2.3:

(a) Suppose (x)s=x

^{-1}defines an automorphism of G. This map is always a bijective, so the hypothesis is really just saying (xy)^{-1}= x^{-1}y^{-1}for all x,y in G. This implies y^{-1}x^{-1}= x^{-1}y^{-1}, which gives x^{-1}= y x^{-1}y^{-1}, then x^{-1}y = y x^{-1}, y = xy x^{-1}, and finally, yx=xy for all x,y in G. Thus $G$ is abelian.(b) Suppose s : G ---> G is an automorphism satisfying s

^{2}= id and (x)s is different from x for x not equal to e. Consider the function G --> G mapping x to x^{-1}x^{s}. We claim this map is injective. Indeed, suppose x^{-1}x^{s}= y^{-1}y^{s}. Manipulating and using that s is a homomorphism, we get yx^{-1}= y^{s}(x^{s})^{-1}= (yx^{-1})^{s}. Then yx^{-1}= e, by hypothesis, so y=x. Since the map is injective, and G is finite, the map is surjective as well. Then, for any y in G, there is an x in G such that y = x^{-1}x^{s}. Applying s, we get that y^{s}= (x^{-1}x^{s})^{s}= (x^{-1})^{s}x^{ss}, which equals (x^{s})^{-1}x since ss = s^{2}, and this is equal to (x^{-1}x^{s})^{-1}, which is y^{-1}. Thus (y)s = y^{-1}for all y in G. Then G is abelian by part a.

This shows, for finite groups, the only possible automorphism of order two with no notrivial fixed points is the inversion mapping, in which cae G is abelian. What about infinite groups?