1. PHN NI DUNG CHNG I. NHM V NHM CON A. L THUYT 1. Nhm 1.1.nh ngha Cho tp X khc rng, * l php ton hai ngi trong X. (X,*) c gi l nhm nu: i) Mi a,b,c X, ta c a*(b*c)= (a*b)*c ii) Tn ti phn t Xe sao cho Xx , ta c e*x = x*e = x iii) Mi phn t Xx lun tn ti Xx , sao cho exxxx == ** ,, Nu (X,*) c tnh giao hon th X c gi l nhm giao hon hay nhm Abel. 1.2. nh l ( v iu kin tng ng vi nhm) Cho X l tp khc rng, * l php ton hai ngi tha: (a*b)*c=a*(b*c), mi Xcba ., . Khi cc pht biu sau l tng ng: i) X l nhm ii) Cc phng trnh a*x=b v x*a=b c nghim trong X, mi a, b X iii)Trong X c phn t n v tri v mi phn t trong X u c nghch o tri iv) Trong X c phn t n v phi v mi phn t trong X u c nghch o phi 1.3. nh l Cho (X,.) l mt nhm th ta c cc khng nh sau: i) Mi phn t ca X ch c mt phn t nghch o ii) Nu xy = xz ( yx = zx) th y = z (lut gin c) iii) Vi mi x, y X , ta c (xy) 111 = xy iv) ( x 1 )-1 = x , vi mi Xx 2. Nhm con 2.1. nh ngha Cho G l nhm, H l mt tp con khc rng ca G. Ta ni rng H l nhm con ca G nu H vi php ton cm sinh ca php ton trong G l mt nhm. K hiu GH . D thy tp hp ch gm mt phn t n v ca nhm G lp thnh mt nhm v c gi l nhm n v . K hiu l 1 hoc {e} Nu H G , H 1 , H G th H c gi l nhm con thc s ca G. K hiu GH < 2.2. nh l ( v iu kin tng ng vi nhm con) Cho GH , H . Khi cc iu kin sau l tng ng: i) GH ii) Mi yx, ,H th xy H v x H1 iii) Mi yx, ,H ta c xy H1 2.3. nh ngha Cho G l nhm, GH < i) H c gi l nhm con ti i ca G nu khng tn ti GN < sao cho GNH >==< aG . Khi xy ra hai trng hp sau i) Tt c cc ly tha am ( m Z ) khc nhau i mt. Trng hp ny G l nhm xiclic v hn. ii) Tn ti nhng ly tha bng nhau. Chng hn ak = al ( )lk th G l nhm xiclic hu hn 6. Cp ca phn t 6.1. nh ngha Cho G l nhm i) Cp ca G chnh l lc lng ca G v k hiu l G . Nu G hu hn th G c gi l nhm hu hn, ngc li G c gi l nhm v hn. ii) Cp ca phn t Ga l cp ca nhm v k hiu l a . Nu a hu hn th a gi l phn t c cp hu hn. Ngc li, a c gi l phn t c cp v hn. iii) Nhm G gi l nhm xon nu mi phn t trong G u c cp hu hn. iv) Nhm G gi l khng xon nu trong G ch c duy nht phn t n v c cp hu hn. B) CC PHNG PHP CHNG MINH THNG GP Bi ton 1. Cho phn t x thuc G. Chng minh |x| = n < Phng php gii: Cch 1. Ta chng minh rng exexexex nn = ,,...,, 132 Cch 2. Ta tin hnh kim tra cc tnh cht sau: (i) xn = e (ii) Gi s tn ti kZ sao cho xk = e. Ta chng minh n|k. Bi ton 2. Cho x thuc nhm G. Chng minh x c cp v hn Phng php gii. Ly mi m, n thuc Z sao cho m n. Ta cn chng minh xm xn . Bi ton 3. Cho G l nhm cp n. Chng minh G xylic Phng php gii: Cch 1. Ly mi yG. Ta cn ch ra phn t a G sao cho y = ak , k Z Cch 2. ta cn chng minh tn ti aG sao cho |a| = n Bi ton 4. Chng minh G l nhm hu hn sinh Phng php gii. Ta cn ch ra tp S G sao cho 12. (i) S hu hn phn t (ii) = G C) MT S BI TP C LI GII Bi 1. Cho G l nhm. Chng minh rng: a) Nu S bng rng th { }eS >=< b) Nu S khc rng th = { kn k nn xxx ...21 21 | ii nkSx ,, Z, i= k,1 } Gii. a) Hin nhin. b) Nu S khc rng , t H = { kn k nn xxx ...21 21 | ii nkSx ,, Z, i= k,1 } Ta s chng minh H=. Tht vy, ly hai phn t x,y bt k thuc H. Khi lk m l mmn k nn yyyyxxxx ...;... 2121 121 == , vi nSyx ji ;, ji m, Z (i= ljk ,1,,1 = ). Suy ra ( )( ) ( )( )12212121 1221 1 2121 1 ............ mmm l n k nnm l mmn k nn yyyxxxyyyxxxxy lklk == . Hay lk m l mmn k nn yyyxxxxy = ...... 2121 2121 1 . Do Hxy 1 , nn GH Hin nhin ta c GS Gi s tn ti , H l nhm con ca G cha S. Ly phn t x bt k thuc H. Khi kn k nn xxxx ...21 21= , vi ii nkSx ,, Z, i= k,1 . V , HSxi v SH , nn , Hx in i . Suy ra , 21 ...21 Hxxx kn k nn hay , Hx . Do , HH . Theo nh ngha ca nhm con sinh bi tp hp th >=< SH . Bi 2. Cho G l nhm, A, B l hai tp con ca G. Chng minh rng: a) Nu BA th >>>>>=>>>=>>< , in , kZ, i= k,1 . Nu Bxi , ki ,1= th >< BAx Nu kiAxi ,1, =>< th >< BAxi (do >>< BAx . Nu 1x , >< + AxxBxx kll ,...,,,..., 12 , vi l< BAxi , vi li ,1= , >< BAxj , vi klj ,1+= (v >>< BAB . Suy ra >< BAx Do >>>>==>>=== < >>< BA c) Ta chng minh >>=>=1). t i k i AA 1= = Suy ra >>=>=>=>>==>==< = i n i n HHHH 1 21 ... v HiHk=HkHi, nkiki ,1,; = c) Nu G l nhm Abel v >==< HKKH . Thy vy, ta c KH khc rng ( do K v H khc rng). Ly k1h1, k2h2 bt k thuc KG vi k1, k2 thuc K; h1, h2 thuc H. Khi ( )( ) ( )1 2 1 212 1 21 1 2 1 211 1 2211 == khhkkkkhhkhkhk . V GK nn Kkk 1 21 . V GH nn HkhhkHhh 1 2 1 212 1 21 , . Do ( )( ) KHhkhk 1 2211 . Vy GKH Ly x bt k thuc HK . Khi x thuc K hoc x thuc H . Nu x thuc K th KHxex = (v e H ). Nu Hx th KHexx = (v )Ke . Suy ra KHx hay KHHK Gi s tn ti M l nhm con ca G cha HK . Ly kh bt k thuc KH vi HhKk , . Khi MHKHh MHKKk nn Mkh ( v GM ). Do MKH Vy >=< HKKH Ta chng minh KH=HK. Tht vy, ly kh bt k thuc KH vi HhKk , th kh=khk-1 k. V GH nn Hkhk 1 , do HKkkhkkh = 1 . Suy ra HKKH . Tng t ta c KHHK . Vy HK=KH . b) Ta chng minh >=< = i k i k HHHH 1 21 ... (*) bng phng php quy np theo n Vi n=1 th (*) hin nhin ng Vi n=2 th (*) ng do chng minh trn 14. Gi s ta lun c >=< = i k i k HHHH 1 21 ... , vi k>1, trong 1,1,,,1, == kjGHkiGH ji . Ta chng minh >=< + = + i k i k HHHH 1 1 121 ... , trong kjGHkiGH ji ,1,,1,1; =+= . t i k i k HSHHHH 1 2 121 ,... + = + == . Theo gi thit quy np th >=< SH . V GHGH ,1 nn theo kt qu cu a) ta c >>>==< = i n i n HHHH 1 21 ... v kiHHHH ikki .,= n,1= c) V G l nhm Abel v niGHi ,1, = nn niGHi ,1, = . Theo kt qu cu b) th >=< nn HHHHHH ...... 2121 . Vy ta c iu phi chng minh. Bi 4. Cho H l nhm con chun tc ca G. Chng minh rng nu H v G/H hu hn sinh th G l nhm hu hn sinh Gii. Gi H= < x1, x2, , xn > ; >=< myyyHG ,...,,/ 21 . Ta chng minh >=< mn yyyxxxG ,...,,,,...,, 2121 . t >=< mn yyyxxxK ,...,,,,...,, 2121 . Ly g bt k thuc G. Nu Hg th Kg ( v KH ) Nu HGg th eg v ( ) ( ) ( ) lm l mm yyyg ... 