Rational points Let $G$ be an affine algebraic group defined over a field of characteristic zero $K$. Suppose $G$ has only one single $K$-point, can we conclude that $G$ does not have more points?
 A: 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 $k$.
