# Lower central series

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.