This might be an easy question for the experts, so I apologise in advance. By a reductive group over a field $k$, I mean a linear algebraic group (not necessarily connected) such that the unipotent radical over $\bar{k}$ is trivial. Let $G$ be a reductive group over a field $k$ (not necessarily algebraically closed) of characteristic zero (this means, in particular, that reductive is the same thing as linearly reductive). I've read that if $G$ is connected, there is a split extension given by $$ 1 \to (G,G) \to G \to G/(G,G) \to 1,$$ where $(G,G)$ is the derived subgroup of $G$. My question is:

For a more general reductive $k$-group $G$ (i.e. with $G$ not necessarily connected), does $G$ fit into a split extension of linear algebraic $k$-groups?

This might be an easy question for the experts, so I apologise in advance. By a reductive group over a field $k$, I mean a linear algebraic group (not necessarily connected) such that the unipotent radical over $\bar{k}$ is trivial. Let $G$ be a reductive group over a field $k$ (not necessarily algebraically closed) of characteristic zero (this means, in particular, that reductive is the same thing as linearly reductive). I've read that if $G$ is connected, there is a split extension given by $$ 1 \to (G,G) \to G \to G/(G,G) \to 1,$$ where $(G,G)$ is the derived subgroup of $G$. My question is:

For a more general reductive $k$-group $G$ (i.e. with $G$ not necessarily connected), does $G$ fit into a split extension of linear algebraic $k$-groups (in a non-trivial way) ?