If a b then the commutator of and is the element aba Of course if and commute then aba Now dene to be the set each is a commutator in In other words is the collection of all nite products of commutators in Then we have Proposition 1 If is any g ID: 69291
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Nownotethat(gaig�1)�1=(g�1)�1a�1ig�1=ga�1ig�1,andwehavetheanalogousstatementifwereplaceaibybi.So,wehavegxig�1=(gaig�1)(gbig�1)(gaig�1)�1(gbig�1)�1;whichisacommutator.Now,from(1),wehavegcg�1isaproductofcom-mutators,andsogcg�12C.ThusCCG. ThesubgroupCofGiscalledthecommutatorsubgroupofG,anditgeneral,itisalsodenotedbyC=G0orC=[G;G],andisalsocalledthederivedsubgroupofG.IfGisAbelian,thenwehaveC=feg,soinonesensethecommutatorsubgroupmaybeusedasonemeasureofhowfaragroupisfrombeingAbelian.Specically,wehavethefollowingresult.Theorem1.LetGbeagroup,andletCbeitscommutatorsubgroup.Sup-posethatNCG.ThenG=NisAbelianifandonlyifCN.Inparticular,G=CisAbelian.Proof.FirstassumethatG=NisAbelian.Leta;b2G.Sinceweareas-sumingthatG=NisAbelian,thenwehave(aN)(bN)=(bN)(aN),andsoabN=baNbythedenitionofcosetmultiplicationinthefactorgroup.Now,weknowabN=baNimpliesab(ba)�12N,whereab(ba)�1=aba�1b�1,andsoaba�1b�12N.Sinceaandbwerearbitrary,anycommutatorinGisanelementofN,andsinceNisasubgroupofG,thenanyniteproductofcommutatorsinGisanelementofN.ThusCN.NowsupposethatCN,andleta;b2G.Thenaba�1b�12N,andsoab(ba)�12N.ThisimpliesabN=baN,orthat(aN)(bN)=(bN)(aN).Sinceaandbwerearbitrary,thisholdsforanyelementsaN;bN2G=N,andthusG=NisAbelian. 2