Difference between revisions of "Nilpotent group"

Latest revision as of 16:55, 5 June 2008

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. $\blacksquare$

@@ Line 5: / Line 5: @@
 == Characterization and Properties of Nilpotent Groups ==
-'''Theorem.''' Let <math>G</math> be a group, and let <math>n</math> be a positive integer.  Then the following three statements are equivalent:
+'''Theorem 1.''' Let <math>G</math> be a group, and let <math>n</math> be a positive integer.  Then the following three statements are equivalent:
 # The group <math>G</math> has nilpotency class at most <math>n</math>;
+# There exists a sequence <cmath> G = G^1 \supseteq G^2 \supseteq \dotsb \supseteq G^{n+1} = \{e\} </cmath> of subgroups of <math>G</math> such that <math>G^{k+1} \subseteq (G,G^k)</math>, for all integers <math>1\le k \le n</math>.
 # For every subgroup <math>H</math> of <math>G</math>, there exist subgroups <math>H^1, \dotsc, H^{n+1}</math>, such that <math>H^1=G</math>, <math>H^{n+1}=H</math>, and <math>H^{k+1}</math> is a [[normal subgroup]] of <math>H^k</math> such that <math>H^k/H^{k+1}</math> is [[commutative]], for all integers <math>1\le k \le n</math>.
 # The group <math>G</math> has a subgroup <math>A</math> in the [[center (algebra) |center]] of <math>G</math> such that <math>G/A</math> has nilpotency class at most <math>n-1</math>.
-''Proof.''  First, we show that (1) implies (2).  Set <math>H_k = H \cdot C^k(G)</math>; we claim that this suffices.  We wish first to show that <math>H \cdot C^k(G)</math> [[normalizer |normalize]]s <math>H \cdot C^{k+1}(G)</math>.  Since <math>H</math> evidently normalizes <math>H^{k+1}</math>, it suffices to show that <math>C^k(G)</math> does; to this end, let <math>g</math> be an element of <math>C^k(G)</math> and <math>h</math> an element of <math>H \cdot C^{k+1}(G)</math>.  Then
+''Proof.'' To show that (1) implies (2), we may take <math>G^k = C^k(G)</math>.
+To show that (2) implies (1), we note that it follows from induction that <math>C^k \subseteq G^k</math>; hence <math>C^{n+1}(G) = \{e\}</math>.
+Now, we show that (1) implies (3).  Set <math>H_k = H \cdot C^k(G)</math>; we claim that this suffices.  We wish first to show that <math>H \cdot C^k(G)</math> [[normalizer |normalize]]s <math>H \cdot C^{k+1}(G)</math>.  Since <math>H</math> evidently normalizes <math>H^{k+1}</math>, it suffices to show that <math>C^k(G)</math> does; to this end, let <math>g</math> be an element of <math>C^k(G)</math> and <math>h</math> an element of <math>H \cdot C^{k+1}(G)</math>.  Then
 <cmath> 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) . </cmath>
 Thus <math>H^k</math> normalizes <math>H^{k+1}</math>.  To prove that <math>H^k/H^{k+1}</math> is commutative, we note that <math>C^k(G)/C^{k+1}(G)</math> is commmutative, and that the canonical homomorphism from <math>C^k(G)/C^{k+1}(G)</math> to <math>H^k/H^{k+1}</math> is [[surjective]]; thus <math>H^k/H^{k+1}</math> is commutative.
-To show that (2) implies (1), we may take <math>H= \{e\}</math>.
+To show that (3) implies (1), we may take <math>H= \{e\}</math>.
-To show that (1) implies (3), we may take <math>A = C^n(G)</math>.
+To show that (1) implies (4), we may take <math>A = C^n(G)</math>.
-Finally, we show that (3) implies (1).  Let <math>\phi</math> be the canonical [[homomorphism]] of <math>G</math> onto <math>G/A</math>.  Then <math>\phi(C^k(G)) = C^k(G/A)</math>.  In particular, <math>\phi(C^n(G))= C^n(G/A)= \{e\}</math>.  Hence <math>C^n(G)</math> is a subset of <math>A</math>, so it lies in the center of <math>G</math>, and <math>C^{n+1}(G)=\{e\}</math>; thus the nilpotency class of <math>G</math> is at most <math>n</math>, as desired.  <math>\blacksquare</math>
