipotent radical is trivial. I feel like you should be able to argue that an element of the unipotent radical of $G/H$ lifts to give you an element of the unipotent radical of $G$, but I'm not sure.


