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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.

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:

  1. The group $G$ has nilpotency class at most $n$;
  2. 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 $1\le k \le n$.
  3. 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 $1\le k \le n$.
  4. 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.

  1. The group $G$ is nipotent;
  2. The group $G$ is a product of $p$-groups;
  3. 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. $\blacksquare$

See also

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