MCA-205: Mathematics II (Discrete Mathematical Structures)

Transcription

1 MCA-05: Mthemts II (Dsrete Mthemtl Strutures) Lesson No: I Wrtten y Pnkj Kumr Lesson: Group theory - I Vette y Prof. Kulp Sngh STRUCTURE.0 OBJECTIVE. INTRODUCTION. SOME DEFINITIONS. GROUP.4 PERMUTATION GROUP.5 SOME RESULTS ON GROUP.6 SUBGROUP.7 COSET.8 KEY WORDS.9 SUMMERY.0 SELF ASSESSMENT QUESTIONS. SUGGESTED READINGS.0 OBJECTIVE: Ojetve of ths hpter s to gn some knowlege out lger struture wth one nry operton.. INTRODUCTION: Let us onser the equton x + =; we see tht the nturl numer n not e soluton of t, whle n nteger s soluton of t. Smlrly rotton of n equlterl trngle out ther xes of symmetry results nto fgure n whh the poston of the ponts hnges whle there s no hnge n shpe of fgure. Therefore, keepng the propertes of ntegers n rotton of some fgures lke lne segment or equlterl trngle out ther MCA-05

2 xes of symmetry, n mn, we ome to know out groups, permuttons groups. In ths hpter, we efne groups, permutton groups, sugroups n osets wth sutle exmples.. SOME DEFINITIONS.. Defnton: For non-empty set A, A A s lle the Crtesn prout of A wth tself. Therefore, A A= (, ), A... Defnton: : A A A, s funton from set A A to A s lle nry operton on A. Exmple () If we two nturl numers then resultnt s gn nturl numer. Hene ton s nry operton on set of nturl numers. () Multplton s nry operton on set of nteger whle vson s not nry operton on set of ntegers... Notton: G, s olleton of set G wth s nry operton on t.. GROUP.. Defnton: For G,, G s lle group f t stsfes the followngs xoms. () s ssotve on G,.e. ( ) = ( ) C ",, G () Ientty exts n G.e., there exst n element e G suh tht e = e =," G, (Here e s lle entty element) () Inverse of every element of G exts n G.e. for G there exst G suh tht = =e. We ll s nverse of. Bese t f = ", G, then G s lle ommuttve group. MCA-05

3 .. Note: In orer to show tht non empty set G s group, we hve to fn n operton whh s nry on G. In other wors we n sy tht G s lose uner tht operton n stsfes ll the three xoms efne ove. Exmple: I (set of ntegers) s group uner +.e. (I, +) s group Soluton: We know tht sum of two ntegers s gn n nteger; therefore, ton s nry operton on I. () Ornry ton s ssotve sne + ( + ) = ( + ) + ",, G () Zero s n nteger suh tht 0+ = + 0 = " I; entty exst n I. () For every I, we hve I suh tht +(-) = (-)+ = 0.e. nverse of every element exst n I. As ll the xoms of group re stsfe y elements of I uner ton, therefore, t s group. Further + = + ", I. Hene I s ommuttve group uner ton... Note: Uner ornry ton zero s lwys entty element n t s lle s tve entty whle s lwys multpltve entty uner ornry multplton. Exmple: Q-{0}, the set of ll non-zero rtonl numers forms group uner multplton. Soluton: Sne multplton of two rtonl numers s lwys rtonl numer, therefore, multplton s nry operton on set Q-{0} () Ornry multplton s ssotve euse () = (),, Q-{0}. () Q- {0}, suh tht. =. = Q-{0}. Here ts s entty element. MCA-05

4 () For Q-{0}, we hve Q-{0} suh tht = =, therefore, nverse of every element exst n Q-{0}. Hene Q-{0} eomes group uner multplton. Exmple: Show tht set G of ll numers of the form +,, I forms group uner the operton ( + ) + ( + ) = ( + ) + ( + ). Soluton: Sne ( + ) n ( + ) re two ntegers, therefore, ( + ) + ( + )= ( + ) + ( + ) G. Hene ove operton s nry operton on G. () Sne ( + ) + (( + ) + ( e + f )) =( + )+( ( + e) + ( + f ) ) = ( + + e) + ( + + f ) () Also (( + ) + ( + )) + ( e + f ) =( ( + ) + ( + ) )+ ( e + f ) = ( + + e) + ( + + f ) () By () n (), we get tht ssotve lw hols n G. () ( + ) + ( ) = ( + 0) + ( + 0) =( + exsts n G n = 0 s entty element. () For ( + ) n G we hve ( ) exst n G suh tht ).e. entty ( + )+ ( ) = ( )+ ( + ) =0, showng tht nverse of every element exst n G. Sne ( + ) + ( + ) = ( + ) + ( + ) = ( + ) + ( + ) MCA-05 4

