SL n (F ) Equals its Own Derived Group

Transcription

1 Internatonal Journal of Algebra, Vol. 2, 2008, no. 12, SL n (F ) Equals ts Own Derved Group Jorge Macel BMCC-The Cty Unversty of New York, CUNY 199 Chambers street, New York, NY 10007, USA macel@cms.nyu.edu Abstract. We prove that SL n (F q ) s equal to ts own commutator group except when n = 2 and q =2orq = 3, by usng the fact that every element n the center Z of SL n (F q ) can be wrtten n a commutator form [A, B], where A, B SL n (F q ). Mathematcs Subect Classfcaton: Prmary 20H20, 15A04 Keywords: Central extenson, Commutator group, General group, Proectve specal lnear group, Specal lnear group Defnton 1.1. A group extenson Introducton F E G of a group G by a group F conssts of a group E, an nectve homomorphsm : F E, and a surectve homomorphsm : E G, such that Im = Ker. Group extensons need only be constructed up to somorphsm. In detal, an equvalence of group extensons F E G F E G of G by F s an somorphsm θ : E E such that the dagram F E G θ F E G commutes, that s, θ = and θ =.

2 586 J. Macel Defnton 1.2. An extenson F E G s called central f the mage (F ) s contaned n the center of E. Ths s possble only f F s Abelan. Defnton 1.3. An element ζ of a feld F s sad to be a root of unty f there exsts an nteger n>0 such that ζ n =1; for every nteger n>0 such that ζ n =1, ζ s called an n-th root of unty. It amounts to say that the roots of unty are the elements of fnte order of the multplcatve group F of non-zero elements of F. The roots of unty form a subgroup μ (F )off, the n-th roots form a subgroup μ n (F )ofμ (F ). We have μ (F )= n>1 μ n(f ) and μ n (F ) μ m (F )fm dvdes n. For every root of unty ζ there exsts a least nteger n 1 such that ζ belongs to μ n (F ), namely the order of ζ n the group F. Defnton 1.4. An n-th root of unty s sad to be prmtve f t s of order n. If there exsts a prmtve n-th root of unty ζ n F, the group μ n (F )sof order n and s generated by ζ. Let F be a feld and n be a postve nteger. We denote by M n (F ) the rng of square matrces of order n over F.By an n n determnant we shall mean a mappng det : M n (F ) F whch, when vewed as a functon of the column vectors A 1,...,A n of a matrx A, s multlnear alternatng, and such that det(i) = 1. It s shown that f determnants exst, they are unque. If A 1,...,A n are the column vectors of dmenson n, of the matrx A =(a ), then det(a 1,...,A n )= σ ɛ(σ)a σ(1),1 a σ(n),n where the sum s taken over all permutatons σ of {1,...,n}, and ɛ(σ) s the sgn of the permutaton. The general lnear group GL n (F ) of nvertble elements of M n (F ) s ust the nverse mage under the mappng det : M n (F ) F of the multplcatve group F of nvertble elements of F. The mappng det : M n (F ) F s moreover surectve and therefore so s the homomorphsm det :GL n (F ) F, snce for all λ F, det(dag(λ, 1,...,1)) = λ.

3 Specal lnear group equals ts derved group 587 The kernel of the surectve homomorphsm det :GL n (F ) F s a normal subgroup of GL n (F ); t s denoted by SL n (F ) and s called the specal lnear group of square matrces of order n over F. We note that f a matrx M commutes wth every element of SL n (F ), then t must be a scalar matrx. Indeed, ust the commutaton wth elementary matrces E (1) = I n +1 show that M commutes wth all matrces 1 (havng 1 n the -component, 0 otherwse), so M commutes wth all matrces, and s a scalar matrx. Takng the determnant shows that the center conssts of μ n (F )I n. Defnton 1.5. Let Z be the center of SL n (F ), that s the group of scalar matrces such that the scalar s an n-th root of unty. We defne the proectve specal lnear group of square matrces of order n over F by the quotent group PSL n (F )=SL n (F )/Z. 2. The group SL n (F ) as a central extenson Let F be a feld and let μ be a prmtve p n -th root of unty n F. Note that SL n (F ) s a central group extenson of PSL n (F )byμ n (F ). Indeed, we have the central extenson μ n (F ) SL n (F ) PSL n (F ) where s the nectve homomorphsm defned by (μ) =μi n, s the surectve homomorphsm defned by (A) =Ā, and Im = Ker. Theorem 2.1. Let n be dvsble by p m where p s prme. Then the scalar matrx μi n can be wrtten n a commutator form [A, B] =ABA 1 B 1, where A, B SL n (F ). Proof. We have the followng possbltes: 1. Let p be an odd prme and A, B be two square matrces of order p m over the feld F such that A =(a, )=(δ, μ 1 ), and B =(b, ) wth b +1, = b 1,p m =1,andb, = 0 otherwse. Then A and B belong to SL p m(f ), and [A, B]=ABA 1 B 1 = μi p m. 2. If p = 2 and n = 1, then the matrces A = ( ) a b b a, where a 2 + b 2 = ( 1, and B =( 1 0 0) 1 satsfy the requred condton. Indeed, [A, B] = a b ) b a ( ) ( ) a b b a ( )=( ) For a fnte feld Fq, wth q odd, such a and b always exst. If F = F q and q =4k + 1 then 1 s a square and we can take a =, where 2 = 1, and b =0.Ifq =4k+3 then the set of all elements of the form x 2 +y 2 concdes wth F q. Indeed, t contans all

