AoPSWiki
Trying to get to the USAMO in 2010? Our AIME Problem Series can help you get there! Click here to enroll today!

Lower central series

From AoPSWiki

Revision as of 05:24, 1 June 2008 by Boy Soprano II (Talk | contribs)
(diff) ← Older revision | Current revision (diff) | Newer revision → (diff)

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.

See also

Want to learn how to tackle those tough AMC/AIME/Olympiad algebra problems? Check out Art of Problem Solving's Intermediate Algebra by Richard Rusczyk and Mathew Crawford. Over 1600 problems!
© Copyright 2008 AoPS Incorporated. All Rights Reserved. • FoundationPrivacyContact Us