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After passing to a finite separable extension of the base field, we may suppose that the reductive group is split. After passing to a finite covering, we may suppose that it is the product of a simply connected semisimple group $G$ and a torus. The torus presents no problem. After passing to a finite separable extension, we may suppose that G is a product of absolutely almost-simple normal subgroups, so it all comes down to looking at the subgroups of the centre of $G$, but this presents no problem.

Added: All the statements used can be found, for example, in this book

After passing to a finite separable extension of the base field, we may suppose that the reductive group is split. After passing to a finite covering, we may suppose that it is the product of a simply connected semisimple group $G$ and a torus. The torus presents no problem. After passing to a finite separable extension, we may suppose that G is a product of absolutely almost-simple normal subgroups, so it all comes down to looking at the subgroups of the centre of $G$, but this presents no problem.

After passing to a finite separable extension of the base field, we may suppose that the reductive group is split. After passing to a finite covering, we may suppose that it is the product of a simply connected semisimple group $G$ and a torus. The torus presents no problem. After passing to a finite separable extension, we may suppose that G is a product of absolutely almost-simple normal subgroups, so it all comes down to looking at the subgroups of the centre of $G$, but this presents no problem.

Added: All the statements used can be found, for example, in this book

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anon
anon
  • 31
  • 2

After passing to a finite separable extension of the base field, we may suppose that the reductive group is split. After passing to a finite covering, we may suppose that it is the product of a simply connected semisimple group $G$ and a torus. The torus presents no problem. After passing to a finite separable extension, we may suppose that G is a product of absolutely almost-simple normal subgroups, so it all comes down to looking at the subgroups of the centre of $G$, but this presents no problem.