# Upper central series may be tight with respect to nilpotency class

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## Statement

Let $c$ be any natural number. Then, we can construct a Nilpotent group (?) $G$ 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 $k \le c$, the upper central series of $Z_k(G)$ is precisely the first $k$ terms of the upper central series of $G$. In particular, we can find a $G$ with the property that each $Z_k(G)$ has nilpotence class precisely $k$: in other words, it is not a group of class $k - 1$.

## Related facts

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

## 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 $k$.