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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

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

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    $\begingroup$ "So it's a quotient by a finite...": you're assuming there that $G$ is reductive. $\endgroup$ – YCor Oct 25 '16 at 22:42
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    $\begingroup$ 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. $\endgroup$ – YCor Oct 25 '16 at 22:43
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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.

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