Nilpotent group

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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.

Characterization and Properties of Nilpotent Groups

Theorem 1. 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 $n$ ;
There exists a sequence $G = G^1 \supseteq G^2 \supseteq \dotsb \supseteq G^{n+1} = \{e\}$ of subgroups of $G$ such that $G^{k+1} \subseteq (G,G^k)$ , for all integers $1\le k \le n$ .
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. To show that (1) implies (2), we may take $G^k = C^k(G)$ .

To show that (2) implies (1), we note that it follows from induction that $C^k \subseteq G^k$ ; hence $C^{n+1}(G) = \{e\}$ .

Now, we show that (1) implies (3). 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 $ghg^{-1} = h \cdot h^{-1}ghg^{-1} = h\cdot (h,g^{-1}) \in h\cdot (G,G^k) = h \cdot C^{k+1}(G) .$ 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 (3) implies (1), we may take $H= \{e\}$ .

To show that (1) implies (4), we may take $A = C^n(G)$ .

Finally, we show that (4) 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. $\blacksquare$

Corollary 2. Let $G$ be a nilpotent group; let $H$ be a subgroup of $G$ . If $H$ is its own normalizer, then $H=G$ .

Proof. Suppose $H\neq G$ ; then there is a greatest integer $k\in [1,n+1]$ for which $H^k \neq H$ . Then $H^k$ normalizes $H$ . $\blacksquare$

Corollary 3. Let $G$ be a nilpotent group; let $H$ be a proper subgroup of $H$ . Then there exists a proper normal subgroup $A$ of $G$ such that $H \subseteq A$ and $G/A$ is abelian.

Proof. In the notation of the theorem, let $k$ be the least integer such that $H^k \neq G$ . Then set $A=H^k$ . $\blacksquare$

Corollary 4. Let $G$ be a nilpotent group; let $H$ be a subgroup of $G$ . If $G = H(G,G)$ , then $G=H$ .

Proof. Suppose that $G \neq H$ . Then let $A$ be the normal subgroup of $G$ containing $H$ as described in Corollary 3. Then $(G,G) \subseteq A$ , so $H(G,G) \subseteq HA = A \subsetneq G,$ a contradiction. $\blacksquare$

Corollary 5. Let $G'$ be a group, let $G$ be a nilpotent group, and let $f: G' \to G$ be a group homomorphism for which the homomorphism $f' : G'/(G',G') \to G/(G,G)$ derived from passing to quotients is surjective. Then $f$ is surjective.

Proof. Let $H$ be the image of $f'$ and apply Corollary 3. $\blacksquare$

Proposition. Let $G$ be a group of nilpotency class at most $n$ , and let $N$ be a normal subgroup of $G$ . Then there exists a sequence $(N^k)_{1\le k \le n+1}$ of subgroups of $G$ such that $N^1=N$ , $N^{n+1}=\{e\}$ , $N^{k+1} \subseteq N^k$ , and $(G,N^k) \subseteq N^{k+1}$ , for all integers $1 \le k \le n+1$ .

Proof. Let $N^k = N \cap C^k(G)$ . Then $(G,N^k) \subseteq (G,G^k) = G^{k+1},$ and $(G,N^k) \subseteq (G,N) \subseteq N,$ since $N$ is a normal subgroup. $\blacksquare$

Corollary 6. Let $G$ be a nilpotent group; let $N$ be a normal subgroup of $G$ , and let $Z$ be the center of $G$ . If $N$ is not trivial, then $N \cap Z$ is not trivial.

Proof. In the proposition's notation, let $k$ be the greatest integer such that $N^k \neq \{e\}$ . The $(G,N^k) \subseteq N^{k+1} = \{e\}$ , so $N^k$ is a nontrivial subgroup that lies in the center of $G$ and in $N$ . $\blacksquare$

Corollary 7. Let $G$ be a nilpotent group, let $G'$ be a group, and let $f$ be a homomorphism of $G$ into $G'$ . If the restriction of $f$ to the center of $G$ is injective, then so is $f$ .

Proof. We proceed by contrapositive. Suppose that $f$ is not injective; then the kernel of $f$ is nontrivial, so by the previous corollary, the intersection of $\text{Ker}(f)$ and the center of $G$ is nontrivial, so the restriction of $f$ to the center of $G$ is not injective. $\blacksquare$

Finite Nilpotent Groups

Theorem 8. Let $G$ be a finite group. Then the following conditions are equivalent.

The group $G$ is nipotent;
The group $G$ is a product of $p$ -groups;
Every Sylow $p$ -subgroup of $G$ is normal in $G$ .

Proof. Since every $p$ -group is nilpotent, condition (2) implies condition (1).

Now we show that (1) implies (3). Let $P$ be a Sylow $p$ -subgroup of $G$ , and let $N$ be its normalizer. Then $N$ is its own normalizer. Then from Corollary 2, $N=G$ , i.e., $P$ is normal in $G$ .

Finally, we show that (3) implies (2). Suppose condition (3) holds for $G$ . For any prime $p$ dividing the order of $G$ , let $P_p$ denote the Sylow $p$ -subgroup of $H$ . Let $p$ and $q$ be distinct primes dividing the order of $H$ . Then $P_p \cap P_q = \{e\}$ , since the order of any element in both of these groups must divide a power of $p$ and a power of $q$ . Since $P_p$ and $P_q$ are both normal, it follows that for any $a\in P_p$ , $b\in P_q$ , the commutator $(a,b)$ is an element both of $P_p$ and $P_q$ . It follows that the canonical mapping of $f : \prod_p P_p \to G$ is a homomorphism, and $P_p$ is in its image, for every prime $p$ . Now, the order subgroup generated by the $P_p$ must be divisible by every power of a prime that divides $G$ , but it must also divide $G$ ; hence it is equal to $G$ . It follows that the order of the image of $\prod_p P_p$ is equal to the order of $G$ ; since $G$ is finite, this implies that $f$ is surjective. Since $G$ and $\prod_p P_p$ have the same size, $f$ is also injective, and hence an isomorphism. $\blacksquare$

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Nilpotent group

From AoPSWiki

Characterization and Properties of Nilpotent Groups

Finite Nilpotent Groups

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