abelian/dedekind/hamiltonian groups

I just proved for homework that if A and B are two normal subgroups of a group G such that the intersection of A and B is only the identity element, than ab = ba for any a in A and any b in B.

If you have a group all of whose subgroups are normal, doesn't that group have to be abelian? Because for any two elements a and b in this group, each generates a cyclic subgroup, which is normal by assumption, and therefore by the result in the previous paragraph, ab=ba and the group is abelian.

However, I googled this and apparently there can exist a nonabelian group whose subgroups are all normal, called a Dedekind group. I don't see how this is possible, could some kind anon point me in the right direction to see the distinction here?

Thank in advance.

>>4
Oh, shit, you're right, simple example, G containing the cyclic subgroups <2> and <4>, <2> is a subset of <4> and their intersection is not just identity, which means that my logic was right, but my assumption for my logic isn't always true.

Thanks for clearing that up, it's much appreciated.