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Difference between revisions of "Commutator (group)"

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== Relations with Commutators ==
 
== Relations with Commutators ==
  
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'''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>
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 +
'''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>.
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# 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>
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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 $a$ and $b$, denoted $(a,b)$ or $[a,b]$, is the element $a^{-1}b^{-1}ab$. If $a$ and $b$ commute, then $(a,b)=e$. More generally, $(a,b) = (ba)^{-1}ab$, or \[ab = ba(a,b) .\] It then follows that \[(a,b)(b,a) = e .\] We also have \[x^y = y^{-1}xy = x(x,y) = (y,x^{-1})x ,\] where $x^y$ denote the image of $x$ under the inner automorphism $\text{Int}(y^{-1})$, as usual.

Relations with Commutators

Proposition 1. For all $x,y,z$ in a group, the following relations hold:

  • $(x,yz) = (x,z)(x,y)^z = (x,z)(z,(y,x))(x,y)$;
  • $(xy,z) = (x,z)^y (y,z) = (x,z)((x,z),y)(y,z)$;
  • $(x^y,(y,z))(y^z,(z,x))(z^x,(x,y)) = e$;
  • $(x,yz)(z,xy)(y,zx) = e$;
  • $(xy,z)(yz,x)(zx,y) = e$.

Proof. For the first equation, we note that \[(x,yz) = x^{-1}z^{-1}y^{-1}xyz = (x^{-1}z^{-1}xz)z^{-1}(x^{-1}y^{-1}xy)z = (x,z) (x,y)^z .\] From the earlier relations, \[(x,y)^z = (z,(x,y)^{-1})(x,y) = (z,(y,x))(x,y) ,\] hence the relation. The second equation follows from the first by passing to inverses.

For the third equation, we define $f(a,b,c) = ca (b)^c$. We then note that \begin{align*} (x^y,(y,z)) &= (x^{-1})^y \cdot (z,y) \cdot x^y \cdot (y,z) \\ &= (x^{-1})^y \cdot z^{-1} (z)^y \cdot x^y \cdot (z^{-1})^y \cdot z &= (x^{-1})^y \cdot z^{-1} \cdot y^{-1}(zxz^{-1})y \cdot z \\ &= \bigl[(x^{-1})^y \cdot z^{-1}y^{-1}\bigr] zx \cdot y^z \\ &= f(z,x,y)^{-1} f(x,y,z) . \end{align*} By cyclic permutation of variables, we thus find \begin{align*} (x^y, (y,z))(y^z, (z,x))(z^x, (x,y)) &= f(z,x,y)^{-1} f(x,y,z) f(x,y,z)^{-1} f(y,z,x) f(y,z,x)^{-1} f(z,x,y) \\ &= e. \end{align*}

For the fourth equation, we have \[(x,yz)(z,xy)(y,zx) = (yzx)^{-1}(xyz)(xyz)^{-1} (zxy) (zxy)^{-1}(yzx) = e .\] The fifth follows similarly. $\blacksquare$

Commutators and Subgroups

If $A$ and $B$ are subgroups of a group $G$, $(A,B)$ denotes the subgroup generated by the set of commutators of the form $(a,b)$, for $a\in A$ and $b\in B$.

The group $(A,B)$ is trivial if and only if $A$ centralizes $B$. Also, $(A,B) \subseteq A$ if and only if $B$ normalizes $A$. If $A$ and $B$ are both normal (or characteristic), then so is $(A,B)$, for if $f$ is an (inner) automorphism, then \[f((a,b)) = (f(a),f(b)).\]

Lemma. Let $A,B$ be a closed subsets of $G$ (not necessarily a subgroups); denote by $(A,B)$ the subgroup of $G$ generated by elements of the form $(a,b)$, for $a\in A$, $b\in B$. Then $(A,B)^a \subseteq$.

Proof. Let $a,a'$ be elements of $A$ and $b$ be an element of $B$. Then \[(a,b)^{a'} = a'^{-1}a^{-1}b^{-1}aba' = (aa',b)b^{-1}a'^{-1}aba'\]

Proposition 2. Let $A,B,C$ be three subgroups of $G$.

  1. The group $A$ normalizes the group $(A,B)$.
  2. If the group $(B,C)$ normalizes $A$, then the set of elements $(a,(b,c))$, for $a\in A$, $b\in B$, $c \in C$, generates the group $(A,(B,C))$.

Proof. For the first, we note that for any $a,a' \in A$ and any $b \in B$, \[(a,b)^{a'} = (aa',b) (a',b)^{-1},\] by Proposition 1.

For the second, we have for any $x\in G$, $a\in A$, $b\in B$, $c \in C$, \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*} Since $(B,C)$ normalizes $A$, the element $((b,c),a)$ lies in $A$. It then follows from induction on $n$ that for all $b_i \in B$, $c_i \in C$, the element \[\biggl( a, \prod_{i=1}^n (b_i,c_i) \biggr)\] lies in the subgroup generated by elements of the form $(a,(b,c))$. Similarly, \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*} lies in the subgroup generated by elements of the form $(a,(b,c))$; it then follows that $(a,(c,b))$ does. Then using the observation \[(a,(c,b)x) = (a,x)(x, ((c,b),a)) (a,(c,b)),\] we prove by induction on $n$ that the element \[\biggl(a, \prod_{i=1}^n (b_i,c_i)^{\pm 1} \biggr)\] lies in the subgroup generated by elements of the form $(a,(b,c))$. This proves the second result. $\blacksquare$

See also

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