Conventions: $ G$ is an algebraic group over an algebraically closed field $ k$ and we identify $ G$ with $ G(k)$.

Consider the algebraic groups $ \mathbb{G}_a$ and $ \mathbb{G}_m$. They are the only one-dimensional connected groups and they are both solvable. What about two-dimensional connected groups? It turns out that if $ \mathrm{dim} G\leq 2$ then $ G$ is solvable.

For $ \mathrm{dim} G= 3$, this is no longer true, for instance $ \mathrm{SL}_2$ is $ 3$-dimensional but not solvable since it is perfect, i.e. equal to its commutator subgroup. So let's prove our theorem:

**Theorem.**Let $ G$ be a connected algebraic group over $ k$. If $ \mathrm{dim} G = 2$ then $ G$ is solvable.

*Proof*. Let's start by considering a Borel subgroup $ B\subseteq G$. Suppose then that $ G$ is not solvable. In this case $ B$ is a proper closed connected subgroup and so has lower dimension than the dimension of $ G$, so $ \mathrm{dim} B = 1$ (it can't be zero because then it would be trivial).

Consider a maximal torus $ T\subseteq G$. There are two possibilities: $ \mathrm{dim} T = 0$ or $ \dim T = 1$. If $ \dim T = 0$ then $ B$ is unipotent, and the conjugates of $ B$ cover $ G$ so $ G$ is unipotent, hence nilpotent, hence solvable.

This leaves the case $ \dim T = 1$. Then $ B$ is equal to its semisimple part, which implies that $ G = B$, and hence $ G$ is solvable. .

In this proof, we used that for a connected group $ G$, if $ B$ is either semisimple or unipotent then actually $ G = B$. We will prove this in the next post.