Unipotent Subgroups Are Triangularizable
Let \(U\) be a unipotent algebraic subgroup of \(\operatorname{GL}(V)\) over an algebraically closed field. Then \(V\) has a complete flag \[ 0=V_0\subset V_1\subset \cdots \subset V_n=V \] stable under \(U\), and \(U\) acts trivially on each quotient \(V_i/V_{i-1}\). Equivalently, after choosing a suitable basis, \(U\) is a closed subgroup of the upper unitriangular group.
Consequently every unipotent algebraic group is nilpotent, hence solvable.
Proof
A unipotent family of linear transformations has a common fixed vector. One way to see this is to use the Lie-Kolchin fixed-vector argument on the closed subgroup generated by the family: every element has only the eigenvalue \(1\), and a minimal nonzero stable subspace must therefore be one-dimensional with trivial action. This gives \(V_1\).
Apply the same argument to \(V/V_1\) and continue by induction. The resulting stable complete flag puts the group inside the upper unitriangular group. The upper unitriangular group has a central series by superdiagonal length, so any closed subgroup is nilpotent; nilpotent groups are solvable.