Lower central series is fastest descending central series

Statement

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

$G=H_{1}\geq H_{2}\geq \dots H_{n}=\{e\}$

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:

$H_{i}\geq G_{i}$

In particular, if $G$ has Nilpotence class (?) $c$, then:

$n\geq c+1$

Related facts

• Upper central series is fastest ascending central series

Facts used

We use the following fact:

• If $H_{1}\leq K_{1}\leq G$ and $H_{2}\leq K_{2}\leq G$, then $[H_{1},H_{2}]\leq [K_{1},K_{2}]$.

Proof

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

$G=H_{1}\geq H_{2}\geq \dots H_{n}=\{e\}$

Consider the lower central series of $G$:

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

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

$H_{i}\geq G_{i}$

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

$n\geq c+1$

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 $H_{i}\geq G_{i}$. Then by the definition of the lower central series:

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

Because $H_{i}\geq G_{i}$, we have, by the fact mentioned above:

$[G,H_{i}]\geq [G,G_{i}]$

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

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

Combining these facts, we see that:

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

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 $n\geq c+1$.