Let (e1,e2,e3) be the canonical basis of K3. Consider the nilpotent Lie algebra structure on K3 such that [e1,e2]=e3,[e1,e3]=[e2,e3]=0. With the associated group law on K3 (no. 5),
(x,y,z)(x′,y′,z′)=(x+x′,y+y′,z+z′+21(xy′−yx′)).
The mapping
(λ,μ,v)↦(expλe1)(expμe2)(expv(e1+e3))
of K3 into G is neither surjective (show that (0,1,1) is not in the image) nor injective (show that (0,1,0) and (1,1,−1) have the same image).