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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) ?

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    $\begingroup$ @anon: what you say is not right: take the torus inside the diagonals in $GL_n$ all of whose entries except the first one are $1$. This maps iso to $GL_n/SL_n=G_m$. $\endgroup$
    – Venkataramana
    Commented Sep 26, 2014 at 0:43
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    $\begingroup$ As to the original question, take $G$ to be a finite group (e.g the group of upper triangular unipotent matrices matrices with entries in $Z/3Z$. Then the commutator is the centre and the extension above does not split. If it did, the group would be abelian. $\endgroup$
    – Venkataramana
    Commented Sep 26, 2014 at 0:47
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    $\begingroup$ Perhaps a better question is whether such a splitting exists if we replace $(G,G)$ by its connected component and similarly for $G/(G,G)$. I think the answer is still no: suppose $G^\circ=\GL_n$ and $G/G^\circ$ is cyclic of order 2, generated by $\theta$ satisfying $\theta g= ({^{t}}g^{-1})\theta$ and $\theta^2=-I$. If $n$ is even then there is a splitting (since $-I$ is the square of an element of $\SO_n$) but if $n$ is odd, then there is no element of $\theta G^\circ=G\setminus G^\circ$ of order 2. However, I think the corresponding statement might be true if we assume $G^\circ$ is simple. $\endgroup$
    – Paul Levy
    Commented Oct 4, 2014 at 11:08
  • $\begingroup$ @user: "I've read .... "? Do you recall where this is written? $\endgroup$
    – Jim Humphreys
    Commented Oct 12, 2014 at 22:43
  • $\begingroup$ What does "in a non-trivial way" mean? If $G$ is connected and semisimple, then your exact sequence is $1 \to G \to G \to 1 \to 1$. $\endgroup$
    – LSpice
    Commented Oct 26, 2016 at 18:22

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