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Let $G$ be an algebraic group over an algebraically closed field. Furthermore, let $G$ be semi-simple, i.e. its radical (viz. its maximal closed, connected, solvable normal subgroup) is trivial. One step in a proof I'm trying to understand seems to use the following fact:

If $G$ is connected and non-trivial, then there is a non trivial maximal torus in $G$.

Why is that true? Thank you!

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up vote 3 down vote accepted

It's a theorem of Grothendieck, even true over any base field. See SGA 3, tome II, Exp. XIV, Theorem 1.1 (it is also available in Linear algebraic groups by Borel, Theorem 18.2).

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In my copy of Borels book, Theorem 18.2 only says that a connected group contains a maximal torus, but not that it is non-trivial if $G$ is non-trivial. Or am I understanding it wrong? – Sh4pe Jan 3 '13 at 9:47

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