Upper central series may be tight with respect to nilpotency class

Statement

Let $c$ be any natural number. Then, we can construct a nilpotent group $G$ of nilpotency class $c$ with the following property.

Let $Z_{k}(G)$ denote the $k^{th}$ member of the Upper central series (?) of $G$: $Z_{1}(G)=Z(G)$ is the center and $Z_{k}(G)/Z_{k-1}(G)$ is the center of $G/Z_{k-1}(G)$ for all $k$. By definition of Nilpotency class (?), $Z_{c}(G)=G$.

We can find a $G$ with the property that for any $k\leq c$, $Z_{k}(G)$ has nilpotency class precisely $k$.

Related facts

Opposite facts for lower central series

The corresponding statement is not true for the lower central series. Some related facts:

• Lower central series is strongly central
• Second half of lower central series of nilpotent group comprises abelian groups
• Penultimate term of lower central series is abelian in nilpotent group of class at least three

Opposite facts for upper central series

It is also true that the upper central series for any member of the upper central series (beyond the center) grows faster than the actual upper central series of the whole group. See:

• Nilpotent of class at least three implies center of second center strictly contains center

Proof

Let $H_{1},H_{2},\dots H_{c}$ be groups such that each $H_{k}$ is a nilpotent group of nilpotency class precisely $k$, i.e., it is not nilpotent of class smaller than $k$. Define $G$ as the external direct product:

$G=H_{1}\times H_{2}\times \dots \times H_{c}$

Now, for each $k$, we have:

$Z_{k}(G)=Z_{k}(H_{1})\times Z_{k}(H_{2})\times \dots \times Z_{k}(H_{c})$

In particular, we obtain that:

$Z_{k}(G)=H_{1}\times H_{2}\times \dots \times H_{k}\times Z_{k}(H_{k+1})\times \dots \times Z_{k}(H_{c})$

From the given data, in particular the fact that $H_{k}$ has nilpotency class exactly $k$, it is clear that $Z_{k}(G)$ has nilpotency class exactly $k$.