Upper central series is fastest ascending central series

Statement

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

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

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

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

Then, we have:

$Z^i(G)\ge H_i$

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

$n\ge c$

Related facts

• Lower central series is fastest descending central series

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 $n\ge c$.