21 21= vi lmHGy jj ,;/ Z, j = l,1 V th lm l mm yyyg ...21 21= . Do ( ) Hyyyg lm l mm 1 21 21 . Nn ( ) kl n k nnm l mm xxxyyyg ...2121 21 1 21 = . Suy ra lk m l mmn k nn yyyxxxg ...... 2121 2121= . Nn Kg . Do G = K hay G l nhm sinh bi tp hp { }mn yyyxxx ,...,,,,...,, 2121 . Vy G l nhm hu hn sinh. Bi 5. Chng minh rng nu H l nhm con thc s ca G th G = . Gii. Ta c HGG nn GHG >< HGgx. V th tn ti x thuc sao cho gx= x . Do vy >< HGG(5.2) T ( 5.1) v ( 5.2) ta suy ra G = < GH >. Vy G = < GH >. Bi 6. Cho X l nhm, Xx . Chng minh rng: = nmx , N, m n th nm xx . Gii. ( ) Gi s tn ti exxxnmnm nmnm ==> :,, t d = m n , d > 0. Ly k ayxy =>< ( k Z). Chia k cho d ta c k = dp + r vi 10 < dr 15. ( ) { } < =< ==== + xx drxx xxxxxy r rrpdrdpk 10 Do = nmx , N sao cho m n th nm xx ( ) Ta c >=< x { xk |kZ } Do mi m ,n N sao cho m n th nm xx xx =>=< T suy ra < nmx , N sao cho m n th nm xx = . Bi 7. Chng minh rng mi nhm c cp v hn u c v hn nhm con. Gii. Nu X = l nhm xiclic c cp v hn th vi mi s t nhin n, ta c l nhm con xiclic ca X v nu mn th >>=< l tp v hn v hu hn. Bi 8. Chng minh rng nu X l nhm ch c cc nhm con tm thng l {e }v X th X l nhm xiclic, hu hn, cp nguyn t. Gii. Ly exXx , . Xt nhm con < x >. V }{ex >< nn < x> = X. Vy X l nhm xiclic Nu x c cp v hn th >< 2 x l nhm con thc s ca X ( tri gi thit ). Vy X phi c cp hu hn n. Nu n khng phi l s nguyn t, tc n = n1n2 ( 21,nn N, n1, n2 1 ), khi nhm con >< 1n x l nhm con thc s cp n2 ca X ( tri gi thit ) Vy X l nhm xiclic, hu hn, cp nguyn t. Bi 9. Chng minh rng nhm con ca nhm xiclic l nhm xiclic Gii. Gi s X l nhm xiclic, X = < a > v A l mt nhm con ca nhm X Nu { }eA = th A = < e > l nhm xiclic Nu { }eA , gi m l s nguyn dng nh nht sao cho Aam Ta chng minh >=< m aA Ta c Aam nn Aam >< (1) Nu Ax , v XA nn >=< aXx nn n Z sao cho x = an , chia n cho m ta c n = mq + r vi mr < m aA (2) T (1) v (2) suy ra >=< m aA 16. Bi 10. Gi s X1, X2 l cc nhm xiclic c cp nguyn t ln lt l n1, n2. Chng minh rng 1X 2X l nhm xiclic khi v ch khi n1, n2 nguyn t cng nhau Gii. Gi s X1 = c cp n1 v X2 = c cp n2. D thy nhm 1X 2X = ( ){ }221121 ,, XxXxxx c cp n1n2 Gi s (n1,n2) = 1 . Ta cn chng minh |(a1a2)| = n1n2 . Tht vy, ta xt ( ) ( ) ( )212121 ,,, 212121 eeaaaa nnnnnn == Gi s tn ti kZ sao cho ( ) ( ) 21 22 11 2121 ,,, nknk ea ea eeaa k k k = = = Do (n1,n2) = 1 nn 21nnk Vy cp ca ( )21,aa trong nhm 1X 2X l n1n2. Do 1X 2X l nhm xiclic sinh bi ( )21,aa Nu (n1,n2) > 1 th [ ] ( ) 21 21 21 21 , , nn nn nn nnk th [ ] kk nnnnnnm ...,...,, 2121 < d n a = H. Vy c duy nht nhm con cp d ca X. Bi 14. Cc nhm sau c bao nhiu nhm con. Tm cp ca chng a) Nhm xiclic cp 12 b) Nhm xiclic cp 17 Gii. a) S 12 c cc c l 1, 2, 3, 4, 6, 12 nn nhm xiclic cp 12 c 6 nhm con c cp 1, 2, 3, 4, 6, 12. b) S 17 c cc c 1, 17 nn nhm xiclic nhm 17 c 2 nhm con c cp 1, 17. Nhn xt. T bi 12 v bi 13 ta c cc kt qu : 18. i) Gi s X l nhm xiclic sinh bi phn t a cp n v b = ak . Khi b l phn t sinh ca nhm con xiclic H ca X cp l d n vi d = (n, k). Hn na >>=< 3 , >< 4 b) 18Z l nhm xiclic vi php cng, c phn t sinh l 1. Tm cc phn t sinh khc ca 18Z Gii. a) Ta c 1.33 = , (12,3) = 3 nn cp ca >< 3 l 4 3 12 = . Do >< 3 = { }9,6,3,0 Ta c 1.44 = , (12,4) = 4 nn cp ca >< 4 l 3 4 12 = . Do >< 4 = { }8,4,0 b) Ta c 5, 7, 11, 13, 17 l cc s nguyn t cng nhau vi 18 nn cc phn t sinh khc ca 18Z l 17,13,11,7,5 . Bi 16. Gi s X l nhm, a v b l hai phn t ca X. a) Chng minh rng cp ca ab bng cp ca ba b) Gi s ab = ba v cp ca a, b l r, s. Khi nu (r, s) = 1 th cp ca ab l rs. Gii. a) Gi s ab c cp l k, khi ( ) eab k = ( ) ( ) ebaeabaaebababababa kk k === 1 ... Nu m l cp ca ba th m l c ca k (1) V ta c ( ) eba m = Ta c ( ) ( ) ( ) eabebabbeabababababeba mm k m ==== 1 ... Suy ra k l c ca m (2) T (1) v (2) ta c m = k Vy cp ca ab bng cp ca ba. b) Do ab = ba nn ( ) ebaab rsrsrs == . Mt khc nu ( ) eab k = th ( ) rkrksaebaeab ksksksks ==== (do (r,s)=1) (1) ( ) skskrbebaeab krkrkrkr ==== (do (r,s)=1) (2). T (1) v (2) suy ra rsk . Vy nu ab = ba v cp ca a, b l r, s v (r, s) = 1 th cp ca ab l rs Bi 17. Tm cp ca cc phn t trong GL2(R) 19. = 10 01 a , = 10 11 b , = 10 11 c Gii. Ta c = 10 01 a , == aaa .2 10 01 10 01 2 10 01 I= = Do a c cp l 2 Ta c = 10 11 b , == bbb .2 10 11 = 10 01 10 11 Do b c cp 2 Ta c = 10 11 c , == ccc .2 10 11 = 10 21 10 11 Gi s = 10 1 m cm , ta chng minh + =+ 10 111 m cm Tht vy ==+ ccc mm .1 10 1 m + = 10 11 10 11 m ,nm = 10 1 m cm = 10 1 n cn Do cp ca c l . Bi 18. Cho G l nhm con ca GL2(C) sinh bi cc phn t = 01 10 A v = 0 0 i i B a) Chng minh rng G l nhm cp 8 b) Chng minh rng G ch c duy nht mt nhm con cp 2 Gii. a) Ta c 4== BA v 22 BA = Nn ABAAB 232 == ; 2 22 IBA = ; BBA =32 ; 33 ABBA = ; ABA =23 ; ABBA =33 Vy { }3332 2 ,,,,,,, ABABBBAAAIG = b) Bng cch kim tra trc tip ta thy A2 l phn t duy nhtc cp 2 Vy >=< 2 AH l nhm con duy nht c cp 2 ( v nhm c cp 2 l nhm xiclic ) Bi 19. Cho G l nhm sinh bi {x,y} tha x2 = y2 . Ch ra nhng phn t trng nhau ca cc phn t sau: ( )3 1 xyu = ; 2 2 yxyxyu = ; 33 3 yxu = ; 32 4 xxyu = ; yxxyxu 312 5 = ; 61 6 yxxu = Gii. Ta bit nu ji uu = th euu ji =1 Ta c 231223112231121 52 === yyxyyxyxyxxyxyxyxxyxxyxyxyxyuu Hay eyyxxyxxyxyyxyyxyyyxyyxyxuu ===== 111122131213121 52 Nn 52 uu = . Tng t ta c 63 uu = Bi 20. Cho G = < a, b > tha a3 = e; b7 = e; a-1 ba = b3 . Chng minh rng G l nhm xiclic cp 3 20. Gii. Gi s a = e th b2 = e. Do b6 = e hay b6 = b7 v th b = e ( mu thun). Do ea nn 3=a . V 31 bbaa = nn ( ) ( )2321 bbaa = . Do 621 baba = . Suy ra 721 babba = . Ko theo eabba = 21 hay ( ) 332 aabb = . V th eabababb =332 nn ( ) ( ) eabbaaabaaab = 112 , suy ra ebabab =23 ( do 931 baba = ). Nn ebaab =210 do ebaab =23 ( do ea =7 ). V th ( ) ebabaaa = 21 suy ra ebba =3 . Ko theo eb =2 nn eb =6 . T ebb == 76 hay b=e Vy G = v 3== aG D) MT S BI TP RN LUYN 1) Cc mnh sau ng hay sai: a) Phn t a ca nhm G c cp l n Z+ nu v ch nu an = e ( e l n v ca G ) b) Nu H v H, l hai nhm con ca nhm xiclic X th H H, l nhm con xiclic ca X c) Tn ti nhm xiclic cp 8 c 5 nhm con phn bit d) Mi nhm xiclic cp 8 u c 4 nhm con phn bit 2) Chng minh rng tp X gm cc ma trn dng 10 1 n , vi nZ cng vi php nhn hai ma trn lp thnh mt nhm xiclic. Tm phn t sinh ca X. 3) Chng minh rng tp X gm tt c cc ma trn dng 100 010 01 x , vi xQ cng vi php nhn hai ma trn lp thnh mt nhm Abel. Hi nhm X ny c phi l nhm xiclic hay khng? Ti sao ? 4) Chng minh rng Z2 Z3 l nhm xiclic nhng Z2 Z4 , Z3 Z6 khng l nhm xiclic . 5) Cho p l s nguyn t.Tm s phn t sinh ca nhm xiclic Z r p , rZ, r 1 6) Tm tt c cc phn t sinh ca cc nhm sau: a) Z12 b) Z7 c) Z9 7) Tm tt c cc nhm con ca cc nhm xiclic cp 12, 17. 21. CHNG III. NG CU NHM A. L THUYT 1. ng cu nhm 1.1 nh ngha. nh x f t nhm X n nhm Y c gi l ng cu nhm nu f bo tn php ton, tc l f(xy)=f(x)f(y) vi mi Xyx , Mt ng cu nhm f t nhm X n nhm X c gi l t ng cu nhm. ng cu nhm YXf : vi f l n nh (ton nh, song nh) c gi l n cu ( ton cu, ng cu). Nu tn ti mt ng cu YXf : ta ni nhm X ng cu vi nhm Y . K hiu YX 1.2. nh l Cho G l nhm, GH . K hiu L(G) l tp tt c cc nhm con ca G, L(G,H) l tp tt c cc nhm con ca G cha H. Khi tng ng HSSf /: l mt song nh t K(H,G) vo L(G/H). Hn na, nu ta k hiu S/H=S* v T/H=T* vi GTSH , th i) ** STST . Khi [ ] [ ]** ,, TSTS = ii) ** STST . Khi ** // TSTS 1.3. Mt s tnh cht i) Nu YXf : l ng cu nhm th f(e)=e v ( ) ( )[ ] 11 = xfxf , Xx ii) Nu cc nh x 21: XXf v 32: XXg l cc ng cu th nh x tch 31: XXfg cng l mt ng cu nhm. c bit tch ca hai n cu ( ton cu, ng cu) l mt n cu ( ton cu, ng cu) iii) Nu YXf : l ng cu nhm th ng cu ngc XYf :1 cng l ng cu nhm 2. nh v to nh ca ng cu nhm 2.1 nh ngha Cho YXf : l ng cu nhm, khi i) nh ca ng cu f ( k hiu l Imf) l tp c xc nh: { } )()(Im XfXxxff == ii) Ht nhn ca ng cu f ( k hiu Kerf) l tp c xc nh { } )()( 1 efexfXxKerf === 2.2 Tnh cht Cho YXf : l ng cu nhm, khi i) Nu XH th YHf )( ii) Nu YK th ( ) XKf 1 iii) Yf Im iv) XKerf v) f l n cu khi v ch khi { }eKerf = vi) f l ton cu khi v ch khi Imf =Y vii) Nu nhm X c sinh bi tp A th Imf c sinh bi tp f(A) 22. viii) Nu XAA ' th )()( ' AfAf ix) Nu YBB ' th ( ) ( )BfBf 1;1 x) X/ Kerf Imf 2.3. nh l. Cho G l nhm, GH v K G. Khi HK/K H/H K 2.4. nh l. Cho G l nhm, H Gv K G, H K. Khi K/H l nhm con chun tc ca G/H v G/K (G/H)/(K/H) B) CC PHNG PHP CHNG MINH THNG GP Bi ton 1. Cho nh x f: X Y. Chng minh f l ng cu nhm Phng php gii. Ta chng minh : (i) X, Y lp thnh nhm vi cc php ton tng ng (ii) Mi x1 , x2 X, ta c f (x1 x2 ) = f (x1) f (x2 ) Bi ton 2. Cho nh x f: X Y. Chng minh f l n cu Phng php gii. Ta chng minh : (i) f l ng cu nhm (ii) f l n nh hoc Kerf = { }e Bi ton 3. Cho nh x f: X Y. Chng minh f l ton cu Phng php gii. Ta chng minh : (i) f l ng cu nhm (ii) f l ton nh Bi ton 4. Cho nh x f: X Y. Chng minh f l ng cu nhm Phng php gii. Ta chng minh (i) f l n cu (ii) f l ton cu Bi ton 5. Chng minh YX , vi X, Y l cc nhm cho trc Phng php gii. Cch 1. Lp nh x YXf : v chng minh f l ng cu nhm Cch 2. Nu X = G/H th ta lp mt ton cu nhm YGf : sao cho Kerf = H Cch 3. Nu X l nhm xylic cp n th ta cn chng minh Y l nhm xylic cp n. C) MT S BI TP C LI GII Bi 1. Cho 1 2,A A ln lt l cc nhm con chnh tc ca nhm 1 2,X X . Chng minh rng: 2121 XXAA v ( ) ( ) ( ) ( )1 2 1 2 1 1 2 2/ / /X X A A X A X A . Gii. Gi iiii AXX /: , 2,1=i l cc ton cu chnh tc. 23. Xt tng ng: ( ) ( )1 2 1 1 2 2: / /X X X A X A ( ) ( ) ( )( )221121 ,, xxxx Ta thy l ton cu nhm v 1 2erK A A = . Do ( ) ( ) ( ) ( )1 2 1 2 1 1 2 2/ /X X A A X A X A . Bi 2. Cho X l nhm xiclic sinh bi phn t a. Chng minh rng: a) Nu cp ca a l v hn th X ng cu vi Z. b) Nu cp ca a l s n hu hn th X ng cu vi Zn Gii. a) Xt nh x f: ZX n an Vi mi m, n Z, ta c ( ) ( ) ( ).m n m n f m n a a a f m f n+ + = = = Nn f l ng cu nhm Hin nhin f l ton nh v X l nhm xiclic sinh bi a. Nu ( ) ( )f m f n= th m n a a= , do ea nm = . Do a c cp v hn nn m n = 0 tc f l n cu. Vy f l ng cu. b) Theo gi thit a n= nn nh x: Zn X c xc nh ( ) r r a = c xc nh tt, khng ph thuc vo vic chn lp r . D dng kim tra c l ng cu nhm. Bi 3. Chng minh rng: a) Mi nhm xiclic cp v hn u ng cu vi nhau. b) Hai nhm xiclic cp hu hn ng vi nhau khi v ch khi chng cng cp. Gii. a) Gi s ,X a Y b= = l cc nhm xiclic cp v hn, Khi nh x :f X Y kk ba R rng l mt ng cu b) Nu hai nhm xiclic hu hn ng cu th hin nhin chng cng cp. Ngc li, gi s ,X a Y b= = l cc nhm xiclic cng cp n. Khi nh x :f X Y k k a b l mt ng cu. Tht vy, ta c k l k l a b k l n b b= = , do f l nh x v l n nh, Hin nhin f l ton nh. Ngoi ra, ta c ( ) ( ) ( ) ( ).k l k l k l k l k l f a a f a b b b f a f a+ + = = = = . Vy f l ng cu. Bi 4. Gi s A v B l hai nhm con chun tc ca nhm X sao cho { }A B e = v X = AB . Chng minh rng X ng cu vi nhm A B Gii . Do AB = X nn vi mi phn t x ca X vit c di dng x = ab vi a A v b B . Gi s c , , x ab a b= = vi , ,a a A v , ,b b B th , 1 , 1 a a b b A B = . V { }A B e = nn , 1 , 1 a a b b e = = do , a a= v , b b= . Vy mi phn t x X c vit mt cch duy nht di dng x = ab, vi ,a A b B . Mt khc cc phn t ca A giao hon c vi cc phn t ca B. 24. Tht vy, vi a, b ty thuc A, B, xt tch 1 1 a b ab . V A v B l nhng nhm con chun tc ca X nn 1 b ab A v 1 1 a b a B . Vy { }1 