2.3.12

a) $$\varphi(x) = x^{-1}$$ is bijective because $$f(x) = f(y) \iff x^{-1} = y^{-1} \iff xx^{-1}y = xy^{-1}y \iff y = x \iff x = y.$$

b) First assume that $$\varphi \in \text{Aut}(G)$$, then $$xy = \varphi^{-1}(\varphi(xy)) = \varphi^{-1}((xy)^{-1}) = \varphi^{-1}(y^{-1}x^{-1})$$$$ = \varphi^{-1}(\varphi(y)\varphi(x)) = \varphi^{-1}(\varphi(y))\varphi^{-1}(\varphi(x)) = yx$$, where we used that $$\varphi^{-1}$$ is an isomorphism (that we know from exercise 2.3.11).

Now assume instead that $$G$$ is Abelian, then $$\varphi(xy) = (xy)^{-1} = y^{-1}x^{-1} = x^{-1}y^{-1} = \varphi(x)\varphi(y).$$