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Every unipotent algebraic subgroup of ${\mathrm{GL}}_n$ is conjugate to a subgroup of the upper unitriangular group. The filtration by vanishing initial superdiagonals then proves that every unipotent algebraic group is a nilpotent group.

The derived subgroup of an affine algebraic group $[G,G]$ is the closed subgroup generated by the commutators $xy{x}^{-1}{y}^{-1}$. If $G$ is connected, the image of every finite product of commutator maps is connected and contains the identity. The closures of these images form an increasing sequence; once their dimensions stabilize, the stable member is closed under products and inverses and equals $[G,G]$. Hence $[G,G]$ is connected.

Now suppose the connected group $G$ is solvable. The Lie-Kolchin theorem conjugates a faithful representation of $G$ into the upper triangular matrices. Every commutator then has all diagonal entries equal to one, so every element of $[G,G]$ is unipotent. Moreover $[G,G]$ lies in the upper unitriangular group, whose superdiagonal filtration is a central series. It is therefore a nilpotent group.

A diagonalizable algebraic group is a closed subgroup of a product of copies of ${\mathbb{G}}_m$. A unipotent algebraic group has only unipotent elements, while a semisimple algebraic group here means one all of whose elements are semisimple. A reductive algebraic group is smooth, connected, affine, and has trivial connected normal unipotent radical.

Solved by gpt-5.6-sol high.

The Kolchin theorem conjugates a faithful representation of a unipotent algebraic group $U$ into the upper unitriangular group $U_n$. Let $U_n^{(r)}$ consist of matrices whose first $r-1$ superdiagonals vanish. Matrix multiplication givesSince $U_n^{(n)}=1$, this filtration is a finite central series, so $U_n$ and every subgroup of it are nilpotent groups. Hence $U$ is nilpotent.

The converse fails: ${\mathbb{G}}_m$ is abelian, hence nilpotent, but its nonidentity points are semisimple rather than unipotent.

Solved by gpt-5.6-sol high.