Let $G$ be an algebraic variety, which is also a group. Suppose that for any $g\in G$ the right-multiplication by $g$ is a morphism of algebraic varieties $G\to G$, and that so is the inverse $G\to G$. Does it imply that $G$ is an algebraic group? (i.e. is the multiplication $G\times G\to G$ a morphism?)

Let $G$ be an algebraic variety over an algebraically closed field $k$ (any characteristic). Suppose that: (1) the set of $k$-points has the structure of a group. (2) for any $g\in G$ the right-multiplication by $g$ is a morphism of algebraic varieties $G\to G$. (3) the inverse map is a morphism $G\to G$. 

Does it imply that $G$ is an algebraic group? (i.e. is the multiplication $G\times G\to G$ a morphism?)