Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

First of all, I would like to call a group immaculate provided that the orders of $G$ and $\Sigma$ (the order of $N$) where $N$ varies over all normal subgroups of $G$, are equal. From here it has been said that any abelianizer of a group whose order is less than that of $D(G)$, where $D(G)$ is the latter function defined above, is cyclic.

Then my question is: Is there any characterization of groups $G$ such that $G^{ab}$ is an imamculate group where $G^{ab}$ is the abelianizer of $G$ which coincides with the previous one in the case that $G$ is a cyclic group of the order a perfect number.

This may be referenced to here, in any case, thanks.

share|improve this question
    
Maybe call them immaculate groups, like in this MO question: goo.gl/Ez8XZ –  Myself Feb 16 '11 at 14:43
    
Thanks, I just cannot find that post again. And this question obviously originates from the proof of the examples given in the link. –  awllower Feb 16 '11 at 14:47
    
The fact that non-cyclic p-groups cannot be immaculate immediately implies non-cyclic nilpotent groups cannot be immaculate. –  user641 Feb 16 '11 at 18:48
    
Well, @Steve D, can you be more specific, please, thanks. –  awllower Feb 17 '11 at 4:03

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.