Normal subgroups and Abelian property

1. Sep 16, 2008

soumyashant

If all subgroups of a group are normal, is the group abelian?
I know that the answer is NO...

Can you give a counter-example..???

Better still can you logically deduce some property the counter-example must have, which will ease our way to finding it...

2. Sep 16, 2008

morphism

Whenever you want to look for a nonabelian counterexample, it's usually good to try the quaternion group Q first (and then the symmetric and dihedral groups!). In the case at hand this strategy is particularly fruitful, because
(1) Q does have this property (all its subgroups are normal), and
(2) every group which has this property contains a copy of Q as a subgroup.

The idea is that if the group G which we're after is nonabelian, then it contains two noncommuting elements x and y in G. If we let Q0 be the subgroup of G generated by x and y, then it can be shown that Q0 is isomorphic to the quaternion group. For a proof, see Rotman's Theory of Groups.

3. Sep 17, 2008

soumyashant

Thanks for the example and the suggestion...

Firs thing I tried with was dihedral groups and was unsuccessful, quaternion groups never crossed my mind...
I tried with the Klein four subgroup (dihedral group with $$x^2=e,y^2=e$$) which satisfies the condition, but is Abelian and thus obviously ahould satisfy the conditions ,of no use...

A little more patience may have helped because the quaternion group is similar to Klein four group...

For example,
In Klein four subgroup, $$x^2=y^2=z^2=e$$ and $$xy=z,yz=x,xz=y$$
And in quaternion,$$x^2=y^2=z^2=-e$$ and $$xy=z,yz=x,zx=y$$
Is there some general quaternion group??
I only know of the quaternion group of order 8...

In that case, how can we have bijection between the subgroup generated by $$x,y$$ and quaternion group if their cardinality is not the same and thus, isomorphism??...

EDIT: In Wikipedia, I see something called the generalised quaternion group, did you mean that??

Last edited: Sep 17, 2008
4. Sep 17, 2008

morphism

Maybe my outline of the proof was overly simplistic. I don't remember much of it, but what I do remember is that you can always choose two appropriate noncommuting elements x and y such that the group generated by {x,y} is isomorphic to the quaternion group of order 8. The proof isn't exactly easy though.