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Nilpotence via central series, the upper central series, and the lower central series

Statement

For a group $G$ and $c \in N$ , the following are equivalent:

$G$ has a central series $1 = H_{0} \leq \dots \leq H_{c} = G$ ;
$Z_{c} (G) = G$ ;
$γ_{c + 1} (G) = 1$ .

Hence $G$ is nilpotent exactly when its lower central series reaches $1$ , and the least such $c$ is its nilpotency class.

Facts & Assumptions

Given: A group $G$ and $c \in N$ .

[F1]

$γ_{1} (G) = G$ and $γ_{r + 1} (G) = [G, γ_{r} (G)]$ (Subgroup commutators and the lower central series).

[F2]

$Z_{0} (G) = 1$ and $Z_{r + 1} (G) / Z_{r} (G) = Z (G / Z_{r} (G))$ (The upper central series).

[F3]

$G$ is nilpotent of class $c$ exactly when $c$ is least with $Z_{c} (G) = G$ (Nilpotent groups and nilpotency class).

[L1]

A chain $1 = H_{0} \leq \dots \leq H_{c} = G$ is central exactly when $[G, H_{i + 1}] \leq H_{i}$ for every $i < c$ (Central factors are equivalent to adjacent commutator containments).

Proof

technique · direct

1.1assume-hypF2L1

Let $1 = H_{0} \leq \dots \leq H_{c} = G$ be central. Inductively, $H_{i} \leq Z_{i} (G)$ : the base is $H_{0} = Z_{0} (G) = 1$ , and [L1] says $[G, H_{i + 1}] \leq H_{i} \leq Z_{i} (G)$ , so the quotient criterion places $H_{i + 1} / Z_{i} (G)$ in $Z (G / Z_{i} (G))$ , hence $H_{i + 1} \leq Z_{i + 1} (G)$ .

1.2F1L1

For any central series as above, descending induction gives $γ_{c - i + 1} (G) \leq H_{i}$ : at $i = c$ , $γ_{1} (G) = G = H_{c}$ ; if $γ_{c - i + 1} (G) \leq H_{i}$ , then $γ_{c - i + 2} (G) = [G, γ_{c - i + 1} (G)] \leq [G, H_{i}] \leq H_{i - 1}$ by [L1].

1.3assume-hypF1L1

Conversely, if $γ_{c + 1} (G) = 1$ , the reversed lower central chain $1 = γ_{c + 1} (G) \leq γ_{c} (G) \leq \dots \leq γ_{1} (G) = G$ is central because $[G, γ_{r} (G)] = γ_{r + 1} (G)$ ; [L1] applies at every adjacent pair.

2.1step 1.1F2L1

At $i = c$ , step 1.1 gives $G = H_{c} \leq Z_{c} (G) \leq G$ , so $Z_{c} (G) = G$ . Conversely, the upper central chain $1 = Z_{0} (G) \leq \dots \leq Z_{c} (G) = G$ is central by [F2] and [L1].

2.2step 1.2

Taking $i = 0$ in step 1.2 gives $γ_{c + 1} (G) \leq H_{0} = 1$ .

3.1step 2.1step 2.2step 1.3F3∎

Steps 2.1 and 2.2 show that a central series is equivalent to both upper-central termination and lower-central termination, while step 1.3 constructs a central series from lower-central termination. Taking least $c$ and using [F3] identifies the nilpotency class with the lower-central termination index.

Nilpotence via central series, the upper central series, and the lower central series

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources