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.

Science & Math, my ass. Someone here has to know what I'm talking about, I'm still undergrad...

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.

>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.

But didn't you assume in the first paragraph that the intersection of the subgroup was only the identity element?  Will that always be true?

>>4
That's meant to be "subgroups", sorry.

>>3
Alright, I understand that example. But, then there must be an error in my proof and I don't see where.

If every subgroup is normal, then every cyclic subgroup generated by any one element is normal. Any two different cyclic subgroups interesect only at the identity element.

Since every cyclic subgroup is normal, then for two such subgroups K and N, if n is in N and k is in K, then (n^-1)kn is in K, let's say (n^-1)kn = b which is in K, and then kn = nb and kn(k^-1) = nb(k^-1) which is in N, and if nb(k^-1) is in N, then b(k^-1) must be identity, so b must be k, and nk = kn for any n and k. Since any two elements of G generate a cyclic subgroup which is normal by assumption, shouldn't every single element of G commute with every other element of G? Obviously not, by your example, but then where is the hole in my logic? I have goneover it about 5 times and I don't see it.

>>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.

>>7

No problem.  Group theory is tricky to get your head around, and we only finished the introductory module last week so for a brief moment after hitting 'Reply' I was worried that I was talking out of my arse.  Sometimes it's not the proof - it's knowing exactly what you can and can't assume...

Another quick thing about >>3

Every subgroup is normal AND cyclic.

{1,-1} is generated by <-1>
{1,-1,i,-i} is generated by <i> or <-i>

Thus, it is also an example where the cyclic subgroups do not intersect at just the identity.

On a side note, you have the reverse proposition : if a group is abelian then all its subgroups are normal. But that's quite trivial...

except where A/0=/A then A={1} is not the same number set as /A={xia}.

Cant divide by zero...think again. divide by zero takes you into an alternate universe numberset unrelated to our own except at superposition.

But I can divide by zero ! Look : 1/0... OH SHI-

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