Upper central series is fastest ascending central series

Statement

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

$\{e\}=H_{0}\leq H_{1}\leq H_{2}\leq \dots H_{n}=G$

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:

$Z^{i}(G)\geq H_{i}$

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

$n\geq c$

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):

$\{e\}\leq H_{0}\leq H_{1}\leq H_{2}\leq \dots H_{n}=G$

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: $Z^{i}(G)\geq H_{i}$

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

$n\geq c$

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 $Z^{i}(G)\geq H_{i}$. We want to show that $Z^{i+1}(G)\geq H_{i+1}$.

Since $[G,H_{i+1}]\leq H_{i}$, and $H_{i}\leq Z^{i}(G)$, 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 $H_{i+1}Z^{i}(G)\leq Z^{i+1}(G)$ so $H_{i+1}\leq Z^{i+1}(G)$ as required.

Since the nilpotence class $c$ is the smallest integer for which $Z^{c}(G)=G$, $H_{n}=G$ must force $n\geq c$.