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.