5 = ( + )+( + nry operton. ), therefore, t s ommuttve group uner ove..4 Theorem: Prove tht n mtrx group uner mtrx multplton, ether ll the mtres re sngulr or non-sngulr. (Sngulr mtrx s mtrx wth zero etermnnt vlue n non-sngulr mtrx hve non-zero etermnnt vlue) Proof: Let M e the mtrx group uner mtrx multplton s nry operton n E e the entty uner multplton then AE =EA =A A M () If E s sngulr mtrx then the equton () s not stsfe y ny nonsngulr mtrx A of M. The reson s tht f A s non-sngulr n E s sngulr then AE s sngulr n so t nnot e equl to nonsngulr mtrx. Therefore f E s sngulr then every mtrx A M must e sngulr. Now suppose E s non sngulr mtrx.let A M n s sngulr then there exsts no mtrx B for whh AB=E (sne AB wll e sngulr whle E s non sngulr. Thus A oes not posses nverse. Ths ontrts the hypothess tht M s group. Therefore, f E s non-sngulr then every mtrx n M must e non-sngulr. Note: By ove theorem we see tht nverse of sngulr mtrx lso exst. The reson s tht f entty element s sngulr mtrx then we n otn the nverse of sngulr mtrx. The followng exmple explns the result. x x Exmple: Show tht the set of ll mtres of the form where x s non x x zero rel numer s group uner mtrx multplton. MCA-05 5

6 MCA-05 6 Soluton: let M = { x x x x / x s non-zero rel numer} lerly etermnnt vlue of ths mtrx x x x x s zero so t s sngulr mtrx. Closure property hols n M euse prout of two sngulr mtres s gn sngulr mtrx. () Sne mtrx multplton s lwys ssotve, ssotve lw hols. () Exstene of entty: Let E = e e e e e the entty element, then AE=EA=A A M, therefore, e e e e x x x x = x x x x ex ex ex ex = x x x x ex =x e =, sne x s not zero. Thus E =. () Exstene of nverse: Let A = x x x x e n rtrry element of M. suppose tht B = y y y y s ts nverse. So AB = E x x x x y y y y =

7 MCA-05 7 xy xy xy xy = xy = y = x 4 Therefore B = x x x x s nverse of A n M. Exmple: Show tht the set of ll rel ( ) mtrx, - 0 s group uner mtrx multplton s nry operton. Soluton: Let G e the set of ll rel ( ) mtrx, - 0. For G to e group t shoul stsfes the followng propertes. () Closure property: Let n re rtrry element of G, therefore, 0 ( ) n 0 ( ). Now = As ) )( ( ) )( ( = ) ( ) ( 0. Hene G. () Assotve property: Let A=, B= n C=, then A(BC) = =

8 MCA-05 8 = ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( = Smlrly (AB)C = =A(BC). () Ientty element: As 0 0 G suh tht 0 0 = 0 0 = G. (4) Exstene of nverse: For G, we hve G suh tht = = 0 0.e. nverse of every element exst n G. Hene G s group. Exmple: Tkng group {e, x, y, z} of orer four, onstrut two fferent tles usng two nry operton uner whh ths set eomes group. We efne two tles s

9 e x y z e e x y z x x e z y y y z e x z z y x e where x = y =z = e, xy =z, (Tle ) e x y z e e x y z x x y z e y y z e x z z e x y where x 4 = e, x s genertor of ths group. (Tle )..5 Defnton: Aton mo m of two ntegers n s wrtten s + m n s the lest non-negtve nteger less then m when sum of n s ve y m. Smlrly we efne m s lest non-negtve nteger less then m when prout of, s ve y m. Ths operton s lle multplton mo m. For exmple =, 5+ 7 = 0 n 5 7=, 5 7=, 5 7=. Exmple: Set of ntegers G = {0,, m -} forms group uner ton mo m. Soluton: - As we know y efnton of vson tht when n nteger s ve y m, ts remner s lwys less thn m, therefore, t s n element of MCA-05 9