1 a b ab A B e = nn 1 1 a b ab e = hay ab=ba. nh x : A B X ( ),a b ab l mt ng cu do ab = ba, l n cu do { }A B e = , l ton cu do X =AB. Vy l ng cu. Bi 5. Gi s X l nhm. a) Chng minh [ ]/ ,X X X l nhm Abel b) Chng minh rng nu H X th X/H l nhm Abel khi v ch khi [ ],X X H Gii. Ta c [ ],X X X a) Vi mi ,x y X ta c [ ]1 1 ,x y xy X X nn [ ] [ ], ,xy X X yx X X= , tc l [ ]/ ,X X X l nhm Abel b) /X H Abel ,x y X th ,xyH yxH x y X= th [ ]1 1 ,x y xy H X X H . Bi 6. Cho X v Y l hai nhm xiclic cp tng ng l s v t v cc phn t sinh l x v y a) Chng minh quy tc cho tng ng vi mi phn t n x X vi phn t ( ) nk y Y ,vi k l mt s t nhin khc khng cho trc l mt ng cu nhm khi v ch khi sk l bi ca t b) Chng minh rng nu l ng cu th (s,k)=1. Gii. a) Nu l ng cu nhm th ( ) ( ) ( ) ss ks Y Xe e x x x = = = = . Vy ks Yy e= nn ks t Ngc li, nu ks t , ta chng minh l ng cu nhm. u tin, ta chng minh l mt nh x. Tht vy, nu n m x x= th n m s do ( )k n m t . Bi vy kn km y y= tc l l nh x. Ngoi ra, ta c ( ) ( ) ( ) ( ) ( ). . n m kn m n m nk mk n m x x x y y y x x ++ = = = = b) Ta c { }2 1 , ,..., ,s s X x x x x e = = . Do l ng cu nn ( ) { }eyyyyY skskk k == ,,...,, 12 Tht vy, do l ng cu nn ts = tc s phn t ca X v Y bng nhau v ji xx th sjiyy ji kk ,1,, = Do k y l mt phn t sinh ca Y nn ( ) 1, =tk . V t = s nn (s, k) = 1 Bi 7. Cho GAB v GC . Chng minh rng: a) ( ) ( )CACB v ( ) ( ) ( ) BCABCBCA // b) Nu GC th ACBC v ( )CABABCAC // 25. Gii. a) V ACA )( v AB . p dng nh l 2.3 vi G c thay bi A, H thay bi CA v K thay bi B, ta c CB = ( )( ) ( )CABCA v ( ) ( ) ( ) ( ) BCABBBCACBCA /// = (Tnh cht 3.2 v chng I ) b) Gi s GC . Theo tnh cht 3.2 v chng I th AC v AB l cc nhm con ca G. Mt khc, GBC ( tnh cht 3.2 vi) chng I ), GACBC ( do BC G, AC G, v BC AC ). Do ACBC p dng nh l 2.3 vi G thay bi AC, H thay bi BC , ta c ( )( ) ABCAA / v A / ( )( ) ( ) BCBCABCA // . Mt khc ( ) ( )CABBCA = . Tht vy, )()( BCACAB v AB . Ly )(BCAa , suy ra a = bc vi b B, c C . Do CAcab =1 . Suy ra )( CABa V vy ( ) )( CABBCA = Vy AC/BC )(/ CABA . Bi 8. Cho A )(GZ v HGf : l ton cu . Chng minh rng )()( HZAf Gii . Ta ch cn chng minh f(A) )(HZ Ly )(Afy . Do f l ton cu nn tn ti Ax sao cho y = f(x) Ta c yh = f(x)f(g) = f(xg) = f(gx) = f(g)f(x) = hy, Hh Do ( )HZy . Vy ( ) ( )HZAf . Do ( ) ( )HZAf Bi 9. a) Cho X l mt nhm giao hon. Chng minh rng: nh x : x X k a a vi k l mt s nguyn cho trc, l mt ng cu. Xc nh Ker . b) Cho X l mt nhm. nh x : X X 1 aa l mt ng cu ca nhm X khi v ch khi X l nhm Abel. Gii. a) Ta c ( ) ( ) k k k ab ab a b = = ( v X l nhm Abel ) nn ( ) ( ) ( )ab a b = Ker { } {k x X x e x X = = = cp x l c ca k } b) Gi s X l mt nhm Abel th theo a) l mt ng cu v {erK x X x = c cp l c ca }1 hay erx K cp x l c ca -1 x e = Do l mt n cu, hn na l mt ton cu v ( ) 11 x x = . Do l mt ng cu . o li, gi s l mt ng cu, khi vi mi ,a b X ta c ( ) ( ) ( ) ( ) 11 1 ab a b a b ba = = = Mt khc ( ) ( ) 1 ab ab = suy ra ( ) ( ) 1 1 ab ba = hay ab = ba. 26. Vy X l mt nhm Abel. Bi 10. Chng minh rng c mt ng cu duy nht t nhm cng cc s hu t Q n nhm cng cc s nguyn Z. T suy ra nhm cng cc s hu t Q khng phi l mt nhm xiclic Gii. Gi s f: Q Z l mt ng cc t nhm cng Q n nhm cng Z. Nu 0f ngha l tn ti mt s hu t ( )0q q sao cho ( ) 0f q . Ta thy rng ( ) ( ) ( ).1 1 0f q f q qf= = , t ( ) 11 0f n= . Vy vi mi s nguyn 0n , ( ) ( ) 01 = nfnf Gi s n > 1, n Z v ( )1, 1n n = . Ta c ( ) 0 1 1 1 1 . .f f n nf n n n n n = = = = vi 0 1 n n f = Nh vy n li l c ca n1. V l Vy ch c mt ng cu 0 t nhm cng cc s hu t Q v nhm cng cc s nguyn Z, Q khng ng cu vi Z nn Q khng l mt nhm xiclic ( bi 3 ) Bi 11. Tm tt c cc ng cu t a) Z6 n Z18 b) Z18 n Z6 c) Mt nhm xiclic cp n n chnh n d) Mt nhm xiclic cp n n mt nhm xiclic v hn Gii. Ta c mi ng cu f : Zn Zm hon ton c xc nh bi ( ) kf =1 ( tc ( ) kxxf = ) Theo bi 6 th f l ng cu khi v ch khi mkn . Bi vy ta c a) Mi ng cu f : Z6 Z18 hon ton xc nh bi kf =)1( vi 180 < k v 3186 kk . Do k = 0, 3, 6, 9, 12, 15 nn c tt c 6 ng cu f : Z6 Z18 l cc ng cu xc nh bi: ( ) 01 =f , ( ) 31 =f 15)1(,12)1(,9)1(,6)1( ==== ffff b) Mi ng cu f : Z18 Z6 hon ton xc nh bi kf =)1( vi 60 < k v 618 k . Do 5,0=k . Nh vy c 6 ng cu f : Z18 Z6 , l 0)1( =f , 1)1( =f , 2)1( =f , 3)1( =f , 4)1( =f , 5)1( =f c) Mi ng cu f : Zn Zn hon ton xc nh bi kf =)1( vi k0 < n v nkn . Do c tt c n ng cu f : Zn Zn c xc nh bi: kf =)1( vi 1,...,2,1,0 = nk d) Gi s baf : l ng cu nhm v gi s k baf =)( . Khi ta c nknn bafafefe ==== ))(()()( . V b c cp v hn nn nk = 0, suy ra k = 0 v do eaf =)( Vy c duy nht mt ng cu t nhm xiclic hu hn v nhm xiclic cp v hn l ng cu tm thng. Bi 12. a) Tm Ker v (25) ca ng cu : Z Z7 bit 4)1( = b) Tm Ker v (18) ca ng cu : Z Z10 . Bit 6)1( = 27. c) Tm Ker v )3( ca ng cu : Z10 Z20 bit 8)1( = Gii. a) : Z Z7, 4)1( = Ker ={xZ| (x)=0 } ={xZ| x (1) = 0 }= { xZ| x4 =0 }={xZ| 4x7} = 7Z ( ) ( ) ( ) 24.251251.2525 ==== b) : Z Z10 , ( ) 61 = Ker ={xZ| (x)=0 }={xZ|x (1)= 0 }={xZ| x6 = 0 }={xZ| 6x10} = 5Z 86.18)1(18)1.18()18( ==== c) : Z10 Z20 , ( ) 81 = Ker ={ x Z10| ( x ) = 0 } ={ x Z10| x (1) = 0 } = { x Z10| x8 = 0 } = { x Z10| 8x 20} = { x Z10| x5}= { 0 , 5, 10 } 424)1(3)13()3( ==== Bi 13. Cho G l nhm hu hn sinh, G l nhm bt k v mi nh x ': GSf , trong S l tp sinh ca G. Chng minh rng tn ti duy nht ng cu ': GGF sao cho fF S = . Ngc li, nu G l nhm Abel v mi nh x ': GSf u tn ti ': GGF sao cho fF S = th G l nhm sinh bi tp S. Gii. Ta bit Gx th kn k nn xxxx ...21 21= , vi ii nSx , Z, i= k,1 Xy dng tng ng ': GGF ( ) kn k nn xfxfxfxFx ...)