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 gives
$$
[U_n^{(r)},U_n^{(s)}]\subseteq U_n^{(r+s)}.
$$
Since $U_n^{(n)}=1$, 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.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 167 / 2 / c / Solution

{wiki=Unipotent_group}

A unipotent algebraic group is an affine algebraic group all of whose elements are unipotent. The <Kolchin theorem> embeds it into an upper unitriangular group.

Unipotent algebraic group

{c}
{wiki=Lie–Kolchin_theorem}

Every unipotent algebraic subgroup of $\operatorname{GL}_n$ 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>.

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 Incoming links: Kolchin theorem