10 the set G. It s ler y efnton of ton mo m tht t s nry operton on G. () Assotve property hols n G sne + m (+ m ) = + m (lest non negtve remner when + s ve y m) = Lest non-negtve remner when +(+) s ve y m. = lest non negtve remner when (+)+ s ve y m.(ornry ton s ssotve ) = (Lest non negtve remner when + s ve y m) + m = (+ m )+ m therefore, ton mo s ssotve. () Exstene of entty: 0 G ts s entty elements sne 0+ m =+ m 0= " G () Exstene of nverse: For " G, we fn m- suh tht + m (m-)=(m-)+ m =0. Therefore nverse of s m-. Sne 0 m, therefore m- m n hene nverse of s n G. Hene G s group. Exmple: Show tht G ={ / I + (set of postve ntegers), (, m) = n < m}.e set of ll postve ntegers whh re less then m( ) n re oprme to m, forms group uner multplton mo m. Note: (, m) s notton for gretest ommon vsor of n m. Soluton: Sne y efnton of multplton mo m, lwys less then m, further, G, therefore, (, m)= n (, m)= n hene ( m, m) = n hene m G.e. multplton mo m s nry operton on G. m MCA-05 0

11 () Multplton s ssotve s m ( m ) = m (lest non-negtve remner when s ve y m. = Lest non negtve remner when () s ve y m Sne ornry multplton s ssotve on set of ntegers, therefore ove expresson s = Lest non negtve remner when () s ve y m = (Lest non negtve remner when s ve y m) m = ( ) ) m m () ts s entty element sne m = m = for ll n G. () Let e n rtrry element of G. Sne (, m) =, we n fn two nteger n n G suh tht +m =(ths s ue to Euler s theorem). Then +m- s ongruent to mo m.e. = mo m =. Hene nverse of every element exsts n G. Also s lest non-negtve remner when s ve y m = lest non-negtve remner when s ve y m =. Therefore G s ommuttve group. m m m..6 Note: - We see tht numer of elements n group G, where G s group uner ton mo m, s extly m. But when we tke G s group uner multplton mo m, then numer of elements n G s equl to?(m),? s Euler s funton n s lulte s f m= p α α α n α α n φ( p p... p n )= φ( p )φ( p ) φ( p α n ) α p α... Where φ( p α α )= p ( ), p re stnt prme numers. p αn p n then φ(m)=..7 Defnton: Let G e group uner * s nry operton, then n element g G s lle genertor of G f every element of G s equl to g t for some MCA-05

12 postve nteger t. Here g t = g*g*g* *g extly t tmes. Suh groups re lle yl group. Exmple: Set of ntegers {0,,,,m-} s yl group uner ton moulo m n s the genertor of ths group.4 PERMUTATION GROUP.4. Defnton: A one-one mppng of G onto tself s lle permutton on G. Generlly we tke G fnte set. For exmple f we tke G ={,}, then numer of permuttons two extly two. Let these re σ n σ where σ s efne s ; σ ()=, σ ()= n σ s efne s;σ ()=, σ ()=. Infet f G s set hvng n elements then t hs n! permuttons efne on t..4. Defnton: If f: X Y n g: Y Z re two funtons, then ther prout or omposton funton s efne s fg whh s funton from X to Z n for x X, (x)f g = ((x)f)g. If G hs n elements sy,,, n then permutton f s... k n represente s gves us tht f( )=, f( )=, f( )= k,... k n f( k )= n, f( n )= n f( t )= t for rest of t fferent from,,,,k,n..4. Note: By permutton on n symols mens tht how mny wys we n rrnge n fferent thngs lnerly. If we tke G ={,, } then numer of fferent rrngement re s follow. = I, = ( ), MCA-05

13 = ( ), = ( ), = ( ), = ( ). These re! Permuttons.e. 6 permuttons. More over f we wrte f s (,,,, k, n ) then mge of uner f s, mge of uner f s, s, s k, k s n n mge of n s..4.4 DEFINITION: A permutton s lle yl permutton f mge of frst element s seon, mge of seon s thr n so on n mge of lst element s frst element. Numer of elements n the yle s lle length of yle. For exmple () s yle of length three..4.5 Remrk () A yl permutton remns sme f we gve yl shft to t elements there fore ()=()=() ut () (). () Every permutton on n symols n e wrtten s prout of ts yl permutton. () A yle of length two s lle trnsposton. (4) A permutton s lle even permutton f t prout of even numer of trnspostons other wse t s lle o permutton. (5) Prout of even permuttons s gn n even permutton. (6) Prout of two o permuttons s gn n even permutton. (7) Prout of n o permutton wth n even permutton s n o permutton. (8) A yle of o length s lwys n even permutton whle tht off even length s n o permutton. MCA-05

