@TimCampion: It does seem to at least be the case if X is an H-group, via multiplication (g,a) |--> g.a in one direction and
b|--> (b(0), b(0)^{-1} . b)
in the other. Do you know where you saw the more general result?
b|--> (b(0), b(0)^{-1} . b)
in the other. Do you know where you saw the more general result?