# 2 Dimensional Connected Algebraic Groups are Solvable

Posted by Jason Polak on 03. August 2013 · Write a comment · Categories: algebraic-geometry, group-theory · Tags: ,

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. QED.

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.