14 .4.6 Note: If we hve two permuttons sy () n () then ther prout s efne s () ()= I. Infet frst permutton tkes to whle seon tke to, therefore ther omposton tkes to. Smlrly goes to n goes to..e. resultnt permutton s =I..4.7 Theorem: Prove tht every permutton on n symols n e wrtten s prout of ts yl permuttons. Proof: Let,,, n re n symols on whh permutton s tken.now hoose frst element sy tke ts mge sy,then fn mge of ontnue ths proess tll we pprohes to whh s possle euse s lso mge of some element sy.in ths wy we otn yle (,,, ). Now tke t whh s element of gven permutton tht oes not elongs to the yle tken ove. Repet the sme proess for t. Contnung n ths wy we get fferent yles whose prout s gven permutton. Hene theorem s prove. Exmple n support of gven theorem: Tke permutton e. permutton on nne symols. It n lso e wrtten s sne we n leve those elements, whh re left unhnge. Now we strt wth. Imge of s, mge of s n tht of s one so we hve yle ( ). As 5 s not n yle ( ) so we strt wth 5. Clerly mge of 5 s 6, mge of 6 s 7, MCA-05 4

15 Imge of 7 s 8 n tht of 8 s 5,we get nother yle ( ). There fore =( ) ( ) Theorem: Prove tht every yle of S n n e wrtten s prout of trnspostons. Proof: Let ( t ) e yle of length t.now onser the permutton ( )( ) ( t ).we see tht uner ths permutton goes to y frst trnsposton, remns unhnge y rest of trnsposton of ths permutton. So uner ths permutton goes to. Now goes to y frst trnsposton, goes to y seon trnsposton n remns unhnge y rest trnsposton of ths permutton.so goes to.contnung n ths wy we get tht prout of ll trnsposton of ths permutton s ( t ) whh s yle of length t.hene ( t ) = ( )( ) ( t ). Exmple n support of ths theorem: Tke ( 4).e. yle of length four, then we n wrte ( 4) =( )( )( 4)..4.9 Corollrry : Prove tht every permutton on n symols n e wrtten s prout of trnspostons. Proof: Let f e permutton on n symols. We n wrte f s prout of t yl permuttons (theorem.4.7). But we know tht every yle n e wrtten s prout of trnsposton (theorem.4.8). So every permutton n e wrtten s prout of trnsposton. Exmple n support of theorem.4.9 Tke permutton we n wrte t s prout of fferent MCA-05 5

16 yles s =( ) ( ). Now we n wrte ( ) = ( )( ) n ( ).= (5 6) (5 7) (5 8). So = ( )( )(5 6) (5 7) (5 8) Note () A permutton s lle entty permutton f mge of every element s sme uner t.e. I( )= for ll..4. Theorem: Prove tht S n.e. set of ll permuttons on n symols eomes group uner omposton of mppng s operton. Proof: As we know tht omposton of mppng whh re one one n onto s gn one one n onto, therefore omposton of mppng or permutton s nry operton on S n () Composton of permutton s ssotve lso s ( )((fg)h)= (( )(fg))h)= (( f)g)h=( )f)(gh)= ( )(f(gh).e. (fg)h= f(gh) () Sne entty permutton ts s entty element, entty exsts n S n We lso know tht f mppng s one one n onto then ts nverse mppng f - exsts n s efne s f - ( )= j ff f( j )=.sne every permutton s oneone n onto,there fore nverse of every element of S n s n S n.hene S n s group.it s lle group of permuttons..4. Theorem: Prove tht set of ll even permuttons s group n ths group s lle lterntng group of egree n. ts orer s n!/ (t s generlly enote y A n ). MCA-05 6

17 Proof: Sne y remrk.4.5 (5), we get tht prout of two even permuttons s n even permutton, therefore prout of permutton s nry operton on A n. () Ths prout s ssotve sne A n S n n elements of S n stsfes ssotve lw () As I s n even permutton so I A n ts s entty element. () Let f A n e n even permutton n g e ts nverse permutton then fg =I. Sne prout s even permutton, so y remrk.4.5 (7) g nnot e o.so g s even, therefore s n element of A n. Hene proof s over. Exmple n support of ths theorem: Tke S ={I, ( ), ( ), ( ), ( ), ( )} then usng remrk.4.5 (8) we get tht A ={I, ( ), ( )} s group s entty exsts n A, nverse of ( ) s ( )..5 SOME RESULTS ON GROUP..5. Theorem: If G s group, then t hs unque entty. Proof: Let f possle e n f e two entty elements of G, then e.f = e (tkng f s entty element ) () n e.f = f (tkng e s entty element ) () From () n () we get e = f.e. entty s unque..5. Theorem: In group nverse of eh element s unque. Proof: Let G hs two nverse n sy, then y efnton of nverse.=.=e ().=.=e () Sne.(.)=(.). (y ssotve lw) () MCA-05 7

