Let $G = (G,\times)$ be a group, and $G^{op} = (G,o), a \ o \ b = b \times a$ be it's opposite group.
I need to prove that there is a group isomorphism $f: G \to G^{op}$ Obviously, $id_G: G \to G$ is a bijection, but it's a group homomorphism if and only if $G$ is commutative.
How might one come up with $g \mapsto g^{-1}$?
It is clear that $g g^{-1}$ must be mapped to $g^{-1} g$ (by definition of the opposite group "reversing multiplication"). Note that $\phi(g g^{-1}) = \phi(g) \phi(g^{-1}) = \phi(g) \phi(g)^{-1}$, which is expected to be $g^{-1} g$. That suggests $\phi: g \mapsto g^{-1}$.
(Incidentally, this is an instance of a much more general phenomenon: that of duality in category theory.)
