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Lower central series

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The lower central series of a group is a particular decreasing sequence of subgroups of that group.

Specifically, let G be a group. The lower central series of G is the sequence (C^n(G))_{n\ge 1} defined recursively as follows: C^1(G)=G, \qquad C^{n+1}(G) = (G,C^n(G)), where (H,K) denotes the commutator group of two subgroups H,K of G. It follows from induction that C^{n+1}(G) is a subgroup of C^n(G).

A group G is called nilpotent if C^n(G) is the trivial group for sufficiently large n.

Theorem 1. Let G and G' be groups, and let f be a group homomorphism mapping G into G'. Then for all positive integers n, f(C^n(G)) = C^n(f(G)) \subseteq C^n(G') . Thus when f is surjective, f(C^n(G)) = C^n(G'). Also, the subgroup C^n(G) is characteristic (and in particular, normal) in G.

Proof. We induct on n to prove the main statement. For n=1, we have C^n(G)=G and the theorem follows.

Now suppose the theorem holds for n. Since the group C^{n+1}(f(G)) is generated by the elements of the form a^{-1}b^{-1}ab, for a\in f(G) and b\in C^n(f(G)) = f(C^n(G)), it follows that f(C^{n+1}(G)) = C^{n+1}f(G). Since f(G) \subseteq G' and C^n(f(G)) \subseteq C^n(G'), it follows similarly that C^{n+1}(f(G)) \subseteq C^{n+1}(G'); equality evidently occurs when f is surjective. By applying the theorem to the automorphisms of G, we see that C^n(G) is a characteristic subgroup of G. \blacksquare

Theorem 2. For all positive integers m,n, (C^m(G),C^n(G)) \subseteq C^{m+n}(G).

Proof. We use strong induction on the quantity m+2n. Our base cases, m=1 and n=1, follow from definition.

Now, suppose that m,n>1, and that the inductive hypothesis holds. Then by properties of commutators, (C^m(G),C^n(G)) = (C^m(G),(G,C^{n-1}(G)) \subseteq (G,(C^m(G),C^{n-1}(G)) \cdot (C^{n-1}(G),(G,C^m(G)) . By inductive hypothesis, (C^m(G),C^{n-1}(G)) \subseteq C^{m+n-1}(G), so (G,(C^m(G),C^{n-1}(G)) \subseteq (G,C^{m+n-1}(G)) = C^{m+n}(G). Also by inductive hypothesis, \begin{align*}(C^{n-1}(G),(G,C^m(G)) &= (C^{n-1}(G),C^{m+1}(G)) = (C^{m+1}(G),C^{n-1}(G)) \\&\subseteq C^{m+n}(G) .\e... Hence (C^n(G),C^m(G)) \subseteq C^{m+n}(G), as desired.

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