# 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 nilpotence 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

Further information: Faithful semidirect product of cyclic p-groups

Let $p$ be an odd prime. Let $H$ be the cyclic group of order $p^c$, and $K$ be the cyclic group of order $p^{c-1}$. Consider the action of $K$ on $H$, where the generator acts via multiplication by $p + 1$. Let $G = H \rtimes K$ be the semidirect product.

The group $Z_k(G)$ in this case is isomorphic to the semidirect product of the cyclic group of order $p^k$ with the cyclic group of order $p^{k-1}$, with the generator acting via multiplication by $p + 1$. This tells us that the upper central series of $Z_k(G)$ is precisely the first $k$ terms of the upper central series of $G$, as required.