Lower central series is strongly central

Statement
The fact about::lower central series of a fact about::nilpotent group is a fact about::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.

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) uses::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: $$[G_m, G_n] \le G_{m+n}$$

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 $$[G_m,G_{n-1}] \le G_{m+n-1}$$. 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:


 * $$[[G_1,G_m],G_{n-1}] = [G_{m+1},G_{n-1}] \le G_{m+n}$$ (by induction assumption)
 * $$[[G_m,G_{n-1}],G_1] \le [G_{m+n-1},G_1] = G_{m+n}$$ (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:

$$[[G_{n-1},G_1],G_m] \le G_{m+n} \implies [G_n,G_m] \le G_{m+n}$$

which is what we require.