Lower central series is strongly central
From Groupprops
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
Contents |
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
Proof
Given: A nilpotent group G, the lower central series of G defined by G1 = G, Gm = [G,Gm − 1]
To prove: ![[G_m, G_n] \le G_{m+n}](/w/images/math/c/b/4/cb4fc948effac1a4ce8b5ade59e3708f.png)
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: [Gm,G] = Gm + 1
Induction step: Suppose we have, for all m, that ![[G_m,G_{n-1}] \le G_{m+n-1}](/w/images/math/5/1/d/51dba9b45265caa31c8674054ddfe51b.png)
. Now, consider the three subgroups:
- A = Gn − 1
- B = G1
- C = Gm
Applying the three subgroup lemma to these yields that [[Gn − 1,G1],Gm] is contained in the normal closure of the subgroup generated by [[G1,Gm],Gn − 1] and [[Gm,Gn − 1],G1].
We have:
-
![[[G_1,G_m],G_{n-1}] = [G_{m+1},G_{n-1}] \le G_{m+n}](/w/images/math/2/b/5/2b5f61e0af94e2d30198080e9d64453d.png)
(by induction assumption)
-
![[[G_m,G_{n-1}],G_1] \le [G_{m+n-1},G_1] = G_{m+n}](/w/images/math/7/3/c/73c51439be72bf54ca262584e38e7fe8.png)
(where the first inequality is by induction assumption)
Since Gm + n is normal, the normal closure of the subgroup generated by both is in Gm + 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}](/w/images/math/6/3/2/632b3a5122d1421ed8762eab24941802.png)
which is what we require.
