# Why is the semisimple rank of a connected reductive group equal to the rank of the commutator?

I am trying to learn the theory of linear algebraic groups over an algebraically closed field. I know that if $R(G)$ denotes it radical, then $G/R(G)$ is semisimple and is therefore equal to its own commutator. So $G/R(G)=[G/R(G),G/R(G)]=[G,G]R(G)/R(G)=[G,G]/(R(G) \cap [G,G])$. So it's a quotient of the commutator by a finite normal subgroup.

But why would the rank (dimension of a maximal torus) of $[G,G]$ be equal to the rank of the quotient of $[G,G]$ by a finite subgroup? In general, is the rank additive in short exact sequences? I know it is true for the sequence

$1 \rightarrow SL_n(k) \rightarrow GL_n(K) \rightarrow \mathbb{G}_m \rightarrow 1$.

• "So it's a quotient by a finite...": you're assuming there that $G$ is reductive. – YCor Oct 25 '16 at 22:42
• Yes, the rank is additive under exact sequences, and it's easier when the kernel in the exact sequence is finite. Anyway, this is an exercise. – YCor Oct 25 '16 at 22:43

If $$G$$ is a semisimple group (such as $$[G,G]$$ in your question) with maximal torus $$T$$, then the rank of $$G$$ is by definition the dimension of $$T$$. If $$S$$ is a finite subgroup of $$G$$ which is contained in the center of $$G$$ (such as $$R(G) \cap [G,G]$$ in your question), the point is that $$G/S$$ is semisimple with maximal torus $$T/S$$. Since $$S$$ is finite, $$\operatorname{Dim}(T/S) = \operatorname{Dim}(T)$$, so the ranks of $$G$$ and $$G/S$$ are the same.