()()( 21 21= Nu Gex = th 1 1 1 1 = xxx , vi Sx 1 . Khi ( )( ) ( )( ) 1 1 1 1)( = xfxfxF hay F(e)=e. Ta chng minh F l nh x. Tht vy, vi Gx th ( ) 'GxF . Ly a, b thuc G tha a = b th eab =1 nn ( ) ( ) eeFabF ==1 (1). V Gba , nn lk m l mmn k nn yyybxxxa ...,... 2121 2121 == , vi lkmnSyx jiji ,,,;, Z, i= k,1 , j = l,1 . V th ( )( ) ( )( )12212121 1221 1 2121 1 ............ mmm l n k nnm l mmn k nn yyyxxxyyyxxxab lklk == . Hay 122 1221 1 ...... mmm l n k nn yyyxxxab lk = . Suy ra ( ) ( )[ ] ( )[ ] ( )[ ] ( )[ ] ( )[ ] ( )[ ] 1221 1221 1 ...... mmm l n k nn yfyfyfxfxfxfabF lk = (2) T (1), (2) suy ra ( )[ ] ( )[ ] ( )[ ] ( )[ ] ( )[ ] ( )[ ] eyfyfyfxfxfxf mmm l n k nn lk = 1221 1221 ...... Hay ( )[ ] ( )[ ] ( )[ ] ( )[ ] ( )[ ] ( )[ ] 12121 1221 ...... mmm l n k nn yfyfyfxfxfxf k = . V th F(a)=F(b). Vi x,y bt k thuc G th lk m l mmn k nn yyybxxxa ...,... 2121 2121 == , vi jiji mnlkSyx ,,,;, Z, i= k,1 , j = l,1 . Ta thy ( ) ( )[ ] ( )[ ] ( )[ ] ( )[ ] ( )[ ] ( )[ ] 1221 1221 ...... mmm l n k nn yfyfyfxfxfxfabF lk = . V th F(ab) = F(a)F(b). Nn F l ng cu. Ly x bt k thuc S th 1 xx = nn ( ) ( )[ ] ( )xfxfxF == 1 . Vy fF S = . Gi s tn ti ng cu ': GGg tha g(x)=f(x), Sx . Ly Ga th = ii n k nn nkSxxxxa k ,;,...21 21 Z, i= k,1 . Khi 28. ( ) ( )[ ] ( )[ ] ( )[ ] ( )[ ] ( )[ ] ( )[ ] kk n k nnn k nn xgxgxgxfxfxfaF ...... 2121 2121 == . V th ( ) ( )kn k nn xxxgaF ...21 21= (v g l ng cu). Nn F(a)=g(a), vi mi a thuc G. Vy F = g hay F l duy nht. Ngc li. Nu G l nhm Abel th G . t >=< Sg . Vy >=< SG . Bi 14. Chng minh rng : a) nh ng cu ca nhm hu hn sinh l nhm hu hn sinh. b)Nhm thng ca nhm hu hn sinh l nhm hu hn sinh. Gii. a) Cho G l nhm hu hn sinh vi tp sinh l S, G l nhm v ' : GGf l ng cu. Ta chng minh = Imf. Tht vy, Imf khc rng v fe Im . Ly 21, yy bt k thuc Imf. Khi ( ) ( )221121 ;:. xfyxfyGxx == . Do ( ) ( )[ ] fxfxfyy Im 1 21 1 21 = . Nn ' Im Gf Ly y bt k thuc f(S) th ( )xfySx = : . Do fy Im , v th ( ) fSf Im . Gi s tn ti ' GH cha f(S) . Ly y bt k thuc Imf th tn ti x thuc G sao cho y = f(x). V Gx nn kn k nn xxxx ...21 21= , vi ii nkSx ,; Z, i= k,1 . Do ( ) ( )[ ] ( )[ ] ( )[ ] Hxfxfxfxfy kn k nn == ...21 21 ( GH v ( ) ( ) )HSfxf i . Nn Imf H . V th >=< )(Im Sff . Vy nh ng cu ca nhm hu hn sinh l nhm hu hn sinh. b) Cho G l nhm, GH . Xt ton cu chnh tc HGG /: xx Theo chng minh trn th ( ) >>==< = SGi n i 1 vi ( ) ( ) ( ) ( ) ( ) ( ){ }nnmnmm xeeexeeeeexeeexeeexeexS ,,...,,;...;,,...,,;...;,...,,,;...;,...,,,;,..,,;...;,...,, 1221111 21 = Tht vy, ly x = ( i n i n Gxxx 1 21 ,...,, = . Trong niSGx iii ,1, =>=< , nn iim i i k im k i k ii xxxx ...2 21= , vi iimiiij kmSx ,; Z, nimj j ,1;,1 == . Do ( )nnm n mmmm k nm k m k m k m kkk m kk xxxxxxxxxx ...,...,...,... 2 2 1 1 22 2 222111 1 1211 2222111211= . V th x= ( ) ( )[ ]( ) ( )[ ] ( ) ( )[ ]nnm n nmm k nm k n k m kk m k xeexeeeexeeexeeexeex ,...,,...,...,,...,...,,,...,...,,,,...,,...,...,, 122 2 2111 1 11 1221111 Do vy : ( ) ( ) ( ) ( ) ( ) ( ) nnm n nmm k nm k n k m kk m k xeexeeexeexeeexeexx ,,...,...,...,,...,..,,...,..,,...,...,,...,...,, 122 2 2111 1 11 1221111= Nn .>< Sx Do G = . Vy i n i G 1= =. D) MT S BI TP RN LUYN 1) Cc pht biu sau l ng hay sai: 30. a) Hay nhm bt k G, G, lun tn ti ng cu t G v G, . b) Mi ng cu nhm l n cu khi v ch khi ht nhn ca n ch cha phn t n v. c) Mt ng cu nhm c th c ht nhn l tp rng. d) Tn ti duy nht ng cu t Z7 Z12. e) C tt c bn ng cu t Z3 Z12 f) Vi X, Y l 2 nhm xiclic cp m, n ( )m n tn ti ng cu nhm t X vo Y. g) Nhm cng cc s hu t Q l mt nhm xiclic. h) Cho :f X Y l ng cu nhm, v X = a . Khi f(x) c th khng l nhm xiclic. 2) Cho G l nhm v :f G G G ( g1, g2 ) g1 g2 Chng minh rng f l ng cu nhm khi v ch khi G l nhm Abel. 3) Chng minh rng nu A l mt nhm con chun tc ca X th tn ti mt song nh t tp hp cc nhm con chun tc ca X cha A n tp hp cc nhm con chun tc ca X/A. 4) Cho G l nhm v 35G = . Chng minh rng G Z35 5) Cho :f X Y l ng cu t nhm X n nhm Y v A l nhm con ca X. Chng minh rng ( )1 f f A AB = , vi erB K f= . 6) Cho :f X Y l ng cu t nhm X n nhm Y v 15A = . Chng minh rng [ ], erX Y K f CHNG IV. NH L LAGRANGE V NHM GII C A. L THUYT 1. nh ngha Cho G l nhm, H G, a G (i) Tp Ha ={ ha|hH } c gi l lp ghp phi ca a i vi nhm con H (ii) Tp aH ={ah| hH } c gi l lp ghp tri ca a i vi nhm con H Nhn xt. Cho G l nhm, H G. Khi , mi a G, ta c |aH| = |Ha| = |H| ( y ta hiu |aH| l lc lng ca tp aH ) 31. 2. nh ngha Cho X l nhm v H X, s cc lp ghp tri ( hoc phi ) ca x theo nhm con H c gi l ch s ca H trong X, k hiu [ ]:X H 3. nh l Lagrange Gi s H l nhm con ca nhm hu hn X. Khi H l c ca X v [ ]:X H = X H 4. Cng thc lp X l hu hn, khi : X = ( )Z X + i I [ ]( )iX C x , ( )ix Z X Ch . Cho X l nhm. Khi (i) Nu H X th X = H . [ ]:X H , ( X ty ) (ii) Nu X- hu hn, H X th /X H = X H (iii) Nu X- hu hn, khi x X, x l c ca X 5. nh ngha Gi s X l nhm, khi : (i) Nu x X c s nguyn dng m sao cho xm =e th m c gi l s m ca x. (ii) S nguyn dng m c gi l s m ca nhm X nu m l s nguyn dng sao cho xm = e, mi x thuc X iii) Cho p l mt s nguyn t. Nhm X c gi l nhm strictly p-closed nu tn ti duy nht H l p-nhm con Sylow ca G v G/H l nhm giao hon vi s m l c ca p-1 6. Mnh Cp ca nhm hu hn X l mt s m ca n. 7. nh ngha Gi s p l s nguyn t, khi : (i) Nhm H c gi l p- nhm nu cp ca n l mt ly tha ca p. (ii) Nhm H c gi l p- nhm con ca nhm X nu H X v H l p- nhm. (iii) Nhm H c gi l p- nhm con Sylow ca nhm X nu H l p- nhm con ca X v H = pn l ly tha cao nht ca p chia ht X 8. Tnh cht. (i) Gi s X l nhm Abel, hu hn cp m v p l s nguyn t chia ht cho m. Khi X c cha mt nhm con cp p. (ii) ( nh l Sylow 1 ) Gi s X l nhm hu hn, p l s nguyn t chia ht cho X . Khi lun lun tn ti p- nhm