+Finally, we show that (4) implies (1).  Let <math>\phi</math> be the canonical [[homomorphism]] of <math>G</math> onto <math>G/A</math>.  Then <math>\phi(C^k(G)) = C^k(G/A)</math>.  In particular, <math>\phi(C^n(G))= C^n(G/A)= \{e\}</math>.  Hence <math>C^n(G)</math> is a subset of <math>A</math>, so it lies in the center of <math>G</math>, and <math>C^{n+1}(G)=\{e\}</math>; thus the nilpotency class of <math>G</math> is at most <math>n</math>, as desired.  <math>\blacksquare</math>
-'''Corollary 1.''' Let <math>G</math> be a nilpotent group; let <math>H</math> be a subgroup of <math>G</math>.  If <math>H</math> is its own [[normalizer]], then <math>H=G</math>.
+'''Corollary 2.''' Let <math>G</math> be a nilpotent group; let <math>H</math> be a subgroup of <math>G</math>.  If <math>H</math> is its own [[normalizer]], then <math>H=G</math>.
 ''Proof.''  Suppose <math>H\neq G</math>; then there is a greatest integer <math>k\in [1,n+1]</math> for which <math>H^k \neq H</math>.  Then <math>H^k</math> normalizes <math>H</math>.  <math>\blacksquare</math>
-'''Corollary 2.'''  Let <math>G</math> be a nilpotent group; let <math>H</math> be a proper subgroup of <math>H</math>.  Then there exists a proper normal subgroup <math>A</math> of <math>G</math> such that <math>H \subseteq A</math> and <math>G/A</math> is abelian.
+'''Corollary 3.'''  Let <math>G</math> be a nilpotent group; let <math>H</math> be a proper subgroup of <math>H</math>.  Then there exists a proper normal subgroup <math>A</math> of <math>G</math> such that <math>H \subseteq A</math> and <math>G/A</math> is abelian.
 ''Proof.''  In the notation of the theorem, let <math>k</math> be the least integer such that <math>H^k \neq G</math>.  Then set <math>A=H^k</math>.  <math>\blacksquare</math>
-'''Corollary 3.''' Let <math>G</math> be a nilpotent group; let <math>H</math> be a subgroup of <math>G</math>.  If <math>G = H(G,G)</math>, then <math>G=H</math>.
+'''Corollary 4.''' Let <math>G</math> be a nilpotent group; let <math>H</math> be a subgroup of <math>G</math>.  If <math>G = H(G,G)</math>, then <math>G=H</math>.
 ''Proof.''  Suppose that <math>G \neq H</math>.  Then let <math>A</math> be the normal subgroup of <math>G</math> containing <math>H</math> as described in Corollary 3.  Then <math>(G,G) \subseteq A</math>, so
@@ Line 34: / Line 39: @@
 a contradiction.  <math>\blacksquare</math>
-'''Corollary 4.'''  Let <math>G'</math> be a group, let <math>G</math> be a nilpotent group, and let <math>f: G' \to G</math> be a group homomorphism for which the homomorphism <math>f' : G'/(G',G') \to G/(G,G)</math> derived from passing to quotients is [[surjective]].  Then <math>f</math> is surjective.
+'''Corollary 5.'''  Let <math>G'</math> be a group, let <math>G</math> be a nilpotent group, and let <math>f: G' \to G</math> be a group homomorphism for which the homomorphism <math>f' : G'/(G',G') \to G/(G,G)</math> derived from passing to quotients is [[surjective]].  Then <math>f</math> is surjective.
 ''Proof.''  Let <math>H</math> be the image of <math>f'</math> and apply Corollary 3.  <math>\blacksquare</math>
@@ Line 46: / Line 51: @@
 since <math>N</math> is a normal subgroup.  <math>\blacksquare</math>
-'''Corollary 5.'''  Let <math>G</math> be a nilpotent group; let <math>N</math> be a normal subgroup of <math>G</math>, and let <math>Z</math> be the [[center (algebra) |center]] of <math>G</math>.  If <math>N</math> is not trivial, then <math>N \cap Z</math> is not trivial.
+'''Corollary 6.'''  Let <math>G</math> be a nilpotent group; let <math>N</math> be a normal subgroup of <math>G</math>, and let <math>Z</math> be the [[center (algebra) |center]] of <math>G</math>.  If <math>N</math> is not trivial, then <math>N \cap Z</math> is not trivial.
 ''Proof.''  In the proposition's notation, let <math>k</math> be the greatest integer such that <math>N^k \neq \{e\}</math>.  The <math>(G,N^k) \subseteq N^{k+1} = \{e\}</math>, so <math>N^k</math> is a nontrivial subgroup that lies in the center of <math>G</math> and in <math>N</math>.  <math>\blacksquare</math>
