# Borel-Morozov theorem

From Groupprops

## Contents

## Statement

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

- Any two Borel subgroups of are conjugate subgroups.
- Any closed connected solvable algebraic subgroup of is contained in a Borel subgroup.

## Related facts

### Breakdown over a non-algebraically closed field

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

### Similar facts

### Facts with similar proofs

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

## Facts used

## Proof

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

**Given**: A linear algebraic group over an algebraically closed field . Borel subgroups of .

**To prove**: is conjugate to .

**Proof**:

Step no. | Assertion/construction | Facts used | Given data used | Previous steps used | Explanation |
---|---|---|---|---|---|

1 | The quotient variety is a complete algebraic variety. | Fact (1) | is Borel in . | ||

2 | is a connected solvable linear algebraic group over the algebraically closed field . | is a Borel subgroup of the linear algebraic group is algebraically closed |
follows from definition of Borel subgroup | ||

3 | Consider the action of on by left multiplication, obtained by restricting the action of . 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 . In other words, . | Fact (2) | Steps (1)-(3) | Step-fact combination direct | |

5 | Step (4) | multiply both sides of the equality in step (4) on the right by , and get . Since both and are subgroups, this gives the subgroup containment. | |||

6 | , so are conjugate subgroups. | are Borel subgroups | Step (5) | Since is Borel, so is . is also Borel. Thus, both are maximal among connected closed solvable subgroups of , and hence, by the containment of Step (5), they must be equal. |