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I think question 2 has a positive answer when $G^{der}$ is simply connected [EDIT: and without anisotropic factor], but not in general. If $G=PGL_d$ (so $G=G^{der}$) then $G(K)/[G(K),G(K)]=M/M^d$ where $M=K^*$, this is not a trivial group in general.

ReleventRelevant refences should be the preliminary chapters in book by Margulis "discrete subgroups of semisimple Lie groups", as well as Platonov-Rapinchuk's book.

I think question 2 has a positive answer when $G^{der}$ is simply connected, but not in general. If $G=PGL_d$ (so $G=G^{der}$) then $G(K)/[G(K),G(K)]=M/M^d$ where $M=K^*$, this is not a trivial group in general.

Relevent refences should be the preliminary chapters in book by Margulis "discrete subgroups of semisimple Lie groups", as well as Platonov-Rapinchuk's book.

I think question 2 has a positive answer when $G^{der}$ is simply connected [EDIT: and without anisotropic factor], but not in general. If $G=PGL_d$ (so $G=G^{der}$) then $G(K)/[G(K),G(K)]=M/M^d$ where $M=K^*$, this is not a trivial group in general.

Relevant refences should be the preliminary chapters in book by Margulis "discrete subgroups of semisimple Lie groups", as well as Platonov-Rapinchuk's book.

Source Link
YCor
  • 63.9k
  • 5
  • 187
  • 286

I think question 2 has a positive answer when $G^{der}$ is simply connected, but not in general. If $G=PGL_d$ (so $G=G^{der}$) then $G(K)/[G(K),G(K)]=M/M^d$ where $M=K^*$, this is not a trivial group in general.

Relevent refences should be the preliminary chapters in book by Margulis "discrete subgroups of semisimple Lie groups", as well as Platonov-Rapinchuk's book.