Normal subgroups and Abelian property

[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...

 

[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.

 

[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

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.

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??

 

[morphism]

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??...

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.