Sylow ca X. (iii) ( H qu Cauchy ) Nu s nguyn t p chia ht cp ca nhm hu hn X th trong X s tn ti phn t cp p. iv) ( nh l Sylow 2 ) Gi s X l nhm hu hn v p l c nguyn t ca X . Khi : Mi p- nhm con H ca X u nm trong p-nhm con Sylow no ca X Nu 1 2,P P l p-nhm con sylow ca X chng lin hp vi nhau, tc l x X sao cho P2 = x. P1. x-1 ( P1 =xP2x-1 ). Nu r l s cc p- nhm con Sylow ca X th r 1 ( mod p ), r X 32. v) Nu X = mpk ( m, p ) = 1, p nguyn t, khi s p- nhm con Sylow ca X l c ca m. (vi) Nu H l p- nhm con Sylow cp t duy nht ca X th H X. (vii) Cho G l nhm hu hn. Nu G c ng mt p- nhm con Sylow vi mi p l c nguyn t ca G th G l tch trc tip ca cc p- nhm con Sylow ca n. 9. nh ngha. Cho dy cc nhm con ca nhm G 0 1 11 ... n nG G G G G= = (*) sao cho ( )1 0, 1i iG G i n+ = . Dy (*) c gi l dy chun tc ca G v k hiu l 0 1 11 ... n nG G G G G= = Vi (*) l dy chun tc ca G, khi i) S n oc gi l di ca chui ii) Cc ( )0,iG i n = c gi l cc s hng ca dy iii) Cc nhm thng ( )1 / 0, 1i iG G i n+ = c gi l cc nhn t ca dy 10. nh ngha. i) Dy chun tc ca G c gi l dy Abel nu tt c cc nhn t ca dy u l nhm giao hon. ii) Dy chun tc ca nhm G c gi l dy xiclic nu tt c cc nhn t ca dy u l nhm xiclic iii) Dy chun tc ca nhm G c gi l dy hp thnh nu tt c cc nhn t ca dy u l nhm n. 11. nh ngha. i) Cho G l nhm, H G H c gi l nhm con chun tc ti i ca G nu H 1 H K = 3 ( u ny khng th v H K H, H K K v H K = H nn H K = H nn H K H K m H khng l c ca K ) Ta c HK = H K H K = 15 = X Mt khc HK X HK = X (2) Ta c H, K l nhm xiclic ( c cp nguyn t ) v ( H , K ) = 1 nn HK = X l nhm xyclic Vy X l nhm Abel Bi 5. Cho G l nhm hu hn, H G. a) Chng minh rng nu H, G/H l p- nhm th G l p- nhm b)Chng minh rng nu H l p- nhm, G l p- nhm th G/H l p- nhm Gii . a) Ta c /G H = G H = pk ( do G/H l p-nhm ) G = pk H = pk pl = pk+l ( H l p- nhm ) G l p- nhm. 35. b) Nu G l p- nhm, H l p- nhm Ta c G = pm H = pn ( H < G, m n ), p nguyn t / m m n n G p G H p H p = = = G/H l p- nhm Bi 6. a) Chng minh rng nu nhm X { }e l p- nhm hu hn th { }( )Z X e b) Chng minh rng mi p- nhm u l nhm gii c. Gii . a) Xt cng thc lp: ( )( ) : ( ) i i I i X Z X X C x x Z X = + Do ( )ix Z X , tc : i ix X x x xx Do ( )iC x l nhm con thc s ca nhm X. Do X l p- nhm nn ( 1)X p = ( )iC x l nhm con thc s ca nhm X nn ( )iC x X ( ) ( ) ( ) : i i i X p X C x C x C x = = l ly tha ca p ( )Z X l mt bi ca p Do ( )Z X { }e b) Gi X l p- nhm v n X p= vi n l s t nhin. Ta chng minh bng quy np theo n. Vi n = 0, 1 n = 0 th { }X e= l nhm Abel nn X l nhm gii c vi chui gii c l X { }e n = 1 th X p= X l nhm xiclic nn X l nhm Abel, do X l nhm gii c Gi s nh l ng cho mi nhm cp pk (1 )k n xt tm giao hon Z(X) ca X . Theo nh l Lagrange th tn ti s t nhin r n ( ) r Z X p= . Theo a th Z(X) khc { }e nn r 0 V th / ( ) ( ) n rX X Z X p Z X = = ( X hu hn, Z(X)< X ) Do gi thuyt quy np nn X/Z(X) l nhm gii c. Theo bi 4 th X l nhm gii c Bi 7. a) Chng minh rng nu X khng phi l nhm Abel th X/Z(X) khng phi l nhm xiclic. b) Chng minh rng mi nhm cp p2 u l nhm Abel vi p l s nguyn t. Gii . a) Gi s X/Z(X) l nhm xiclic z0 X, X/Z(X) = oz Mi x, y X, tn ti k, m Z sao cho 0 0 k k x z z= = , 0 0 m m y z z= = . Suy ra tn ti z1, z2 Z(X) sao cho x = z k 0 z1, y = z m 0 z2 36. Ta c ( )( )0 1 0 2 0 1 0 2 k m k m xy z z z z z z z z= = = ( )( )0 0 1 2 0 0 1 2 0 2 0 1 yxk m m k m k z z z z z z z z z z z z= = = Do X l nhm Abel Vy nu X khng phi l nhm Abel th X/Z(X) khng th l nhm xiclic b) Gi s 2 G p= p l s nguyn t Ta c 2 G p= l p- nhm theo 6a) th Z(G) {e}. Do ( )Z G p= hoc 2 ( )Z G p= Nu Z(G) = p 2 / ( ) ( ) G p G Z G p Z G p = = = G/Z(G) l nhm xiclic G l nhm Abel 7a) Nu 2 ( )Z G p= / ( ) 1 / ( ) ( ) G G Z G G Z G Z G = = l nhm xiclic G l nhm Abel Bi 8. Nu nhm X khc { }e khng c nhm con chun tc no ngoi X c gi l nhm n. Chng minh rng nhm gii c l nhm n khi v ch khi n l nhm gii c cp nguyn t. Gii. ( ) Gi s X l nhm n gii c vi dy Abel l: { }0 1 ... nX X X X e= = Khng mt tnh tng qut gi s 1i iX X + vi mi i= n,1 Do 1X X< , nhng X l nhm n v 1X X= nn { }1X e= . V X gii c nn 0 1/X X X= Abel. Do nu H l nhm con ca X th H cng l nhm con chun tc ca X. M X ch c hai nhm con chnh tc tm thng { }e v X nn X ch c hai nhm con tm thng. Tht vy, nu X c thm mt nhm con khc X v { }e l K th K cng l nhm con chun tc ca X tri gi thit X l nhm n. Theo bi 8 chng II th X l nhm xiclic cp nguyn t nn X l nhm gii c cp nguyn t. ( ) Gi s X l nhm gii c cp nguyn t th { }X e ; X ch c hai nhm con X v{ }e nn X khng c nhm con chun tc no ngoi X v { }e . Do X l nhm n. Vy X l nhm n gii c Bi 9. Chng minh rng H, K l nhm gii c th H K l nhm gii c. Gii. H l nhm gii c nn ta c chui gii c: { }0 1 ... nH H H H e= = K l nhm gii c nn ta c chui gii c : { }0 1 ... mK K K K e= = Khng mt tnh tng qut gi s n m { } { }0 1 1... ...m m nK K K K e K K e+= = = { }0 0 1 1 ... ,n n H KH K H K H K H K e e = = 37. Ta chng minh 1 1i i i iH K H K < v 1 1 /i i i iH K H K l nhm Abel Tht vy, mi 1 1( , ) i iz x y H K = , 1 2( , ) i il l l H K= , vi 1 1 1 2, , ,i i i ix H y K l H l K V H gii c nn 1 1 ixl x H K gii c nn 1 2 iyl y K Ta c 1 1 1 1 2( , )( , )( , )zlz x y l l x y = 1 1 1 2( , ) i ixl x yl y H K = Do 1 1i i i iH K H K < Mi 1 2 1 2 1 1( , ), ( , ) i im x x n y y H K = = Ta c 122111 ,,, ii KyxHyx Do H gii c nn 1 1 1 1 1 1 ix y x y H ( 1,i n= ) K gii c nn 1 1 2 2 2 2 ix y x y K ( 1,i n= ) Ta c 1 1 1 1 1 1 1 1 2 2 1 1 2 2( , )( , )mnm n x y x y x y x y = 1 1 1 1 1 1 1 1 2 2 2 2( , y x ) i ix y x y x y H K = Do 1 1 /i i i iH K H K l nhm Abel Vy H K l nhm gii c Bi 10. a) Chng minh rng mi nhm cp 6 u gii c b) Hi mi nhm cp pq vi pq l cc s nguyn t c phi l nhm gii c khng ? Ti sao ? Gii . a) 2.3X = Trong X tn ti 3-nhm con Sylow Gi r l s 3-nhm con Sylow (mod3) 1 2 r r r = Trong X tn ti duy nht 3-nhm con Sylow H do , 3H X H =< , / 2 X X H H = = Vy H gii c, /X H gii c do X gii c b) Vi 2 p q X p X= = gii c. Khng mt tnh tng qut gi s p > q Khi theo nh l Sylow th trong X tn ti q nhm con Sylow Gi r l s q nhm con Sylow Khi 1(mod ) 1 r q r r p = ( )q p Tn ti duy nht q nhm con Sylow H nn ,H X H q=< , / X pq X H p H q = = = Nn H, X/H l nhm gii c do X l nhm gii c. Bi 11. Chng minh rng mi cp 30 u gii c Gii. Ta c 30 2.3.5X = = Trong X tn ti 5-nhm con Sylow Gi r l s 5-nhm con Sylow, theo nh l Sylow 38. 