Let $k$ be a finite field and $G$ a finite type smooth $k$-group scheme. Let $G^0$ and $\Gamma$ be the connected component of identity and the component group of $G$, so there is an exact sequence $1 \rightarrow G^0 \rightarrow G \rightarrow \Gamma \rightarrow 1$. Is the induced $H^1(k, G) \rightarrow H^1(k, \Gamma)$ surjective?

By Lang's theorem, $H^1(k, G^0) = *$, so the map in question is injective. It is surjective if $G^0$ is central (in particular, if $G$ is commutative), since for commutative $G^0$ one has $H^2(k, G^0) = 0$ for cohomological dimension reasons. It is also surjective if the short exact sequence splits as a semi-direct product. Thus, my question is: does the surjectivity hold in general for noncommutative smooth $G$?

every$k$-form $\mathscr{G}$ of $G^0$; just for $G^0$ is insufficient. Classifying $K$-forms of ${\rm{SL}}_n$ for a general $K$ uses $G={\rm{Aut}}_{{\rm{SL}}_n/K}$, a semi-direct product of $\mathbf{Z}/(2)$ against $G^0 = {\rm{PGL}}_n$. Here, ${\rm{H}}^1(K,G^0)=1$ but the fiber of ${\rm{H}}^1(K,G)\rightarrow {\rm{H}}^1(K,\mathbf{Z}/(2))$ over separable quadratic $K'/K$ is ${\rm{SU}}(h)$'s for $(-1)^{n+1}$-hermitian $h:{K'}^n\times{K'}^n\rightarrow K'$; one is quasi-split but more can arise! – user27920 Jun 29 '14 at 21:22