Nilpotent group

A nilpotent group can be thought of a group that is only finitely removed from an abelian group. Specifically, it is a group $G$ such that $C^{n+1}(G)$ is the trivial group, for some integer $n$ , where $C^m(G)$ is the $m$ th term of the lower central series of $G$ . The least integer $n$ satisfying this condition is called the nilpotency class of $G$ . Using transfinite recursion, the notion of nilpotency class can be extended to any ordinal.

All abelian groups have nilpotency class at most 1; the trivial group is the only group of nilpotency class 0.

Theorem. Let $G$ be a group, and let $n$ be a positive integer. Then the following three statements are equivalent:

The group $G$ has nilpotency class at most $n$ ;
For every subgroup $H$ of $G$ , there exist subgroups $H^1, \dotsc, H^{n+1}$ , such that $H^1=G$ , $H^{n+1}=H$ , and $H^{k+1}$ is a normal subgroup of $H^k$ such that $H^k/H^{k+1}$ is commutative, for all integers $1\le k \le n$ .
The group $G$ has a subgroup $A$ in the center of $G$ such that $G/A$ has nilpotency class at most $n-1$ .

Proof. First, we show that (1) implies (2). Set $H_k = H \cdot C^k(G)$ ; we claim that this suffices. We wish first to show that $H \cdot C^k(G)$ normalizes $H \cdot C^{k+1}(G)$ . Since $H$ evidently normalizes $H^{k+1}$ , it suffices to show that $C^k(G)$ does; to this end, let $g$ be an element of $C^k(G)$ and $h$ an element of $H \cdot C^{k+1}(G)$ . Then $ghg^{-1} = h \cdot h^{-1}ghg^{-1} = h\cdot (h,g^{-1} \in h\cdot (G,G^k) \in h \cdot C^{k+1}(G) .$ Thus $H^k$ normalizes $H^{k+1}$ . To prove that $H^k/H^{k+1}$ is commutative, we note that $C^k(G)/C^{k+1}(G)$ is commmutative, and that the canonical homomorphism from $C^k(G)/C^{k+1}(G)$ to $H^k/H^{k+1}$ is surjective; thus $H^k/H^{k+1}$ is commutative.

To show that (2) implies (1), we may take $H= \{e\}$ .

To show that (1) implies (3), we may take $A = C^n(G)$ .

Finally, we show that (3) implies (1). Let $\phi$ be the canonical homomorphism of $G$ onto $G/A$ . Then $\phi(C^k(G)) = C^k(G/A)$ . In particular, $\phi(C^n(G))= C^n(G/A)= \{e\}$ . Hence $C^n(G)$ is a subset of $A$ , so it lies in the center of $G$ , and $C^{n+1}(G)=\{e\}$ ; thus the nilpotency class of $G$ is at most $n$ , as desired. $\blacksquare$

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Nilpotent group

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