Borel-Morozov theorem

Statement
Suppose $$G$$ is a linear algebraic group over an algebraically closed field $$K$$. Then, the following are true:


 * 1) Any two  Borel subgroups of $$G$$ are conjugate subgroups.
 * 2) Any closed connected solvable algebraic subgroup of $$G$$ is contained in a Borel subgroup.

Breakdown over a non-algebraically closed field
See Borel-Morozov theorem fails for non-algebraically closed field.

Similar facts

 * Lie-Kolchin theorem
 * Kolchin's theorem
 * Lie's theorem
 * Engel's theorem

Facts with similar proofs

 * Sylow implies order-dominating: This shows that any $$p$$-Sylow subgroup of a finite group contains some conjugate of any $$p$$-subgroup. It therefore show that both Sylow implis order-conjugate and that every $$p$$-subgroup is contained in some $$p$$-Sylow subgroup.

Facts used

 * 1) uses::Equivalence of definitions of Borel subgroup
 * 2) uses::Borel fixed-point theorem

Proof
We prove version (1). Version (2) is equivalent to it.

Given: A linear algebraic group $$G$$ over an algebraically closed field $$K$$. Borel subgroups $$B_1, B_2$$ of $$G$$.

To prove: $$B_1$$ is conjugate to $$B_2$$.

Proof: