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Say $G$ is a reductive group over a field $k$. I usually take $k = \mathbb{C}$ so assume what you want about the field except maybe that its finite. If $X$ is a scheme over $k$ then a principal $G$ bundles over $X$ is a scheme $P$ together with a right action of $G$ and an equivariant projection to $X$ (with trivial action on $X$) such that $P$ is locally trivial in the etale topology. For some groups like $GL_n,SL_n$ and solvable groups principal bundles are locally trivial even in the Zariski topology. These are called special groups and Grothendieck classified them.

I'm curious if $G',G''$ are special groups and $G$ fits into an exact sequence $1 \to G' \to G \to G'' \to 1$, then is it the case that $G$ is special?

There is a paper by Serre that claims this at least for $G',G''$ commutative and its supposed to be a consequence of the exact sequence $\check H^1(X,G') \to \check H^1(X,G) \to \check H^1(X,G'')$. This is \check Cech cohomology in the etale topology. You have $\check H^1(X,G') \cong \check H^1(X_{zar}, G')$ and $\check H^1(X,G'') \cong \check H^1(X_{zar}, G'')$ and a map $\check H^1(X_{zar},G') \to \check H^1(X, G)$ but it seems you are still short of being able to use e.g. the 5-lemma. This can probably be deduced from Grothendieck's thm but I'm wondering if there is a direct argument.

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Let $f\colon X_{et}\to X_{zar}$ be the identity map. For a special group $G$, $R^1f_*G=0$ and so the Leray spectral sequence shows that the map from $H^2_{zar}$ to $H^2_{et}$ is injective in the commutative case, which is what you want in order to apply the 5-lemma. It is surely also true in the noncommutative case, but writing down a proof will be more complicated. –  anon Aug 2 '11 at 23:51
    
Dear anon -- What? –  Jason Starr Aug 3 '11 at 1:44
    
Dear solbap -- There is a long exact sequence, of sorts, for a central extension of a group $G''$ by a group $G'$, cf. Section I.5.7 of Serre's "Galois Cohmology". This is stated only in the case that $X$ is the spectrum of a field (so that \'etale cohomology is Galois cohomology). The generalization to schemes may be contained in Giraud's thesis. –  Jason Starr Aug 3 '11 at 1:48
    
@Jason, sorry, I don't understand your question. The argument Angelo gives is essentially that in my comment, but with the cohomology removed. –  anon Aug 3 '11 at 12:58
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If you say so. I do know something about non-Abelian cohomology. I certainly understand Angelo's answer. I do not see how what you wrote is "essentially" what he wrote. –  Jason Starr Aug 3 '11 at 20:59
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up vote 3 down vote accepted

The answer is more or less as Jason says, but the proof is very easy, and does not require any cohomological machinery. If $P \to X$ is a $G$-torsor, then $P/G' \to X$ is a $G''$-torsor, hence it is Zariski-locally trivial. By passing to a cover, we may assume that it is trivial; hence $P$ has a reduction of structure group to $G'$, that is, it comes from a $G'$-torsor $P' \to X$. Such a torsor is Zariski-locally trivial, and this completes the proof.

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