-'''Corollary 6.'''  Let <math>G</math> be a nilpotent group, let <math>G'</math> be a group, and let <math>f</math> be a homomorphism of <math>G</math> into <math>G'</math>.  If the restriction of <math>f</math> to the center of <math>G</math> is [[injective]], then so is <math>f</math>.
+'''Corollary 7.'''  Let <math>G</math> be a nilpotent group, let <math>G'</math> be a group, and let <math>f</math> be a homomorphism of <math>G</math> into <math>G'</math>.  If the restriction of <math>f</math> to the center of <math>G</math> is [[injective]], then so is <math>f</math>.
 ''Proof.''  We proceed by contrapositive.  Suppose that <math>f</math> is not injective; then the [[kernel]] of <math>f</math> is nontrivial, so by the previous corollary, the intersection of <math>\text{Ker}(f)</math> and the center of <math>G</math> is nontrivial, so the restriction of <math>f</math> to the center of <math>G</math> is not injective.  <math>\blacksquare</math>
+== Finite Nilpotent Groups ==
+'''Theorem 8.'''  Let <math>G</math> be a [[finite]] group.  Then the following conditions are equivalent.
+# The group <math>G</math> is nipotent;
+# The group <math>G</math> is a product of [[p-group |<math>p</math>-groups]];
+# Every [[Sylow p-subgroup |Sylow <math>p</math>-subgroup]] of <math>G</math> is normal in <math>G</math>.
+''Proof.''  Since every <math>p</math>-group is nilpotent, condition (2) implies condition (1).
+Now we show that (1) implies (3).  Let <math>P</math> be a Sylow <math>p</math>-subgroup of <math>G</math>, and let <math>N</math> be its normalizer.  Then <math>N</math> is its own normalizer.  Then from Corollary 2, <math>N=G</math>, i.e., <math>P</math> is normal in <math>G</math>.
+Finally, we show that (3) implies (2).  Suppose condition (3) holds for <math>G</math>.  For any prime <math>p</math> dividing the [[order (group theory) |order]] of <math>G</math>, let <math>P_p</math> denote the Sylow <math>p</math>-subgroup of <math>H</math>.  Let <math>p</math> and <math>q</math> be distinct primes dividing the order of <math>H</math>.  Then <math>P_p \cap P_q  = \{e\}</math>, since the order of any element in both of these groups must divide a power of <math>p</math> and a power of <math>q</math>.  Since <math>P_p</math> and <math>P_q</math> are both normal, it follows that for any <math>a\in P_p</math>, <math>b\in P_q</math>, the [[commutator (group) |commutator]] <math>(a,b)</math> is an element both of <math>P_p</math> and <math>P_q</math>.  It follows that the canonical mapping of <math>f : \prod_p P_p \to G</math> is a homomorphism, and <math>P_p</math> is in its image, for every prime <math>p</math>.  Now, the order subgroup generated by the <math>P_p</math> must be divisible by every power of a prime that divides <math>G</math>, but it must also divide <math>G</math>; hence it is equal to <math>G</math>.  It follows that the order of the image of <math>\prod_p P_p</math> is equal to the order of <math>G</math>; since <math>G</math> is finite, this implies that <math>f</math> is surjective.  Since <math>G</math> and <math>\prod_p P_p</math> have the same size, <math>f</math> is also injective, and hence an [[isomorphism]].  <math>\blacksquare</math>
 == See also ==
+* [[Lower central series]]
 * [[Derived group]]
 * [[Solvable group]]
-* [[Lower central series]]
 * [[Derived series]]
+* [[Sylow Theorems]]
+* [[p-group |<math>p</math>-group]]
+* [[Sylow p-subgroup |Sylow <math>p</math>-subgroup]]
 [[Category:Group theory]]

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Difference between revisions of "Nilpotent group"

Latest revision as of 16:55, 5 June 2008

Characterization and Properties of Nilpotent Groups

Finite Nilpotent Groups

See also