1(mod5) 1 6 6 r r r r = = r = 1, khi trong X tn ti duy nht 5-nhm con Sylow H 5, = HXH < nn H gii c. Mt khc / X X H H = = 6 X/H gii c Vy X gii c r = 6, khi trong X c 6 nhm 5-nhm con Sylow Tp hp 6 nhm 5-nhm con Sylow c 25 phn t (mi nhm c 5 phn t, trong mi nhm u c chung phn t n v v mi nhm u l nhm xiclic ( do cp nguyn t ) nn ngoi phn t n v cc phn t mi nhm l khc nhau nn bt k 2 nhm 5-nhm con Sylow c chung mt phn t khc ngoi n v th chng phi trng nhau ) Do X = 2.3.5 nn trong X tn ti 3-nhm con Sylow Gi t l s 3- nhm con Sylow ( ) = = 10 1 10 3mod1 t t t t Vi t =10 loi do trong X khng c phn t t = 1 Vy tn ti 3-nhm con Sylow K trong X , 3K X K =< nn K gii c. Mt khc, / 10 2.5 X X K K = = = X/K gii c Vy X/K gii c v K gii c nn X gii c Vy mi nhm cp 30 u gii c . Bi 12. Chng minh rng mi nhm cp pqr vi p, q, r l cc s nguyn t r pq u l nhm gii c. Gii . Ta c X pqr= Trong X tn ti r nhm con Sylow Gi a l s r nhm con Sylow. 1(mod )a r a pq Do r pq nn a = 1 Trong X tn ti duy nht r-nhm con Sylow H ,H X H r=< . Mt khc / X qpr X H qp H r = = = Vy H gii c, X/H gii c nn X gii c Vy mi nhm cp pqr ( p, q, r l s nguyn t ), r pq u l nhm gii c. Bi 13. a)Chng minh rng mi cp 12 u gii c. b) Chng minh rng mi nhm cp 588 l nhm gii c Gii . a) Gi s 2 12 2 .3X = = Trong X tn ti 3- nhm con Sylow Gi r l s 3- nhm con Sylow, khi 39. 1(mod3) 1 4 r r r = Trong X tn ti duy nht 3- nhm con Sylow H nn H < X v |H| = 3. Mt khc |X/H| = 2 2 2 3 3.2 || || == H X Ta c 3H H= gii c, 2 / 2X H = nn X/H gii c. Do X gii c. b) Gi s 2 2 588 2 .3.7X = = Trong X tn ti 7-nhm con Sylow Gi r l s 7-nhm con Sylow 1(mod7) 1 12 r r r = Tn ti duy nht 7-nhm con Sylow H ca X 2 , 7H X H =< nn H gii c. Mt khc 2 / 2 .3 12 /X H X H= = gii c. Do X gii c Vy mi nhm cp 588 l nhm gii c D. MT S BI TP RN LUYN 1. Xt tnh ng sai ca cc mnh sau: a) Mi nhm cp 9 u gii c b) Mi nhm cp nguyn t u gii c c) Mi nhm hu hn c cp chia ht cho 5 u cha nhm xiclic cp 5 d) Cho G l nhm cp 8, trong G tn ti nhm con cp 3 e) Mi nhm cp 45 u c cha nhm con chun tc cp 9 f) Mi nhm cp 27 u c cha phn t cp 3 2) Chng minh rng mi nhm cp 28 u gii c 3) Cho H, K l hai nhm con gii c ca nhm X v XH < Chng minh rng HK l nhm gii c 4) Chng minh rng mi nhm cp l b hn 60 u gii c 5) Chng minh rng Sn l nhm gii c vi n < 5 6) Chng minh rng Sn khng l nhm gii c vi 5n 40. CHNG V. NHM LY LINH A. L THUYT 1. nh ngha Ta ni rng nhm con K chun ha H nu ( )GK N H ( ) { }( )1 :aHaGN H a G H = = Nh vy K chun ha H khi v ch khi [ ],H K H= 2. nh ngha Cho H l nhm con ca G, tm ca H trong G l CG(H) = { xG |xh = hx, mi h H }= {x G| [x, h] = e, mi h H } Ta ni nhm con K tm hon H nu ( )GK C H Nh vy ( )GK C H khi v ch khi [ ], 1H K = 3. Tnh cht (i) Nu K G< v K H G th [ ],H G K khi v ch khi: ( )/ /H K Z G K (ii) Nu ,H K G v :f G L l ng cu th f ([H, K]) = [f(H), f(K)] (iii) Cho G l mt nhm hu hn v H G< . Nu P l mt nhm con Sylow ca H th . ( )GG H N P= v [ ]:G H l c ca ( )GN P (iv) Cho H G< . Khi G/H l nhm giao hon khi v ch khi , G H v) Nu P l mt p-nhm con Sylow ca mt nhm hu hn G th NG(P) bng chun ha ca n trong G 4. nh ngha Nhm con c trng ( )i G ca G c xc nh bng quy np nh sau: ( ) GG =1 , ( ) ( )[ ]GGG ii ,1 =+ Nhn xt (i) ( ) ( )[ ] [ ] , 12 ,, GGGGGG === (ii ) 1( ) ( )i iG G + T [ ]1 1( ) ( ); ( ), ( )i i i iG G G G G + + = v tnh cht 3 (i) ta c 1 1( ) / ( ) ( / ( )i i iG G Z G G + + 5. nh ngha Chui tm di ( hay chui tm gim ) ca G l chui: 1 2( ) ( ) ...G G G = 41. 6. nh nghi Tm trn ( )i G l nhm con c trng ca G xc nh bng quy np nh sau: ( )0 1 ( ) 1, ( ) / ( ) / ( )i i i G G G Z G G + = = Tc l nu : / ( )i iv G G G l ng cu t nhin th 1 ( )i G + l nghch nh ca tm ( )1 1 ( / ( ); ( ) ( / ( )i i i iZ G G G v Z G G + = Ta c 1 ( ) ( )G Z G = v 1 0 0 ( ) / ( ) ( / ( ))G G Z G G = Suy ra 1 ( ) /1 ( /1)G Z G = hay 1 ( ) ( )G Z G = 7. nh ngha (i) Chui tm trn ( hay chui tm tng ) ca G l chui 0 1 2 1 ( ) ( ) ( )G G G = Nu khng c g nhm ln ta c th vit gn li 2 thay cho 2 ( )G v i thay cho ( )i G (ii) Chui chun tc 0 11 ... nG G G G= =< < < vi iG G< v ( )1 1/ /i i iG G Z G G c gi l chui trung tm. (iii) Nhm con hon t trn ca G c nh ngha bng quy np ( ) ( ) ( )10 ; , i i i G G G G G + = = vi mi 0i Vi 0i = ta k hiu ( ) [ ]1 , , ,G G G G G= = Ta c ( )i G G< vi mi i Chui ( ) ( ) ( )0 1 2 ...G G G G= c gi l chui dn xut ca G. 8. Tnh cht (i) Cho G l nhm. Nu tn ti c N sao cho ( )c G G = th: ( ) ( )GG ic i + 1 vi mi i v khi 1( ) 1c G + = Ngc li, nu tn ti c N sao cho 1( ) 1c G + = th ( ) ( )GG ic i + 1 vi mi i v ( )c G G = . (ii) Chui chun tc ca G 0 11 ... nG G G G= =< < < l chui trung tm khi v ch khi [ ] 1,i iG G G vi mi i = 1,, n. (iii) Nu 0 1 2... 