18 Also.(.)=.e= (4) n (.).=e.= (5) By (),(4),(5) we get =..e. Inverse of n element s unque n G..5. Theorem: For group G, prove tht for every n G, ( ) = when operton s multpltve n -(-) = when operton s tve. Proof: Let - e nverse of n ( ) e nverse of - then y efnton of nverse -. =. - =e () n -. ( ) = ( ). - =e () Now usng (), () n theorem.5. we get ( ) =. Smlrly we n show seon result..5.4 Theorem: Prove tht for ll n n group G we hve () - = - - n uner ton t n e wrtten s (+) =(-) + (-). Proof: Let () - e nverse of, then y efnton of nverse (). () - = () -.() = e () n (). - - = ( - ) - =.e. - =e () - - ()= - ( - ). = - e. = e () By () n (), - - s nverse of. Now usng (), (), () n theorem.5. we get () - = - -. Smlrly we n show seon result..6 SUBGROUP.6. Defnton: Let H e suset of group G. If H lso eomes group uner the sme nry operton s n G, then H s lle s sugroup of G. MCA-05 8

19 Exmple: Tke I, the set of even ntegers, then t s group uner ornry ton. Sne I I, the set of ntegers whh s lso group uner ornry ton, therefore, I s sugroup of I. Exmple: Tke G={,,,, 4, 5 }suh tht 6 =, then G eomes group multplton. Now tke H={, }; 6 =. Clerly H s lso group uner multplton n s sugroup of G..6. Theorem: A suset H of group G wll e sugroup of G f n only f - H, H (Equvlently, for suset H of G, HH - = H f n only f H s sugroup of G). Proof: Let H e sugroup of group G, therefore, for, H, - H n y losure property - H. Conversely let us suppose tht H s suset of G suh tht - H, H. Sne H s suset of G, therefore, elements of H stsfes ssotve property. Also for H,, H whh gves us tht - =e H.e. entty exst n G. Agn, e H - e= - H H.e nverse of every element exst n G. Fnlly H - H, therefore, ( - ) -. = H, H.e. losure property s stsfe y the element of H. Sne H stsfes ll the xoms of group, t eomes sugroup of G..6. Theorem: Let G s fnte group, then suset H of G wll e sugroup of G f n only f t s lose uner nry operton efne on G.e. H, H (Equvlently, for fnte group G, suset H of G wll e sugroup of G f n only f HH = H). MCA-05 9

20 Proof: Let H e sugroup, then for, H, H (y efnton of group).conversely, let us suppose tht H, H, we ssert tht H s sugroup of G. Sne H s suset of G, elements of H stsfy ssotve lw. For H, y ssumpton, = H. Smlrly k H. But H s fnte eng suset of fnte set, therefore, there exst postve ntegers s n t suh tht s = t s-t =e H, showng tht entty exst n H. Also we n wrte s-t = s-t- =e s-t- s nverse of exst n H. Smlrly we n show tht nverse of every element of H exst n H. Hene H s sugroup of G. Exmple: Let G e the group of ll rel ( ) mtrx, - 0 wth mtrx multplton s nry operton. Show tht the set H ={ G / 0} s sugroup of G. 0 Soluton: Let h n h e rtrry elements of H, then h = 0 n h = 0. Now (h ) - = 0 = 0. MCA-05 0

21 n h (h ) - = 0 0 = 0 + H euse =/ 0. Exmple: Show tht the set K = s sugroup of the group H 0 ={ / 0,,, re ll rel numers}. H s group uner mtrx 0 multplton s nry operton. Soluton: let k n k re two elements of K. we wll show tht k.k K. let k = 0 n k = 0, then (k ) - = 0 n k (k ) - = 0 Hene K s sugroup H. = whh s n element of K..6.4 Note: The result n Theorem.6. s true when G s fnte. For exmple f we tke G n nfnte set sy N (set of nturl numers) then G s lose uner multplton ut t s not group..6.5 Theorem: Prove tht nterseton of two sugroups of group G s gn sugroup of G. MCA-05

