- Any two Borel subgroups of are conjugate subgroups.
- Any closed connected solvable algebraic subgroup of is contained in a Borel subgroup.
Breakdown over a non-algebraically closed field
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.
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 .
|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.|