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.

Related facts

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.

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:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 The quotient variety $G/B_1$ is a complete algebraic variety. Fact (1) $B_1$ is Borel in $G$.
2 $B_2$ is a connected solvable linear algebraic group over the algebraically closed field $K$. $B_2$ is a Borel subgroup of the linear algebraic group $G$
$K$ is algebraically closed
follows from definition of Borel subgroup
3 Consider the action of $B_2$ on $G/B_1$ by left multiplication, obtained by restricting the action of $G$. This is an algebraic group action (i.e., it is regular in the algebraic group sense). The action is obtained by restricting a group multiplication, which is algebraic.
4 The action in Step (3) has a fixed point, say a coset $gB_1$. In other words, $B_2(gB_1) = gB_1$. Fact (2) Steps (1)-(3) Step-fact combination direct
5 $B_2 \le gB_1g^{-1}$ Step (4) multiply both sides of the equality in step (4) on the right by $g^{-1}$, and get $B_2(gB_1g^{-1}) = gB_1g^{-1}$. Since both $B_2$ and $gB_1g^{-1}$ are subgroups, this gives the subgroup containment.
6 $B_2 = gB_1g^{-1}$, so $B_1,B_2$ are conjugate subgroups. $B_1,B_2$ are Borel subgroups Step (5) Since $B_1$ is Borel, so is $gB_1g^{-1}$. $B_2$ is also Borel. Thus, both are maximal among connected closed solvable subgroups of $G$, and hence, by the containment of Step (5), they must be equal.