Extensions of Tori by Tori are Tori

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

$1\to G_u\to G\to T\to 1$

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.