22 Proof: Let H n K re two sugroups of group G. If H K = {e} then we hve nothng to prove, so suppose tht h, k H K. We wll show tht hk - elongs to H K. As h H K h H n h K. Smlrly k H K k H n k K. Now h H n k H hk - H (H s sugroup of G) Smlrly h K n k K hk - K (K s sugroup of G) Ang oth results we get tht hk - H K..6.6 Theorem: Prove tht rtrry nterseton of sugroups of group G s sugroup of G. Proof: Let G e group n let {H t : t T} e ny fmly of sugroups of G. here T s n nex set n s suh tht for ll t T, H t s sugroup of G. Let H = I ={x G : x H t t T. H t t T Now let n e ny two elements of H. then I H H t T n t T t I H H t T. t T t t t But for ll t T, H t s sugroup of G. therefore H t, H t - H t t T. Consequently I H t T t t T. There fore I H t t T s sugroup..6.7 Theorem: Prove tht unon of two sugroups of group G s gn sugroup of G f n only f one s ontne n the other. MCA-05

23 Proof: Suppose H n K re two sugroups of group G. Let us suppose tht H K or K H then H K s equl to H or K. But H n K oth re sugroup of G, therefore H K= H or K s lso sugroup. Conversely, suppose tht H K s sugroup of G. Let us ssume tht H s not suset of K n K s not suset of H. Now H s not suset of K h H suh tht h K () Also K s not suset of H k K suh tht k H () But y (), h H K n y (), k H K so hk H K. Let hk = t, therefore t H or K. If t elongs to H then k = h - t elongs to H, ontrton. Hene t oes not elong to H. Smlrly we n show tht t oes not elong to K. It shows tht t oes not elongs to H K, ontrton n hene ontrton to our ssumpton tht H s not suset of K n K s not suset of H. Hene ether H s suset of K or K s suset of H..7 COSET.7. Defnton: For group G n sugroup H of t we efne *H ={*h/ h H}. Ths set s lle oset of H n G generte y. *H s lle left oset of H n G. Smlrly H* s lle rght oset of H n G.* s nry operton on G. Exmple Tke G=(I,+) n H ={I,+} then 0+Hn +H re two fferent left osets gven s 0+H = {0+h/h H} = H whh s set of even ntegers n +H = {+h/h H} whh s set of o ntegers. We lso note tht I s unon of H n +H..7. Theorem: If H, then H = H n H, then H = H Proof: Sne H = h for some h H, But then h = h h = h H. MCA-05

24 As h s rtrry element of H, therefore H H. Now =h = (h ) - h = (h ) - h H, Therefore H H. Hene the result s over. Smlrly we n prove seon prt..7. Theorem: For sugroup H of G, f H n H re two left osets of H n G, then ether H = H or H H = φ. Proof: If H H = φ then we hve nothng to prove. So suppose tht H H. Ths mples tht H n H. But H H = H n H H = H (y theorem.7.). Therefore H = H..7.4 Theorem: If n re ny two element s of group G n H ny sugroup of G, then H = H f n only f - H n H = H f n only f - H. Proof: Sne s n element of H then H = H H - H - = H. Conversely let us suppose tht - H - H H, ut then y theorem.7. H = H. Smlrly we n prove other prt. MCA-05 4

25 .7.5 Theorem :If we efne tht mo H f n only f - H. then ths s n equvlene relton n lss of mo H s rght oset of H n G generte y. Proof: We know relton s lle equvlene relton f t s reflexve, symmetr n trnstve. Sne - = e H.e. relton s reflexve. Also mo H tht - H whh further gves us tht ( - ) - = - H mo H.e. relton s symmetr. Further mo H n mo H Tht - H n - H - - H - H.e. mo H. Hene ths relton s n equvlene relton. Now lss of s enote s []={ H suh tht mo H }. Let [] then mo H whh mples tht - H.e. - H H there fore [] H. Conversely let suppose tht h H s n rtrry element of H. But h H h - h.e. h mo H. Hene h [] gvng us tht H [].there fore [] =H. Hene the result s prove..7.6 Note: We know tht for every g G we hve oset gh suh tht g gh. Hene we n wrte G = U gh = U Hg g G g g Exmple: Fn ll left osets of V 4 ={I, ( )( 4),( )( 4), ( 4)( )} n S 4 Soluton: Frst we wll see tht V 4 s group. Beng suset of S 4 ssotve lw s followe y elements of V 4. Also nverse of ( )( 4) s ( )( 4) s ( ) ( 4). ( )( 4) =I. Smlrly eh element of V 4 s nverse of tself. Now V 4 s frst left oset whh s {I, ( )( 4), ( )( 4), ( 4)( )} MCA-05 5

