Difference between revisions of "Commutator (group)"
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== Relations with Commutators == | == Relations with Commutators == | ||
− | '''Proposition.''' For all <math>x,y,z</math> in a group, the following relations hold: | + | '''Proposition 1.''' For all <math>x,y,z</math> in a group, the following relations hold: |
* <math>(x,yz) = (x,z)(x,y)^z = (x,z)(z,(y,x))(x,y)</math>; | * <math>(x,yz) = (x,z)(x,y)^z = (x,z)(z,(y,x))(x,y)</math>; | ||
* <math>(xy,z) = (x,z)^y (y,z) = (x,z)((x,z),y)(y,z)</math>; | * <math>(xy,z) = (x,z)^y (y,z) = (x,z)((x,z),y)(y,z)</math>; | ||
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The group <math>(A,B)</math> is [[trivial group | trivial]] if and only if <math>A</math> [[centralizer | centralizes]] <math>B</math>. Also, <math>(A,B) \subseteq A</math> if and only if <math>B</math> [[normalizer | normalizes]] <math>A</math>. If <math>A</math> and <math>B</math> are both [[normal subgroup | normal]] (or [[characteristic subgroup | characteristic]]), then so is <math>(A,B)</math>, for if <math>f</math> is an ([[inner automorphism |inner]]) [[automorphism]], then | The group <math>(A,B)</math> is [[trivial group | trivial]] if and only if <math>A</math> [[centralizer | centralizes]] <math>B</math>. Also, <math>(A,B) \subseteq A</math> if and only if <math>B</math> [[normalizer | normalizes]] <math>A</math>. If <math>A</math> and <math>B</math> are both [[normal subgroup | normal]] (or [[characteristic subgroup | characteristic]]), then so is <math>(A,B)</math>, for if <math>f</math> is an ([[inner automorphism |inner]]) [[automorphism]], then | ||
<cmath> f((a,b)) = (f(a),f(b)). </cmath> | <cmath> f((a,b)) = (f(a),f(b)). </cmath> | ||
+ | |||
+ | '''Lemma.''' Let <math>A,B</math> be a closed subsets of <math>G</math> (not necessarily a subgroups); denote by <math>(A,B)</math> the subgroup of <math>G</math> generated by elements of the form <math>(a,b)</math>, for <math>a\in A</math>, <math>b\in B</math>. Then <math>(A,B)^a \subseteq </math>. | ||
+ | |||
+ | ''Proof.'' Let <math>a,a'</math> be elements of <math>A</math> and <math>b</math> be an element of <math>B</math>. Then | ||
+ | <cmath> (a,b)^{a'} = a'^{-1}a^{-1}b^{-1}aba' = (aa',b)b^{-1}a'^{-1}aba' </cmath> | ||
+ | |||
+ | '''Proposition 2.''' Let <math>A,B,C</math> be three subgroups of <math>G</math>. | ||
+ | # The group <math>A</math> normalizes the group <math>(A,B)</math>. | ||
+ | # If the group <math>(B,C)</math> normalizes <math>A</math>, then the set of elements <math>(a,(b,c))</math>, for <math>a\in A</math>, <math>b\in B</math>, <math>c \in C</math>, generates the group <math>(A,(B,C))</math>. | ||
+ | |||
+ | ''Proof.'' For the first, we note that for any <math>a,a' \in A</math> and any <math>b \in B</math>, | ||
+ | <cmath> (a,b)^{a'} = (aa',b) (a',b)^{-1}, </cmath> | ||
+ | by Proposition 1. | ||
+ | |||
+ | For the second, we have for any <math>x\in G</math>, <math>a\in A</math>, <math>b\in B</math>, <math>c \in C</math>, | ||
+ | <cmath> \begin{align*} | ||
+ | (a,(b,c)x) &= a^{-1}x^{-1}(c,b)a(b,c)x = (a,x)x^{-1}a^{-1}(c,b)a(b,c)x \ | ||
+ | &= (a,x) (x,((b,c),a)) (a,(b,c)) . | ||
+ | \end{align*} </cmath> | ||
+ | Since <math>(B,C)</math> normalizes <math>A</math>, the element <math>((b,c),a)</math> lies in <math>A</math>. It then follows from induction on <math>n</math> that for all <math>b_i \in B</math>, <math>c_i \in C</math>, the element | ||
+ | <cmath> \biggl( a, \prod_{i=1}^n (b_i,c_i) \biggr) </cmath> | ||
+ | lies in the subgroup generated by elements of the form <math>(a,(b,c))</math>. Similarly, | ||
+ | <cmath> \begin{align*} | ||
+ | (a,(b,c)) &= (a, (c,b) \cdot (b,c)^2) \ | ||
+ | &= (a,(b,c)^2)\Bigl( (b,c)^2,\bigl((c,b),a \bigr) \Bigr) \cdot (a,(c,b)) , | ||
+ | \end{align*} </cmath> | ||
+ | lies in the subgroup generated by elements of the form <math>(a,(b,c))</math>; it then follows that <math>(a,(c,b))</math> does. Then using the observation | ||
+ | <cmath> (a,(c,b)x) = (a,x)(x, ((c,b),a)) (a,(c,b)), </cmath> | ||
+ | we prove by induction on <math>n</math> that the element | ||
+ | <cmath> \biggl(a, \prod_{i=1}^n (b_i,c_i)^{\pm 1} \biggr) </cmath> | ||
+ | lies in the subgroup generated by elements of the form <math>(a,(b,c))</math>. This proves the second result. <math>\blacksquare</math> | ||
== See also == | == See also == |
Revision as of 16:31, 28 May 2008
In a group, the commutator of two elements and
, denoted
or
, is the element
. If
and
commute, then
. More generally,
, or
It then follows that
We also have
where
denote the image of
under the inner automorphism
, as usual.
Relations with Commutators
Proposition 1. For all in a group, the following relations hold:
;
;
;
;
.
Proof. For the first equation, we note that
From the earlier relations,
hence the relation. The second equation follows from the first by passing to inverses.
For the third equation, we define . We then note that
By cyclic permutation of variables, we thus find
For the fourth equation, we have
The fifth follows similarly.
Commutators and Subgroups
If and
are subgroups of a group
,
denotes the subgroup generated by the set of commutators of the form
, for
and
.
The group is trivial if and only if
centralizes
. Also,
if and only if
normalizes
. If
and
are both normal (or characteristic), then so is
, for if
is an (inner) automorphism, then
Lemma. Let be a closed subsets of
(not necessarily a subgroups); denote by
the subgroup of
generated by elements of the form
, for
,
. Then
.
Proof. Let be elements of
and
be an element of
. Then
Proposition 2. Let be three subgroups of
.
- The group
normalizes the group
.
- If the group
normalizes
, then the set of elements
, for
,
,
, generates the group
.
Proof. For the first, we note that for any and any
,
by Proposition 1.
For the second, we have for any ,
,
,
,
Since
normalizes
, the element
lies in
. It then follows from induction on
that for all
,
, the element
lies in the subgroup generated by elements of the form
. Similarly,
lies in the subgroup generated by elements of the form
; it then follows that
does. Then using the observation
we prove by induction on
that the element
lies in the subgroup generated by elements of the form
. This proves the second result.