Continuing our previous series, $ G$ is an algebraic group over an algebraically closed field $ k$ and we identify $ G$ with $ G(k)$.

Here is an interesting fact:

**Theorem.**In a connected solvable group the unipotent part $ G_u$ is a closed connected normal subgroup of $ G$ and contains the commutator subgroup $ [G,G]$.

Why is this interesting? It will allow us to prove that if $ T\subseteq G$ is a normal torus in a connected group $ G$, and $ G/T$ is also a torus then $ G$ itself is a torus!

But first let us prove the theorem. Here, $ G$ being connected solvable really does the trick because then $ G$ is a subgroup of $ T_n$, the $ n\times n$ upper triangular matrices for some $ n$. The short exact sequence $ 1\to U_n\to T_n\to D_n\to 1$ where $ U_n$ is the upper unipotent and $ D_n$ are the diagonal matrices give a short exact sequence

where $ T$ is a torus. Thus we get that $ G/G_u$ is abelian, but $ G/[G,G]$ is abelian and is the largest abelian quotient so $ [G,G]\subseteq G_u$.

### Extensions of Tori by Tori

Suppose we have a short exact sequence $ 1\to T\to G\to G/T\to 1$ where $ T$ is a normal torus and $ G/T$ is also a torus. In this case any commutator is trivial since $ G/T$ is commutative, and so the commutator subgroup lies in the torus $ T$, so $ G$ is solvable and the commutator subgroup $ [G,G]$ consists of semisimple elements, being a subgroup of a torus. Thus $ G_u = \{1\}$, i.e. is trivial. However, a connected solvable group is the semidirect product of a torus and its unipotent group, so $ G$ is necessarily a torus.