# Lower central series is fastest descending central series

## Statement

Suppose $G$ is a Nilpotent group (?) with a Central series (?):

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

Then, if we consider the 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 Nilpotence class (?) $c$, then:

$n \ge c + 1$

## 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 $n \ge c + 1$.