4 588 J. Macel quadratc resdues x 2, and t s nvarant under the multplcaton by an arbtrary element z 2, where z F q. Therefore, f t contans at least one non-quadratc element, then t concdes wth F q. If not, then quadratc resdues form an addtve subgroup n F q. However, F q does not have an addtve subgroup of ndex 2 for odd q. Therefore, F q s the set of elements of the form x 2 + y 2. In partcular, there are a and b such that a 2 + b 2 = 1, and they provde entres for A. 3. If p = 2 and m>1, we select A = σ 1 X and B = Yσ 2 where X and Y are dagonal matrces and σ 1,σ 2 are commutng permutaton matrces. In ths case [A, B]=ABA 1 B 1 = σ 1 XYσ 2 X 1 σ1 1 σ2 1 Y 1 = σ 1 XYσ 2 X 1 σ2 1 σ1 1 Y 1 snce σ 1 and σ 2 commute. Therefore the equatons (2.1) and ABA 1 B 1 = μi (2.2) XY(σ 2 X 1 σ2 1 )(σ 1 1 Y 1 σ 1 )=μi are equvalent, we note also that the matrces X, Y,(σ 2 X 1 σ2 1 ), and (σ1 1 Y 1 σ 1 ) are dagonal. Suppose now that σ 1 has order 2 k and σ 2 has order 2 m k, where k 1 and m k 1. Then the correspondng lnear space has a specal coordnate system z,, where 1 2 k and 1 2 m k wth the property that σ 1 cyclcally permutes coordnates z, wth the same ndex and σ 2 cyclcally permutes coordnates z, wth the same ndex. Therefore, f we denote by x, and y, the dagonal elements of the matrces X and Y respectvely, then equaton (2.2) becomes equvalent to a seres of equatons (2.3) x, y, x 1 +1(mod 2 k ), y 1 = μ,+1(mod 2 m k ) for the dagonal elements. If we denote by and u, =: x, x 1 +1(mod 2 k ), v, =: y, y 1,+1(mod 2 m k ),

5 Specal lnear group equals ts derved group 589 then u, = Equaton (2.3) above becomes v, =1. u, = μv 1, and hence we have obtaned equatons only for the parameters u,. Thus our ntal matrx equaton (2.1) has been reduced to equatons for u, : u, =1, u, = μ 2m k. These parameters u, defne the complementary set of parameters v,. Notce that for any u, and v, satsfyng u, = v, = 1, we can fnd x, and y, so that u, = x, x 1 +1(mod 2 k ),, v, = y, y 1,+1(mod 2 m k ), and hence we can obtan solutons of the equaton (2.1). Thus there are many matrx pars A, B that satsfy equaton (2.1). 4. If n s dvsble by p m, then matrces A and B, consstng of n/p m dagonal blocks of matrces A and B respectvely, also satsfy the relaton [A, B] =ABA 1 B 1 = μi n. The Theorem follows. Ths fact appeared prevously n [BM]. Example 2.2. Let F q be a feld and let μ be a prmtve p n -th root of unty n F q, where n>1 and 4 q 1. We decompose 2 n coordnates nto two groups of order 2 n 1 and take the dagonal matrx X = dag([1, 1,...,], [1, 1,..., ]), where brackets show the boundares of each block. The element, wth 2 = 1, s contaned n F q snce, by our assumpton, 4 dvdes q 1. Smlarly, take the dagonal matrx Y = dag([1,μ,...,μ 2n 1 1 ], [μ, μ 2,...,μ 2n 1 ]). Let σ 1 be the permutaton whch permutes varables cyclcally wthn each block, and σ 2 be the permutaton of order 2 whch nterchanges these two blocks of varables. Recall that μ 2n 1 = 1. Then X(σ 2 X 1 σ 1 2 )=dag([ 1, 1,...,1], [ 1, 1,...,1])

