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? 



Groups in which lower central series and upper central series coincide
I thought you might find this interesting:



> 
> **Claim 1**: For a [UL-equivalent group](http://groupprops.subwiki.org/wiki/UL-equivalent_group), $\Gamma$, of [rank](http://en.wikipedia.org/wiki/Rank_of_a_group) $k$, 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](http://en.wikipedia.org/wiki/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, $H(\mathbb{Z}/p\mathbb{Z})\times \mathbb{Z}/p\mathbb{Z}$.


