$\textbf{Some definitions:}$ Let $G$ be an algebraic group (for me that is the complex points of an affine algebraic group). We say $G$ is reductive if its unipotent radical (maximal connected normal unipotent subgroup) $R_u(G)$ is trivial and we say $G$ is semisimple if its radical (maximal connected normal solvable subgroup) $R(G)$ is trivial. The derived group $\mathcal{D}(G)$ of $G$ is the intersection of the normal subgroups $N$ such that $G/N$ is commutative. If there are algebraic subgroups $H$ and $K$ of $G$ such that the product morphism $H\times K\rightarrow G$ is surjective with finite kernel we say that $G$ is an almost-direct product of $H$ and $K$. (edit: as pointed out in the comments/answers I forgot to specify that the images of $H$ and $K$ commute in $G$.)

$\textbf{A theorem:}$ When $G$ is a connected reductive complex algebraic group, it is the almost-direct product of a central torus $R(G)$ and a semisimple group $\mathcal{D}(G)$. (See page 181 in Borel's book, or page 168 in Humphreys' book "Linear Algebraic Groups".)

$\textbf{My question:}$ I would like to know to which extent this decomposition fails to generalize in the case where $G$ is disconnected. More precisely, I would be interested to know if there are similar decomposition results for a reductive group $G$ (not necessarily connected) that happens to be nilpotent or solvable.

anyfinite group against a nontrivial action on a torus) is a counterexample, if I'm not misunderstanding the question. How would you like to rule these out? $\endgroup$ – user29283 Apr 25 '13 at 5:47