These notes on affine algebraic groups mention the following theorem.
Let $G$ be a connected nilpotent affine algebraic group (over an algebraically closed field $k$), and denote $G_s$ and $G_u$ its respective semisimple and unipotent parts. Then, $G_s$ and $G_u$ are closed normal subgroups of $G$, and the natural map $G_s \times G_u \to G$ is a isomorphism of algebraic groups.
Then it makes the remark (page 24) that in such a $G$, the semisimple part $G_s$ is then a torus (1). Which I don't get.
I get that with Lie-Kolchin's theorem, $G_s \simeq G/G_u$ is identified with a subgroup of $D_n$ (the diagonal matrices of $\mathrm{GL}_n(k)$, which is torus). But I don't see why it must be a torus. The remark seems to point out that it comes from the connectedness of $G_s$ (as a quotient of the connected group $G$), but it does not help me. Any hint ?
(1) A torus is a direct power of the multiplicative group $\mathbb G_m = (k^\times,\cdot)$.