gr.group theory - Rational points

Question (edited here)

Let $G$ be an affine algebraic group defined over a field $k$ of characteristic zero. Is it possible for $|G(k)|=1$, even if $G$ is not trivial?

As shown by David Speyer in the comments, if $dim G=0$ then yes. For example, let $G$ be the solutions to $z^3-1$. Then over $|G(mathbb{Q})|=1$, but $|G(mathbb{C})|=3$ and hence $G$ is not trivial.

On the other hand, the comments by Brian Conrad show that if $dim G geq 1$, then $|G(k)|not=1$.

I think this proves it: Since the identity component of $G$ is a connected affine algebraic group over $k,$ it suffices to prove this for $G$ connected. Then, since we are in characteristic 0, $G$ is isomorphic (as a variety, but not as an algebraic group) to $(G/G_u) times G_u$ where $G_u$ is its unipotent radical. The unipotent radical is likewise isomorphic to an affine space, and $G/G_u$ is reductive. By the Bruhat-decomposition $G/G_u$ contains an affine open subset whose $overline{k}$-points are isomorphic to $(overline{k}^*)^n times overline{k}^m$ where $overline{k}$ is an algebraic closure of $k$.