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 $k \le c$, $Z_k(G)$ has nilpotency class precisely $k$.

Related facts

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

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 $k$.