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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.

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Facts used


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.