26 For seon left oset we tke ( ) V 4 = {( ) I, ( )( )( 4), ( ) ( )( 4), ( )( 4)( )} = {( ), ( 4 ), ( 4 ), ( 4)} For thr left oset we tke ( 4) V 4 = {( 4) I, ( 4)( )( 4), ( 4) ( )( 4), ( 4)( 4)( )} = {( 4), ( 4), ( 4 ) ( )}. For forth-left oset we tke ( 4) V 4 = {( 4) I, ( 4)( )( 4), ( 4) ( )( 4), ( 4)( 4)( )} = {( 4), ( 4), ( 4 ), ( )} For ffth left oset we tke ( 4 ) V 4 = ( 4 ) I, ( 4 )( )( 4), ( 4 ) ( )( 4), ( 4 )( 4)( )} ={( 4 ), ( ), ( 4), ( 4 )}. For sxth left oset we tke ( 4) V 4 = {( 4) I, ( 4)( )( 4), ( 4) ( )( 4), ( 4)( 4)( )} = {( 4), ( 4 ), ( 4), ( )}. Here we see tht unon of ll left osets s {I, ( ), ( ), ( 4), ( ), ( 4), ( 4), ( ), ( 4), ( 4), ( 4), ( ), ( 4 ), ( 4 ), ( 4 ), ( 4), ( 4 ), ( 4), ( 4 ), ( 4 ), ( 4 ), ( )( 4), ( )( 4), ( 4)( )} whh s S 4..e. S 4 = G = U gh. g G Now V 4 s frst rght oset whh s {I, ( )( 4), ( )( 4), ( 4)( )} MCA-05 6

27 For seon rght oset we tke V 4 ( ) = {I ( ), ( )( 4) ( ), ( )( 4) ( ), ( 4)( ) ( )} = {( ), ( 4), ( 4 ), ( 4 )} For thr rght oset we tke V 4 ( 4) = {I ( 4), ( )( 4) ( 4), ( )( 4) ( 4), ( 4)( ) ( 4) } = {( 4), ( 4 ), ( ),( 4)}. For forth rght oset we tke V 4 ( 4) = { I ( 4), ( )( 4) ( 4), ( )( 4) ( 4), ( 4)( ) ( 4) } = {( 4), ( ), ( 4 ), ( 4)}. For ffth rght oset we tke V 4 ( 4 ) = {I ( 4 ), ( )( 4) ( 4 ), ( )( 4) ( 4 ), ( 4)( ) ( 4 ) } ={( 4 ), ( 4), ( ), ( 4 )}. For sxth rght oset we tke V 4 ( 4) = {I ( 4), ( )( 4) ( 4), ( )( 4) ( 4), ( 4)( ) ( 4) } = {( 4), ( 4 ), ( ), ( 4) }. Here we see tht unon of ll rght osets s {I, ( ), ( ), ( 4), ( ), ( 4), ( 4), ( ), ( 4), ( 4), ( 4), ( ), ( 4 ), ( 4 ), ( 4 ), ( 4), ( 4 ), ( 4), ( 4 ), ( 4 ), ( 4 ), ( )( 4),( )( 4), ( 4)( )} whh s S 4..e. S 4 = U Hg g g.7.7 Note: Aove result hs very fne ppltons, whh re MCA-05 7

28 () Orer of every sugroup of group ves orer of group. (Orer of group mens the numer of elements t hs. It s lso lle Lgrnge s theorem) () Orer of every element of G ves orer of group G. (Orer of element g G mens smllest postve nteger t suh tht g t =e, entty of G.) () Numer of stnt left or rght osets of H n G s lle nex of H n G.It s gven y O(G). O(H).8 KEY WORDS Bnry operton, groups sugroups, permutton, entty, nverse..9 SUMMERY: Ths hpter ontns efnton of groups, sugroups, permutton group, osets n some theorem on groups.0 SELF ASSESSMENT QUESTIONS () Proves tht set of ll ( ) non-sngulr mtres s group uner multplton of mtres. () Let G ={ 0,,,, 4, 5, 6 } where. j = +j f +j< 7. j = +j-7 f +j 7.( for exmple 4. 5 = =.) () Fn ll the osets of H={, } n G ={,,,, 4, 5 }suh tht 6 =. (4) Prove tht group n sugroup hve sme entty. MCA-05 8

29 (5) Prove tht unon of sugroup of G my or my not e sugroup of G. (6) Prove or sprove whether the followng s group or not B A Where = =, =. Also fn out =,,,(). (7) Prove tht those elements of group G whh ommute wth the squre of gven element of G forms sugroup H of G n those whh ommute wth t self form sugroup of H. (8) () Cn n eln group hve non-eln sugroup. () Cn non eln group hve n eln sugroup. () Cn non eln group hve non-eln sugroup. Gve n exmple n support of your nswer. (9) Prove tht followng tle on relton of elements of set G = {0 4 5} multplton mo 6 s not group MCA-05 9

