> Since G is abelian, the map x -> -x is an automorphism. > Since this must be trivial, we have x + x = 0 for all > x. Hence G is a vector space over Z_2. And now as > above, if dim(G) = 0 or 1 then |G| = 1 or 2, while > if dim(G) > 1 then G has a non-trivial automorphism.
Is this necessarily true without the axiom of choice? With it, yes, it is true: you just take a base of G over Z_2 and then you use it to get a non-trivial automorphism. But without the axiom of choice, I don't see why is it still possible to get such an automorphism.