Vaughan says -- correctly I think -- that "you can't define sets in the internal logic of Ab." I suspect that the proof will have to use the dualities that reside in Ab. Let Gr be the category of all groups, abelian or not. Then: The category of co-groups in Gr is equivalent to Set. Cf: The category of co-groups in Ab is equivalent to Ab. "Co-group" means, of course, "group in the opposite category". The second of the two assertions above holds for any additive category: it's easy to show that each object in any additive category has a unique co-group (indeed, a unique co-monoid) structure (the co-multiplication is the diagonal map). I know no citation for the first assertion. To prove it, one shows that the only groups with co-group structures are free groups and that the only co-multiplication on Free(X) is such that X -> Free(X) -> Free(X+X) is the point-wise product of the two functions of the form X -> X+X -> Free(X+X).