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Difference between revisions of "Nilpotent group"

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All [[abelian group]]s have nilpotency class at most 1; the [[trivial group]] is the only group of nilpotency class 0.
 
All [[abelian group]]s have nilpotency class at most 1; the [[trivial group]] is the only group of nilpotency class 0.
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== 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.''' Let <math>G</math> be a group, and let <math>n</math> be a positive integer.  Then the following three statements are equivalent:
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''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.''  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
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<cmath> 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) . </cmath>
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<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.
 
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.
  
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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 (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>
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'''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>.
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''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>
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'''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.
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''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>
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'''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>.
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''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
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<cmath> H(G,G) \subseteq HA = A \subsetneq G, </cmath>
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a contradiction.  <math>\blacksquare</math>
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'''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.
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''Proof.''  Let <math>H</math> be the image of <math>f'</math> and apply Corollary 3.  <math>\blacksquare</math>
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'''Proposition.'''  Let <math>G</math> be a group of nilpotency class at most <math>n</math>, and let <math>N</math> be a normal subgroup of <math>G</math>.  Then there exists a sequence <math>(N^k)_{1\le k \le n+1}</math> of subgroups of <math>G</math> such that <math>N^1=N</math>, <math>N^{n+1}=\{e\}</math>, <math>N^{k+1} \subseteq N^k</math>, and <math>(G,N^k) \subseteq N^{k+1}</math>, for all integers <math>1 \le k \le n+1</math>.
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''Proof.''  Let <math>N^k = N \cap C^k(G)</math>.  Then
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<cmath> (G,N^k) \subseteq (G,G^k) = G^{k+1}, </cmath>
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and
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<cmath> (G,N^k) \subseteq (G,N) \subseteq N, </cmath>
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since <math>N</math> is a normal subgroup.  <math>\blacksquare</math>
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'''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.
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''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>
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'''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>.
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''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>
  
 
== See also ==
 
== See also ==

Revision as of 17:08, 1 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. 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. 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 $1\le k \le n$.
  3. 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 \[ghg^{-1} = h \cdot h^{-1}ghg^{-1} = h\cdot (h,g^{-1}) \in h\cdot (G,G^k) = h \cdot C^{k+1}(G) .\] 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. $\blacksquare$

Corollary 1. 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 2. 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 3. 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 4. 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 5. 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 6. 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. $\blacksquare$

See also

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