Any two Borel subgroups of a linear algebraic group are conjugate: working backwards in Springer's book.

Question

This question can be too broad if we're not careful. It's my hope that an answer would consist of a short list of theorems and their dependencies. If this is too much to ask or is in some way inappropriate, then I'm sorry.

In Springer's book, "Linear Algebraic Groups (Second Edition)", we have the following theorem:

Theorem 6.2.7(iii): Two Borel subgroups of [a linear algebraic group] are conjugate.

NB: We're working over an algebraically closed field.

The proof given is:

If $B$ and $B'$ are two Borel subgroups then $B'$ is conjugate to a subgroup of $B$ and $B$ to a subgroup of $B'$. Hence $\dim B=\dim B'$, which implies (iii).

Context:

I'm learning the basics of linear algebraic group theory, to facilitate research in the area. My supervisor suggested working backwards from a couple of theorems (to lighten the load), one of which being Theorem 6.2.7(iii).

Currently, I am halfway through Chapter 3 of Springer's book. I have copies of Humphreys' and Borel's books, which I dive into to strengthen my understanding of what's covered in Springer's.

My supervisor and I discuss the general theory once a week. We're building up my knowledge & intuition slowly but surely.

The Question:

What are the main theorems Theorem 6.2.7(iii) relies upon, and what do they rely upon, and so on until the basics are covered? Where can these theorems be found in Springer's book?

Thoughts:

The proof given above can be broken down into three statements I don't yet understand:

1. If $B$ and $B'$ are two Borel subgroups, then $B$ is conjugate to a subgroup of $B'$.

2. If $B$ and $B'$ are conjugate to subgroups of each other, then $\dim B=\dim B'$.

3. If $\dim B=\dim B'$, then $B$ and $B'$ are conjugate to each other.

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The one that makes the most sense to me is 3. However, I don't know where any of them is in Springer's book.

Further Context:

To get some idea of my ability with linear algebraic groups, see the following questions of mine:

• Show, using a specific approach, that $\dim \Bbb P^n=\dim\Bbb A^n=n$.

• Show that $({\rm id}\otimes \Delta)\circ\Delta=(\Delta\otimes{\rm id})\circ\Delta$ "translates" to associativity of linear algebraic groups

• Showing the linear algebraic subgroup $\Bbb U_n$ of $\Bbb{GL}_n(k)$ is closed.

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This feels like trying to run before I can walk; however, I am sure many professional mathematicians, when reading lengthy papers/books, work backwards from theorems they deem useful, therefore avoiding having to read entire documents. I hope to develop this skill. I just need a leg-up for now, please.

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I have included the proof-explanation tag because it approximates what I am after nicely.

Answer 1

Here is how you prove 2 and 3 (assuming only that the subgroups are Zariski-closed):

1. Iterating the conjugations $f: B\to B', f': B'\to B$ you obtain a sequence of closed subgroups $$ B=B_1\supset B_2\supset B_3.... $$ all conjugate to each other. But such a sequence has to stabilize. From this you see that $f(B)=B'$.

2. Even though it is an overkill, there is a more general fact that you should know (the Ax–Grothendieck theorem): If $X$ is a variety over an algebraically closed field then every injective morphism $X\to X$ is also surjective.