6 590 J. Macel and and hence Y (σ1 1 Y 1 σ 1 )=dag([ μ,μ,...,μ], [ μ,μ,...,μ]) XY (σ 2 X 1 σ 1 2 )(σ 1 1 Y 1 σ 1 )=μi. If we take A = σ 1 X, B = Yσ 2 SL 2 n(f ), then [A, B] =ABA 1 B 1 = μi 2 n. Example 2.3. Let F q be a feld and let μ be a p n -th root of unty n F q of order m. Ifp = 2 and n = 2, consder the permutaton matrces σ 1 = σ 2 = of the group SL 4 (F q ). Snce σ 1 and σ 2 commute, t s seen that, for every par of dagonal matrces x y X = 0 x x 3 0 and Y = 0 y y x y 4 n SL 4 (F q ), the commutator [Xσ 1,Yσ 2 ] s a dagonal matrx n SL 4 (F q ). Takng account of ths, t s mmedately seen that the scalar matrx μi 4 can be expressed n the form of commutator [Xσ 1,Yσ 2 ] for some dagonal matrces X, Y SL 4 (F q ). For, the defnton of product of matrces gves x 1 y 2 x y [Xσ 1,Yσ 2 ]= 0 x 2 y 1 x y x 3 y 4 x y x 4 y 3 x y 4 Then necessarly

7 Specal lnear group equals ts derved group 591 (2.4) x 1 y 2 x 1 3 y 1 1 = μ (2.5) x 2 y 1 x 1 4 y 1 2 = μ (2.6) x 3 y 4 x 1 1 y 1 3 = μ (2.7) x 4 y 3 x 1 2 y 1 4 = μ Multplyng condtons (2.4) and (2.5) gves x 1 x 2 (x 3 x 4 ) 1 = μ 2. Smlarly, multplyng (2.6) and (2.7) gves x 3 x 4 (x 1 x 2 ) 1 = μ 2. Note that the addtonal hypothess mples that det X = x 1 x 2 x 3 x 4 =1 (x 1 x 2 ) 2 = μ 2 and (x 3 x 4 ) 2 = μ 2. Then the above three relatons show that when μ 2 = 1 and x 1 x 2 = ±μ and x 3 x 4 = ±μ x 1 x 2 = ±μ and x 3 x 4 = μ when μ 2 = 1. On the other hand, multplyng condtons (2.4) and (2.6) gves y 2 y 4 (y 1 y 3 ) 1 = μ 2. Smlarly, multplyng condtons (2.5) and (2.7) gves Note that the hypothess y 1 y 3 (y 2 y 4 ) 1 = μ 2. det Y = y 1 y 2 y 3 y 4 =1 on det Y therefore mples that

8 592 J. Macel Then (y 2 y 4 ) 2 = μ 2 and (y 1 y 3 ) 2 = μ 2. when μ 2 = 1 and y 2 y 4 = ±μ and y 1 y 3 = ±μ y 2 y 4 = ±μ and y 1 y 3 = μ when μ 2 = 1. In partcular we derve from these results that the matrces satsfy the relaton X = 0 μ and Y = 0 μ μ μ when m =2. Smlarly, the matrces satsfy the relaton [Xσ 1,Yσ 2 ]=μi X = 0 μ and Y = 0 μ μ μ μ [Xσ 1,Yσ 2 ]= 0 μ 2 μ μ 2 μ 0 = μi μ when m =4. In conformty wth the above results, we shall dstngush two cases: (a) If m = 2, let A and B be two square matrces of order 4 over the feld F q whch can be wrtten n the form

9 Specal lnear group equals ts derved group A = μ and B = μ μ μ wth respect to the same bass. It follows from the method of calculatng a determnant that det A = 1 and det B = 1. The defnton of product of matrces gves μ ABA 1 B 1 = 0 μ μ μ (b) If m = 4, let A and B be two square matrces of order 4 over the feld F q whch can be wrtten n the form A = μ and B = μ μ μ wth respect to the same bass. The method of calculatng a determnant and the hypothess on μ mply that det A = 1 and det B = 1. Now, t s mmedate that A and B satsfy the desred condton μ ABA 1 B 1 = 0 μ μ μ Theorem 2.4. Let p be a prme number and F be a fnte feld of order q = p k. If n 3 (n 2 f q 5), then SL n (F ) s equal to ts own commutator group. Proof. The commutator subgroup SL c n(f )=[SL n (F ),SL n (F )] s the smallest normal subgroup N of SL n (F ) such that SL n (F )/N s abelan. Theorem 2.1 readly mples that the center Z of SL n (F ) s a normal subgroup of SL c n(f ). By the Frst Isomorphsm Theorem, and SL c n (F )/Z PSL n(f ) PSL n (F )/(SL c n(f )/Z)) = SL n (F )/SL c n(f ).

10 594 J. Macel Snce PSL n (F ) s a fnte smple group for every fnte feld F of order q and n 3(n 2fq 5), t follows that SL c n (F ) s ether Z or SL n(f ). On the other hand, a non-cyclc smple group s not solvable. Then SL c n(f )=SL n (F ). References [1] Bourbak N. Elements of Mathematcs, Algebra I. Addson-Wesley publshng company (1973). [2] Bogomolov F. A., Macel J., Thomr P. Unramfed Brauer Groups of Fnte Smple Groups of Le Type A l. Amercan Journal of Mathematcs 126 (2004), [3] Grllet P. A., Algebra. John Wley & sons, Inc. (1999). Receved: December 14, 2007