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, $\begin{align*}(C^{n-1}(G),(G,C^m(G)) &= (C^{n-1}(G),C^{m+1}(G)) = (C^{m+1}(G),C^{n-1}(G)) \\&\subseteq C^{m+n}(G) .\end{align*}$ Hence $(C^n(G),C^m(G)) \subseteq C^{m+n}(G),$ as desired.

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Lower central series

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