Lower central series is strongly central

This fact is an application of the following pivotal fact/result/idea: three subgroup lemma
View other applications of three subgroup lemma OR Read a survey article on applying three subgroup lemma

Statement

The lower central series of a nilpotent group is a strongly central series.

Explanation

Intuitively, what we're saying is that the slowest way to make commutators fall is by bracketing them completely to one side. Thus, for instance, doing a bracketing like:

$[[[G,G],G],G]$

is bigger than the subgroup:

$[[G,G],[G,G]]$

This is closely related to the fact that the property of being a nilpotent group, which is characterized by the lower central series reaching the identity, is substantially stronger than the property of being a solvable group, which is characterized by the derived series reaching the identity.

Related facts

Stronger facts

• Centralizer relation between lower and upper central series: This states that members of the lower central series centralize corresponding members of the upper central series.

Applications

• Second half of lower central series of nilpotent group comprises Abelian groups
• Solvable length is logarithmically bounded by nilpotence class
• Penultimate term of lower central series is Abelian in nilpotent group of class at least three
• Nilpotent and every Abelian characteristic subgroup is central implies class at most two

Breakdown for upper central series

• Upper central series not is strongly central: There are groups where the upper central series is not a strongly central series.

Facts used

1. Three subgroup lemma

Proof

Given: A nilpotent group $G$, the lower central series of $G$ defined by $G_{1}=G$, $G_{m}=[G,G_{m-1}]$

To prove: $[G_{m},G_{n}]\leq G_{m+n}$

Proof: We prove the result by induction on $n$ (letting $m$ vary freely; note that we need to apply the result for multiple values of $m$ for the same $n$ in the induction step).

Base case for induction: For $n=1$, we have equality: $[G_{m},G]=G_{m+1}$

Induction step: Suppose we have, for all $m$, that $[G_{m},G_{n-1}]\leq G_{m+n-1}$. Now, consider the three subgroups:

• $A=G_{n-1}$
• $B=G_{1}$
• $C=G_{m}$

Applying the three subgroup lemma to these yields that $[[G_{n-1},G_{1}],G_{m}]$ is contained in the normal closure of the subgroup generated by $[[G_{1},G_{m}],G_{n-1}]$ and $[[G_{m},G_{n-1}],G_{1}]$.

We have:

• $[[G_{1},G_{m}],G_{n-1}]=[G_{m+1},G_{n-1}]\leq G_{m+n}$ (by induction assumption)
• $[[G_{m},G_{n-1}],G_{1}]\leq [G_{m+n-1},G_{1}]=G_{m+n}$ (where the first inequality is by induction assumption)

Since $G_{m+n}$ is normal, the normal closure of the subgroup generated by both is in $G_{m+n}$, hence the three subgroup lemma yields:

$[[G_{n-1},G_{1}],G_{m}]\leq G_{m+n}\implies [G_{n},G_{m}]\leq G_{m+n}$

which is what we require.