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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

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Statement

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

Facts used

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.