Lower central series is fastest descending central series

Statement
Suppose $$G$$ is a fact about::nilpotent group with a fact about::central series:

$$G = H_1 \ge H_2 \ge \dots H_n = \{ e \}$$

Then, if we consider the fact about::lower central series of $$G$$:

$$G_1 = G, G_{i+1} = [G_i,G]$$

Then, for every $$i$$, we have:

$$H_i \ge G_i$$

In particular, if $$G$$ has fact about::nilpotence class $$c$$, then:

$$n \ge c + 1$$

Related facts

 * Upper central series is fastest ascending central series

Facts used
We use the following fact:


 * If $$H_1 \le K_1 \le G$$ and $$H_2 \le K_2 \le G$$, then $$[H_1,H_2] \le [K_1,K_2]$$.

Proof
Given': $$G$$ is a nilpotent group with a central series:

$$G = H_1 \ge H_2 \ge \dots H_n = \{ e \}$$

Consider the lower central series of $$G$$:

$$G_1 = G, G_{i+1} = [G_i,G]$$

To prove: For every $$i$$, we have:

$$H_i \ge G_i$$

In particular, if $$G$$ has nilpotence class $$c$$, then:

$$n \ge c + 1$$

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

Base case for induction: For $$i = 1$$, $$H_1 = G_1 = G$$ so it's true.

Induction step: Suppose $$H_i \ge G_i$$. Then by the definition of the lower central series:

$$[G,G_i] = G_{i+1}$$

Because $$H_i \ge G_i$$, we have, by the fact mentioned above:

$$[G,H_i] \ge [G,G_i]$$

Finally, by the definition of central series, we have:

$$H_{i+1} \ge [G,H_i]$$

Combining these facts, we see that:

$$H_{i+1} \ge G_{i+1}$$

Thus, if $$H_n$$ is trivial, $$G_n$$ must be trivial. Since the smallest $$n$$ for which $$G_n$$ is trivial is $$c + 1$$, we have $$n \ge c + 1$$.