In Highlights 12, we used some of the equivalent conditions for a connected algebraic group $ G$ over a field $ k=\overline{k}$ to have semisimple rank 1 in the study of reductive groups (these are the groups whose unipotent radical $ R(G)_u$ is trivial).

Precisely, we showed that such a $ G$ must have a semisimple commutator $ [G,G]$ subgroup whose dimension is three, and that we can write

$ G = Z(G)^\circ\cdot [G,G]$

where the $ -\cdot-$ denotes that this is an almost direct product: in other words, the multiplication map $ Z(G)^\circ\times[G,G]\to G$ is surjective with finite kernel.

Let $ T\subseteq G$ be a maximal torus. We will show in this post that $ C_G(T) = T$ for a connected reductive group $ G$ of semisimple rank 1.
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Borel subgroups are an important type of subgroup that will allow us to gain insight into the mysterious structure of algebraic groups. We shall look at the definition and some basic examples in this post. As usual, algebraic group means some linear algebraic group defined over an algebraically closed field $ k$.

A Borel subgroup $ B\subseteq G$ of an algebraic group $ G$ is a maximal connected solvable subgroup amongst the solvable subgroups of $ G$. Notice the absense of the adjective "closed" here: although $ G$ may contain solvable groups that are not closed, one that is maximal amongst the connected solvable ones must be closed. Indeed, if $ B$ is a subgroup maximal amongst the connected solvable subgroups of $ G$ then its closure $ \overline{B}$ is also connected and solvable.

Before we go any further, it's helpful to have an example. Since all linear algebraic groups are closed subgroups of some $ \mathrm{GL}_n(k)$, let's do $ \mathrm{GL}_n(k)$. We claim that a Borel subgroup of $ \mathrm{GL}_n(k)$ is $ B = \mathrm{T}_n(k)$, the closed subgroup of upper triangular matrices. Let's now sketch the proof that $ B$ is actually a Borel.
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