Difference between revisions of "Nilpotent group"
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'''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: | ||
# 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>. | ||
# 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>. | # 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>. | # 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.'' | + | ''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> | <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. | ||
− | To show that ( | + | To show that (3) implies (1), we may take <math>H= \{e\}</math>. |
− | To show that (1) implies ( | + | To show that (1) implies (4), we may take <math>A = C^n(G)</math>. |
− | Finally, we show that ( | + | 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 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>. |
Revision as of 22:42, 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 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. 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 1. 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 2. 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 3. 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 4. 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 5. 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 6. 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.