1nG G G G G= => > < l chui gii c th ( )i iG G vi mi i N iv) Nhm G l nhm gii c khi v ch khi tn ti n N sao cho ( ) 1 n G = Nhn xt. Chui dn xut l chui gii c khi v ch khi G l nhm gii c. 9. nh ngha (i) Cho M G , nhm con ca M ca G c gi l nhm con cc i ca G nu M G v G khng c nhm con ca H no M H G . Nu. G v hn th G c th khng c nhm con cc i no. Nhn xt. Cho K G< , khi K l nhm con cc i ca /G G K c cp nguyn t (ii) Cho G l nhm. Nhm con ca Frattini ca n k hiu l ( )G c nh ngha bng giao ca tt c cc nhm con cc i ca G. Nu mt nhm G (v hn ) khng c nhm con cc i, ta nh ngha ( ) GG = 10. nh ngha 42. Cho H G , A l mt tp khc rng cc t ng cu ca G. H c gi l A con bt bin nu ( ) , ,h H h H A Nhn xt. Nu A = 1, mi nhm con ca G l bt bin Nu H l mt G nhm con bt bin th H c gi l nhm con c trng ca G, k hiu H char G. Vy H char G nu ( )H H = vi t ng cu ca G 11. Tnh cht i) Nu ( )H H vi mi t ng cu th H char G. ii) Nu H char G th H G< iii) Cho GH , nu ( ) HHf , f AutG th H char G iv) Nu H char G th GH < v) Nu H char K, K char G th H char G vi) Nu H char K, GK < th GH < vii) Cho G l nhm xiclic hu hn, nu GH th H char G vii) Cho GK < . Khi K l nhm con cc i ca G khi v ch khi G/K c cp nguyn t 12. nh ngha Phn t Gx c gi l phn t khng sinh nu n c th c loi ra khi tp sinh. Tc l nu ,G x Y= th G Y= . 13. Tnh cht (i) Vi mi nhm G, mhm con Frattini ( )G l tp tt c cc phn t khng sinh. (ii) Cho G l mt nhm. Khi , ( )G l nhm con c trng ca G. (iii) Cho G l nhm. Khi ( )G l nhm con chun tc ca G iv) Nu H l mt nhm con ca G sao cho ( )HGG = th H = G 14 . Nhm ly linh Mt nhm G c gi l nhm ly linh nu tn ti s t nhin c sao cho ( )1 1c G + = S c nh nht nh vy gi l nhm lp ca nhm ly linh G. Nh vy c th G l nhm ly linh nu tn ti cN sao cho c G = Nhn xt. (i) i vi nhm ly linh, chui cc tm gim v chui cc tm tng l cc chui chun tc c cng chiu di. (ii) Mt nhm l ly linh lp 1 nu v ch nu n l nhm Abel. Tht vy, 1 ( )G G = nu v ch nu ( )Z G G= nu v ch nu G Abel. Vy mi nhm Abel u l ly linh (iii) Mt nhm ly linh lp 2 c th biu din di dng ( ) ( ) ( ), 1 2 G G Z G G = Mi nhm khng Abel bc 3 p l nhm ly linh lp 2.( xem bi 16 phn bi tp) B. CC PHNG PHP CHNG MINH THNG GP Bi ton. Chng minh nhm G l ly linh Phng php gii. 43. Cch 1. Ta tin hnh cc bc sau: (i) Lp chui tm di ca G ( ) ( ) ...21 = GGG (ii) Tm s t nhin c sao cho 1( ) 1c G + = Cch 2. Ta tin hnh cc bc sau: (i) Lp chui tm di ca G ( ) ( ) ...1 10 = GG (ii) Tm s t nhin c sao cho c G = C. MT S BI TP C LI GII Bi 1. Chng minh rng mi nhm ly linh G u l nhm gii c. Gii. Ta c ( ) ( ), i iG G i () Tht vy, ta s chng minh ( ) bng quy np. Vi ( ) [ ]1 1, ,i G G G= = 1( )G G = , m [ ],G G G nn ( ) ng vi i =1 Gi s ( ) ng vi i = k, tc ( ) ( )k kG G . Ta cn chng minh ( ) ( )1 1 k kG G+ + Ta c ( ) , k G G k m ( ) ( )k kG G ( theo gi thit quy np ) Do ( ) ( ) ( ), , k k kG G G G , suy ra ( ) ( )1 1 k kG G+ + Vy ( ) ng vi mi i Do G l nhm ly linh nn tn ti c N sao cho ( )1 1c G + = Do ( )1 1 c G + Suy ra ( )1 1 c G + = Vy G l nhm gii c. Bi 2. Chng minh rng mi p- nhm hu hn l nhm ly linh Gii. Nu 1G = th G ly linh Nu 1G . Gi s G l p- nhm hu hn Ta c |G| = |Z(G)| + [ G: C(xi) ] , xi Z(G) V |G| = pr , G 1 nn Z(G) 1 Z(G) = ps Do p |G| v p |Z(G)| nn p ( [ G: C(xi) ] ) Do 1G nn ( ) 1GZ Ta c ( )1 1Z G = Nu 1 / 1G th ( )1 / 1Z G Ta c ( )2 1 1 / /Z G = , do nu 1 G th 2 1 , suy ra 1 2 44. Tng t nu 2 G th 3 2 Nu 2 G th 1i i + . Do ta c chui 0 1 11 ..... i i += V G hu hn nn tn ti k sao cho k = G Vy G ly linh. Bi 3. Chng minh rng nu 1G l nhm ly linh th ( ) 1Z G . Gii . Gi s 1G l nhm ly linh lp c, ta c ( )1 1c G + = v ( ) 1c G Theo tnh cht 8i) ta c ( ) ( ) ( )GZGGc = 1 1 Vy ( ) 1Z G Bi 4. Chng minh rng nhm con H ca mt nhm ly linh G l nhm ly linh. Hn na nu G l nhm ly linh lp c th H l nhm ly linh lp c Gii. Ta c H G nn ( ) ( )i iH G vi mi i () Ta s chng minh ( ) bng quy np theo i Vi i = 1 th ( )1 1( ) ;H H G G = = m H G nn ( ) ( )1 1H G Vy ( ) ng vi i = 1 Gi s ( ) ng vi i k= , tc l ta c ( ) ( )k kH G Ta s chng minh ( ) ng vi i = k + 1 Tc chng minh ( ) ( )1 1k kH G + + Ta c ( ) ( )1 ,k kH H H + = [ ]1( ) ( ),k kG G G + = M ( ) ( )k kH G ( gi thit quy np ) H G Do ( ) [ ], ( ),k kH H G G Hay 1 1( ) ( )k kH G + + Vy ( ) ng vi mi i Mt khc, G l nhm ly linh nn tn ti s t nhin c sao cho 1( ) 1c G + = Do 1( ) 1c H + Vy ( )1 1c H + = . Hay H l nhm ly linh lp c Bi 5. a) Chng minh rng nu G l nhm ly linh v :f G H l mt ton cu th H ly linh. b) Chng minh rng nu G l nhm ly linh lp c v H G< th G/H l nhm ly linh lp c Gii. a) Gi s :f G H l mt ton cu. Ta c ( )( ) ( )i iH f G vi mi i ( ) Tht vy ta chng minh ( ) bng quy np Vi i = 1 th 1( )H H = 45. 1( )G G = v ( )1( )f G H = Do 1 1( ) ( )H G Gi s () ng vi i = k, tc l ( )( ) ( )k kH f G Ta chng minh ( ) ng vi 1i k= + Tc chng minh ( )1 1( ) ( )k kH f G + + Ta c [ ]1( ) ( ),k kH H H + = [ ]1( ) ( ),k kG G G + = ( ) ( )1( ) ( ) , ( )k kf G f G f G + = ( )( ) ,kf G H= M ( ) ( )( )k kH f G Do ( )1 1( ) ( )k kH f G + + Vy ( )( ) ( )i iH f G , vi mi i Do G ly linh nn tn ti c N sao cho 1( ) 1c G + = Khi ( ) ( )1 1( ) ( ) 1 1c cH f G f + + = = Do 1( ) 1c H + = Vy H l nhm ly linh lp c b) p dng cu a) vi ton cu : /f G G H , ta c iu phi chng minh. Bi 6. Chng minh rng nu H v K l cc nhm ly linh th tch trc tip H K l mt nhm ly linh Gii. Ta c ( ) ( ) ( )1 1 1H K H K H K = = Gi s ( ) ( ) ( )i i iH K H K Ta c ( ) ( )1 ,i iH K H K H K + = ( ) ( )[ ] ( )[ ] ( )[ ] ( ) ( )KHKKHHKHKH iiiiii 11, ++ == Vy ( ) ( ) ( ) ,i i iH K H K i Do H, K l nhm ly linh nn tn ti a, b Z sao cho ( ) ( ) = = + + 1 1 K H ib ia Ly { }ax ,c m a b= , ta c ( ) ( ) ( )1 1 1 1 1c c cH K H K + + + = . Do ( )1 1c H K + = Vy H K l nhm ly linh Bi 7. Chng minh rng nu G l nhm ly linh th G tha iu kin chun ha. Tc l nu H < G th ( )GH N H Gii. Do G l nhm ly linh nn trong G tn ti chui: ( ) ( ) ( )1 2 1.... 1cG G G G += = M H G nn t