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
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:
is bigger than the subgroup:
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.
- 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.
- 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.
Given: A nilpotent group , the lower central series of defined by ,
Proof: We prove the result by induction on (letting vary freely; note that we need to apply the result for multiple values of for the same in the induction step).
Base case for induction: For , we have equality:
Induction step: Suppose we have, for all , that . Now, consider the three subgroups:
Applying the three subgroup lemma to these yields that is contained in the normal closure of the subgroup generated by and .
- (by induction assumption)
- (where the first inequality is by induction assumption)
Since is normal, the normal closure of the subgroup generated by both is in , hence the three subgroup lemma yields:
which is what we require.