Upper central series is fastest ascending central series

Statement
Suppose $$G$$ is a fact about::nilpotent group with a fact about::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 fact about::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 fact about::nilpotence class of $$G$$, we have:

$$n \ge c$$

Related facts

 * Lower central series is fastest descending central series

Proof
Given: $$G$$ is a fact about::nilpotent group with a fact about::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 fact about::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 fact about::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 $$n \ge c$$.