Order of general- and special linear groups over finite fields.

Here GL is the general linear group, the group of invertible n×n matrices, and SL is the special linear group, the group of n×n matrices with determinant 1. MathematicsStackExchangeisaquestionandanswersiteforpeoplestudyingmathatanylevelandprofessionalsinrelatedfields.Itonlytakesaminutetosignup. Signuptojointhiscommunity Anybodycanaskaquestion Anybodycananswer Thebestanswersarevotedupandrisetothetop Home Public Questions Tags Users Unanswered Teams StackOverflowforTeams –Startcollaboratingandsharingorganizationalknowledge. CreateafreeTeam WhyTeams? Teams CreatefreeTeam Teams Q&Aforwork Connectandshareknowledgewithinasinglelocationthatisstructuredandeasytosearch. LearnmoreaboutTeams Orderofgeneral-andspeciallineargroupsoverfinitefields. AskQuestion Asked 11years,5monthsago Modified 1year,11monthsago Viewed 41ktimes 62 $\begingroup$ Let$\mathbb{F}_3$bethefieldwiththreeelements.Let$n\geq1$.Howmanyelementsdothefollowinggroupshave? $\text{GL}_n(\mathbb{F}_3)$ $\text{SL}_n(\mathbb{F}_3)$ HereGListhegenerallineargroup,thegroupofinvertiblen×nmatrices,andSListhespeciallineargroup,thegroupofn×nmatriceswithdeterminant1. linear-algebraabstract-algebramatricesfinite-groups Share Cite Follow editedDec15,2014at14:01 TobiasKildetoft 20.3k11goldbadge5959silverbadges8989bronzebadges askedApr21,2011at11:15 user9656user9656 $\endgroup$ 4 10 $\begingroup$ Let$q=3$,andtake,say,$n=4$.Thefirstrowofthematrixcanbeanythingbutthe$0$-vector,$q^4-1$possibilities.Foranyoneofthese,thesecondrowisanythingbutamultipleofthefirstrow,sothereare$q^4-q$possibilities.Foranyspecificchoiceoffirsttworows,thethirdrowisanythingbutlinearcombinationsofthefirsttworows.Thenumberoflinearcombinations$au+bv$oflinearlyindependent$u$,$v$isjustthenumberofchoicesforthepair$(a,b)$,namely$q^2$.Soforeverychoiceoffirsttworows,thereare$q^4-q^2$choicesofthirdrow.Continue. $\endgroup$ – AndréNicolas Apr21,2011at12:42 7 $\begingroup$ Continuing...user6312'sobservationandthemultiplicationprincipleofcountingwillgetyouonlytheanswerto1).Theeasiestwaytodo2)istouse1)andalittlegrouptheory,ifyouknowsome.Thedeterminantfunction$\det\colon\mathrm{GL}_n(\mathbb{F}_3)\rightarrow\mathbb{F}_3^{\times}$isagrouphomomorphismwhosekernelis$\mathrm{SL}_n(\mathbb{F}_3)$.Nowusethefactthatallcosetsofasubgroupofafinitegrouphavethesamecardinality. $\endgroup$ – BarrySmith Apr21,2011at13:06 $\begingroup$ @BarrySmithSothisthenfollowsfromthefirstisomorphismtheorem? $\endgroup$ – AnthonyPeter Nov2,2016at22:08 $\begingroup$ Ifby"this",youmeantheanswertotask2,thenyes. $\endgroup$ – BarrySmith Nov3,2016at0:36 Addacomment  |  2Answers 2 Sortedby: Resettodefault Highestscore(default) Datemodified(newestfirst) Datecreated(oldestfirst) 117 $\begingroup$ Firstquestion:Wesolvetheproblemfor"the"finitefield$F_q$with$q$elements.Thefirstrow$u_1$ofthematrixcanbeanythingbutthe$0$-vector,sothereare$q^n-1$possibilitiesforthefirstrow. Foranyoneofthesepossibilities,thesecondrow$u_2$canbeanythingbutamultipleofthefirstrow,giving$q^n-q$possibilities. Foranychoice$u_1,u_2$ofthefirsttworows,thethirdrowcanbeanythingbutalinearcombinationof$u_1$and$u_2$.Thenumberoflinearcombinations$a_1u_1+a_2u_2$isjustthenumberofchoicesforthepair$(a_1,a_2)$,andthereare$q^2$ofthese.Itfollowsthatforevery$u_1$and$u_2$,thereare$q^n-q^2$possibilitiesforthethirdrow. Foranyallowedchoice$u_1$,$u_2$,$u_3$,thefourthrowcanbeanythingexceptalinearcombination$a_1u_1+a_2u_2+a_3u_3$ofthefirstthreerows.Thusforeveryallowed$u_1,u_2,u_3$thereare$q^3$forbiddenfourthrows,andtherefore$q^n-q^3$allowedfourthrows. Continue.Thenumberofnon-singularmatricesis $$(q^n-1)(q^n-q)(q^n-q^2)\cdots(q^n-q^{n-1}).$$ Secondquestion:Wefirstdealwiththecase$q=3$ofthequestion.Ifwemultiplythefirstrowby$2$,anymatrixwithdeterminant$1$ismappedtoamatrixwithdeterminant$2$,andanymatrixwithdeterminant$2$ismappedtoamatrixwithdeterminant$1$. Thuswehaveproducedabijectionbetweenmatriceswithdeterminant$1$andmatriceswithdeterminant$2$.Itfollowsthat$SL_n(F_3)$hashalfasmanyelementsas$GL_n(F_3)$. Thesameideaworksforanyfinitefield$F_q$with$q$elements.Multiplyingthefirstrowofamatrixwithdeterminant$1$bythenon-zerofieldelement$a$producesamatrixwithdeterminant$a$,andallmatriceswithdeterminant$a$canbeproducedinthisway.Itfollowsthat $$|SL_n(F_q)|=\frac{1}{q-1}|GL_n(F_q)|.$$ Share Cite Follow editedOct9,2020at12:58 Thissitehasbecomeadump. 58.8k77goldbadges6969silverbadges154154bronzebadges answeredOct19,2011at9:16 AndréNicolasAndréNicolas 494k4444goldbadges522522silverbadges956956bronzebadges $\endgroup$ 5 $\begingroup$ Excellentlyexplained.Thiswasofgreathelp. $\endgroup$ – Cauchy Jul1,2017at23:39 1 $\begingroup$ @andrewelldone.SoitmeansSpeciallineargroupsarrnotnormalsubgroupsofGenerallineargroup,right? $\endgroup$ – PrinceKhan Oct28,2017at14:26 1 $\begingroup$ Greatandsimplesolution!+1forthat:)Thanksalotforthat:) $\endgroup$ – ZFR Apr17,2018at15:10 4 $\begingroup$ @PrinceThomasThespeciallineargroup\emph{is}anormalsubgroupofthegenerallineargroup.Thespeciallineargroupisthekernelofthedeterminant,whichisahomomorphismfromthegenerallineargrouptotheunderlyingfieldofunits. $\endgroup$ – frito_mosquito Aug5,2018at14:07 $\begingroup$ Howdoweknowthateachpairofcoefficients$(a_1,a_2)$givesadistinctlinearcombination$a_1u_1+a_2u_2$? $\endgroup$ – Anakhand May17at18:14 Addacomment  |  32 $\begingroup$ Determinantfunctionisasurjectivehomomorphismfrom$GL(n,F)$to$F^*$withkernel$SL(n,F)$.Hencebythefundamentalisomorphismtheorem$\frac{GL(n,F)}{SL(n,F)}$isisomorphicto$F^*$,themultiplicativegroupofnonzeroelementsof$F$. Thusif$F$isfinitewith$p$elementsthen$|GL(n,F)|=(p-1)|SL(n,F)|$. Share Cite Follow editedMar31,2020at16:20 cqfd 11.5k66goldbadges2020silverbadges4343bronzebadges answeredApr3,2018at5:30 RENJITHTRENJITHT 32133silverbadges44bronzebadges $\endgroup$ 1 $\begingroup$ Thisisrathershortbutrigorousproof.Thankyou. $\endgroup$ – HenryChoi Sep1,2020at2:05 Addacomment  |  YourAnswer ThanksforcontributingananswertoMathematicsStackExchange!Pleasebesuretoanswerthequestion.Providedetailsandshareyourresearch!Butavoid…Askingforhelp,clarification,orrespondingtootheranswers.Makingstatementsbasedonopinion;backthemupwithreferencesorpersonalexperience.UseMathJaxtoformatequations.MathJaxreference.Tolearnmore,seeourtipsonwritinggreatanswers. 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