Take the 2-minute tour ×  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?)
share|improve this question you might specify if you have in
mind characteristic zero or arbitrary characteristic. –  YCor
Nov 17 '12 at 18:01 What set are you using to be the elements of
the group? –  Will Sawin Nov 17 '12 at 18:03 The multplicative
group of quaternions (presented as $\mathbb{C}\times \mathbb{C}
\setminus \{0\}$ yields a non-algebraic example of your (1)+(2);
yet it fails (3). –  Mikhail Bondarko Nov 17 '12 at 19:02
@Mikhail: In char. $p$ such near-misses (losing (3)) abound. Let
$H$ be a linear algebraic group and $H'$ an infinitesimal non-
normal closed subgro