Talk:2.3.12/@comment-160.39.88.238-20120608190643/@comment-5149782-20120609204932

I don't know if I agree, because as far as I can see your version is essentially the same as mine but with some steps not written out explicitly. Your first equality, $$xy = \varphi((xy)^{-1})$$, only holds because $$\varphi^{-1}(x) = x^{-1}$$, so it seems strange not writing that out explicitly. But starting with $$\varphi^{-1}(\varphi(xy))$$ rather than $$\varphi(\varphi^{-1}(xy))$$ probably results in a cleaner, as you did.