Commutator group

In mathematics , the designated commutator (or commutator subgroup ) to a group that subset , of the commutators in the group is generated: ${\displaystyle G}$ ${\displaystyle [a,b]:=aba^{-1}b^{-1}}$ ${\displaystyle G}$

${\displaystyle K(G):=\left\langle \,\left\{[a,b]\mid \ a,b\in G\right\}\,\right\rangle .}$

The commutator group is also referred to with and with (or ) and called the derived group (of ). ${\displaystyle [G,G]}$ ${\displaystyle G'}$ ${\displaystyle G^{(1)}}$ ${\displaystyle G}$

In general, the set of all commutators is not a group, so the phrase "generated by" in the definition (equivalent to the angle brackets in the formula) cannot be left out. ${\displaystyle \left\{[a,b]\mid \ a,b\in G\right\}}$ ${\displaystyle \left\langle \,\right\rangle }$

The order of the commutator group is a measure of how far a group is from commutativity . A group is commutative ( Abelian ) if and only if its commutator group consists only of the neutral element called . In this case applies to everyone . In contrast, groups in which the commutator group includes the whole group are called perfect groups . ${\displaystyle e}$ ${\displaystyle [a,b]=e}$ ${\displaystyle a,b\in G}$

properties

The equations apply

• ${\displaystyle ab=[a,b]ba}$ .
• ${\displaystyle [a,b]^{-1}=[b,a]}$ .
• ${\displaystyle [a,b]^{s}=[a^{s},b^{s}]}$with as the conjugate of under . ${\displaystyle a^{s}:=s^{-1}as}$ ${\displaystyle a}$ ${\displaystyle s}$

For every homomorphism is . ${\displaystyle f\colon G\to H}$ ${\displaystyle f([a,b])=[f(a),f(b)]}$

Since the amount of the commutators under each automorphism of is mapped to itself, the commutator is a characteristic subgroup of , and thus a normal subgroup of the group. ${\displaystyle G}$ ${\displaystyle G}$

The factor group is always abelian, it is called the abelization of the group. The following applies to every normal divider : ${\displaystyle G/K(G)}$ ${\displaystyle N}$

${\displaystyle G/N}$ is Abelian if and only if ${\displaystyle K(G)\subseteq N.}$

That is, the commutator group is the smallest normal divisor for which the factor group is Abelian.

Examples

Let it be the symmetrical group and the alternating group . Then: ${\displaystyle S_{n}}$ ${\displaystyle A_{n}}$

• ${\displaystyle K(S_{n})=A_{n}}$ For ${\displaystyle n\geq 2.}$
• ${\displaystyle K(A_{n})=A_{n}}$ For ${\displaystyle n\geq 5.}$
• ${\displaystyle K(A_{4})=V}$, where the Klein group of four denotes. ${\displaystyle V}$
• ${\displaystyle K(A_{3})=K(A_{2})=\{e\}.}$

Higher commutator groups

The formation of the commutator group can be iterated; the -th commutator group (or also the -th derived group) is called . The recursive definition is: ${\displaystyle n}$ ${\displaystyle n}$ ${\displaystyle G^{(n)}=K_{n}(G)}$

1. ${\displaystyle G^{(0)}:=K_{0}(G):=G.}$
2. ${\displaystyle G^{(n+1)}:=(G^{(n)})'=K_{n+1}(G):=K\left(K_{n}(G)\right).}$

A group is called solvable if and only if there is a descending chain of subnormal divisors ${\displaystyle G}$

${\displaystyle G=G_{0}\triangleright G_{1}\triangleright \ldots \triangleright G_{n}=\{e\}}$( Subnormal series )

exists such that the factor groups are Abelian. The construction of the iterated commutator group provides a criterion for the solvability of : ${\displaystyle G_{k}/G_{k+1}}$ ${\displaystyle G}$

${\displaystyle G}$is resolvable if and only if there is a with ${\displaystyle n\in \mathbb {N} }$ ${\displaystyle G^{(n)}=\{e\}.}$

Either the descending series of subgroups resulting from the continued formation of the commutator or a refinement of this series is equivalent to each such subnormal series or a refinement of the same.

The relationship between the two equivalent definitions of solvability, via continued commutator formation on the one hand and via a subnormal series on the other, as well as the concept of the subnormal series itself are explained in more detail in the article " Series (group theory) ".

example

The symmetrical group or the alternating group can be resolved if and only if . For you see the immediately above example one. The following applies to: ${\displaystyle S_{n}}$ ${\displaystyle A_{n}}$ ${\displaystyle n<5}$ ${\displaystyle n\in \{2,3\}}$ ${\displaystyle n=4}$

${\displaystyle K(S_{4})=A_{4}}$, , As is Abelian. ${\displaystyle K(A_{4})=V}$ ${\displaystyle K(V)=\{e\}}$ ${\displaystyle V}$

For the chain of the iterated commutator groups becomes stationary at , so it is then neither resolvable nor . ${\displaystyle n\geq 5}$ ${\displaystyle A_{n}\neq \{e\}}$ ${\displaystyle S_{n}}$ ${\displaystyle A_{n}}$

Individual evidence

1. ↑ Dr. Ludwig Baumgartner Group Theory Collection Göschen Volume 837 / 837a p. 99
2. ^ Robert M. Guralnick: Commutators and commutator subgroups ADVANCES IN MATHEMATICS 45, 319-330 (1982)
3. ↑ There is no commutator in the above free group . Proof: Suppose there were with ${\displaystyle a,b}$ ${\displaystyle F:=\left\langle a,b\right\rangle }$ ${\displaystyle [a,b]^{2}}$
   ${\displaystyle x,y\in F}$
   

   ${\displaystyle [a,b]^{2}=aba^{-1}b^{-1}aba^{-1}b^{-1}=xyx^{-1}y^{-1}=[x,y],}$
   

   then would be the word
   

   ${\displaystyle w:=y^{-1}x^{-1}aba^{-1}b^{-1}aba^{-1}b^{-1}yx,}$
   

   through a clever choice of the variables in the empty word "transferable". Conversions can be carried out one after the other and also reversed (application of the abatement rules anyway, and turning back a substitution must result in a correct substitution), so that "transferable" is an equivalence relation . Now the choice can be converted into the word which, however, cannot be converted into the empty word. ${\displaystyle x,y\in F}$ ${\displaystyle e}$
   ${\displaystyle w}$ ${\displaystyle x:=a,y:=b}$ ${\displaystyle a^{-1}b^{-1}ab=[a^{-1},b^{-1}],}$

literature

• Thomas W. Hungerford: Algebra (= Graduate Texts in Mathematics. Vol. 73). 5th printing. Springer, New York NY et al. 1989, ISBN 0-387-90518-9 .

See also

• Whitehead's Lemma
• Perfect group