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.

Wikipedia says to look at the quaternion group, but then doesn't really expand upon that at all. So instead I will.

Let Q={1,-1,i,-i,j,-j,k,-k}. This is the quaternion group of order 8. Wikipedia has some more information here:http://en.wikipedia.org/wiki/Quaternion_group

anyway, the idea is that ij=k jk=i ki=j and yet ij=-k kj=-i ik=-j.
So, ijk=kk=-1 jki=ii=-1, etc. The inverse of i is -i, similar for j and k.

The only proper subgroups are {1,-1} {1,-1,i,-i} {1,-1,j,-j} and {1,-1,k,-k}. These are also all abelian, although Q is not.

{1,-1} is obviously normal.
Let's test {1,-1,i,-i}. We already know about 1 and -1, so let's look at i, -i.

Basically, we want to show ji(-j) is in {1,-1,i,-i) for normal. And ji=-k, -k(-j)=-i, which is in the subgroup.
The same is true for (-j)ij and ki(-k),(-k)ik and everything with (-i) as well. You can also check this with the other two subgroups.

You can check all of this and it would probably be best to go through it yourself, but this is a good example of a non-abelian group, yet every subgroup is normal.