Upper central series is fastest ascending central series

Statement

Suppose $G$ is a Nilpotent group (?) with a Central series (?) (written in ascending order as):

$\{ e \} = H_0 \le H_1 \le H_2 \le \dots H_n = G$

Denote by $Z^i(G)$ the $i^{th}$ member of the Upper central series (?) of $G$, i.e.:

$Z^0(G) = \{ e \}, Z^1(G) = Z(G), Z^i(G)/Z^{i-1}(G) = Z(G/Z^{i-1}(G))$

Then, we have:

$Z^i(G) \ge H_i$

In particular, if $c$ is the Nilpotence class (?) of $G$, we have:

$n \ge c$

Proof

Given: $G$ is a Nilpotent group (?) with a Central series (?) (written in ascending order as):

$\{ e \} \le H_0 \le H_1 \le H_2 \le \dots H_n = G$

Denote by $Z^i(G)$ the $i^{th}$ member of the Upper central series (?) of $G$, i.e.:

$Z^0(G) = \{ e \}, Z^1(G) = Z(G), Z^i(G)/Z^{i-1}(G) = Z(G/Z^{i-1}(G))$

To prove: $Z^i(G) \ge H_i$

In particular, if $c$ is the Nilpotence class (?) of $G$, we have:

$n \ge c$

Proof: We prove this by induction on $i$.

Base case for induction: For $i = 0$, $Z^0(G) = H_0 = \{ e \}$ so we're okay.

Induction step: Suppose $Z^i(G) \ge H_i$. We want to show that $Z^{i+1}(G) \ge H_{i+1}$.

Since $[G,H_{i+1}] \le H_i$, and $H_i \le Z^i(G)$, we see that under the projection $G \to G/Z^i(G)$, the image of elements of $H_i$ are in the center of $G/Z^i(G)$. Thus, $H_{i+1}Z^i(G)/Z^i(G)$ is in the center of $G/Z^i(G)$, so $H_{i+1}Z^i(G) \le Z^{i+1}(G)$ so $H_{i+1} \le Z^{i+1}(G)$ as required.

Since the nilpotence class $c$ is the smallest integer for which $Z^c(G) = G$, $H_n = G$ must force $n \ge c$.