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