Upper central series is fastest ascending central series

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Statement

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

\{ e \} = 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

Definitions used

Nilpotent group

Central series

Lower central series

Nilpotence class

Related facts

Facts used

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.