Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $G$ a finite two-generated $p$-group in which lower and upper central series coincide. Clearly we obtain that the upper central series become strongly central, we have also that at least half of the members of the upper central series are abelian. Are there not immediate results that involve this kind of group?

share|improve this question
    
The folk at the group props call a nilpotent group in which the lower and upper central series coincide UL-equivalent. The relevant page is groupprops.subwiki.org/wiki/UL-equivalent_group –  Nick Gill Apr 24 at 15:31
    
Some of the answers to this question, mathoverflow.net/questions/30750/…, might be of interest to you. –  Khalid Bou-Rabee Apr 24 at 15:34
    
@NickGill Thank you for the comment. I've seen on group props before write the question, unfortunately i'd not able to find anything useful since the only weaker property listed is the nilpotence, but i'm studying a problem on a $p$- group, and all $p$-groups are nilpotents. –  Marco Ruscitti Apr 24 at 15:47
1  
@KhalidBou-Rabee Thank you for the comment. I've seen yesterday the thread that you linked to me, there are listed in it examples of groups which have the UL-equivalence, unfortunately i was searching results that assume the UL-equivalence like an hypothesis. –  Marco Ruscitti Apr 24 at 15:50
    
@MarcoRuscitti, It might help if you described the kind of results you're interested in. Do you have specific questions / conjectures? –  Nick Gill Apr 24 at 16:13

1 Answer 1

up vote 1 down vote accepted

I thought you might find this interesting:

Claim 1: For a UL-equivalent group, $\Gamma$, of rank $k$, we have, for any natural number $i$, for which $\Gamma_{i+1} \neq 1$, $$ | \Gamma_i / \Gamma_{i+1} | \leq |\Gamma_{i+1}/\Gamma_{i+2}|^k. $$

Proof: Since $\Gamma$ has rank $k$, we can fix a generating set $x_1, \ldots, x_k$ for $\Gamma$. Consider the map $\varphi: \Gamma_i \to (\Gamma_{i+1}/\Gamma_{i+2})^k$ given by $g \mapsto ([x_1,g], [x_2, g], \ldots, [x_k,g] )$. This map is a homomorphism since in any group $[x,yz] = [x,z][x,y]^z$. Further, its kernel is $\Gamma_{i+1}$, as for any $g \in \Gamma_i \setminus \Gamma_{i+1}$, if $\varphi(g) = 1$, then $g$ is in the center of $\Gamma/\Gamma_{i+2}$. But this is impossible, as the center of $\Gamma/\Gamma_{i+2}$ is contained in $\Gamma_{i+1}/\Gamma_{i+2}$ since $\Gamma$ is a UL-equivalent group. QED

Some remarks:

  1. Claim 1 is sharp. For the Heisenberg group, $H(\mathbb{Z}/p\mathbb{Z})$, over $\mathbb{Z}/p \mathbb{Z}$, which has rank 2, we have that the center has order $p$ and the abelianization has order $p^2$.
  2. Claim 1 is not true for all nilpotent groups, as the example $\left< x, y, z \;|\; x^4 = y^4 = z^4 = 1, [x,y] = z^2, x^2,y^2,z \text{ are central} \right>$, demonstrates.
  3. Claim 1 is true for some groups that are not UL-equivalent groups. That is, the conclusion of Claim 1 is not equivalent to the property of being UL-equivalent. Consider, for instance, $H(\mathbb{Z}/p\mathbb{Z})\times \mathbb{Z}/p\mathbb{Z}$.
share|improve this answer
    
I'm grateful for this answer. –  Marco Ruscitti Apr 26 at 14:55

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.