30 (0) Let R e group uner multplton n Q e suset of R n s group uner ton. Cn t e sugroup of R? () Let G e the group of nteger uner ton.let H n e the set of ll nteger whh re multple of fxe nteger n. Show tht t s sugroup of G.s t norml sugroup. Determne the nex of H n n G n wrte ll the osets of H n n G. lso fn out H n H m. () Prove tht set of ll n th root of unty forms group uner multplton Answer () Yes t s group. () Yes t s group. () It hs three osets gven s H = {, }, H = {, 4 }n H= {, 5 } t s ler tht ther unon s G. (6) Yes t s group n =, =, =,() =. MCA-05 0

31 (0) No t nnot e sugroup euse f t s sugroup of R then Q n R must hve nry operton. () H n H m wll ontn set of ll ntegers, whh re multple of t, where t s lest ommon multple of n n m.. SUGGESTED READINGS () I.N. Hersten, Tops n Alger, Wley estern Lt., New Delh, 975. () Surjeet Sngh n Quz Zmeerun., Moern Alger. MCA-05

32 MCA-05: Mthemts II (Dsrete Mthemtl Strutures) Lesson No: Wrtten y Pnkj Kumr Lesson: Group theory - II Vette y Prof. Kulp Sngh STRUCTURE.0 OBJECTIVE. INTRODUCTION. NORMAL SUBGROUP. SEMI GROUP AND FREE SEMI GROUP.4 APPLICATION OF ALGEBRAIC STRUCTURES GROUP IN MODULAR AIRTHMATRIC.5 SOME DEFINITION AND RESULTS ABOUT CODE WORDS AND CODES.6 COSET LEADER DECODING.7 LANGUAGES.8 FINITE-STATE MACHINES.9 KEYWORDS AND SUMMARY.0 SELF ASSESSMENT QUESTION. SUGGESTED READINGS.0 OBJECTIVE : Ojetve of ths hpter s to efne some more lger struture n to fn ther pplton n ommunton n omputer.. INTRODUCTION : In Chpter, we see tht there re susets of gven group eomes group n tself uner the sme operton s n G. We ll these susets s su-groups of G. In ths hpter we efne some other onton on sugroup of G. We hve efne some more lger struture. MCA-05

33 At lst we hve shown the ppltons of these lger strutures n moulr rthmet, ong theory n fnte stte mhne.. N0RMAL SUBGROUP.. Defnton: A sugroup H of group G s lle norml sugroup of G. If gh = Hg g G... Theorem : If G s ommuttve group, then prove tht every sugroup of G s norml n G. Proof: Let H e sugroup of G. Now for some C gh, C = gh, h H. Sne h n g oth re elements of G n G s ommuttve, therefore, gh=hg. As hg s element of Hg, therefore C Hg. But C s rtrry element of gh, therefore gh Hg. Smlrly we n show tht Hg gh. Hene gh = Hg g G. It proves the result... Theorem: A sugroup H of group G wll e norml n G f n only f g hg H for ll g G n h H. Proof: Let H e norml sugroup of G, then y efnton of norml sugroup gh = Hg g G. On multplyng oth ses y g -, we get tht g - gh = g - Hg g G.e g - Hg = H g G whh mples tht g - hg H h H n g G. Conversely let us suppose tht g - hg H g G n h H. Now g - hg H tht mples gg - hg gh hg gh. As hg s generl element of Hg, hene Hg gh. MCA-05

34 Moreover numer of elements n Hg s equl to numer of elements n gh. Hene gh = Hg,.e. H s norml n G. Hene the result s prove...4 Remrk: Numer of elements n Hg s equl to numer of elements n gh n t n e shown y efnng mppng from gh to Hg s f: gh Hg y f(gh) = h'g for some h' H...5 Theorem: Let G e group. If nex of sugroup H n G s two, then H s norml sugroup of G. Proof: As we know tht H s oset of H tself gven y e G. If gh s nother oset [Q nex of H n G s two] then G = H gh () [Q G s unon of ll stnt osets of H n G]. Smlrly we n wrte G = H Hg () By () n () we get tht Hg = gh g G..e. H s norml n G...6 Theorem: A sugroup N of group G s norml n G f n only f the prout of two-left oset of N n G s gn left oset n G. (or prout of two rght osets of N n G s gn rght oset). Proof: frst we suppose tht N s norml sugroup of G. let N n N re ts two left osets of N n G gven y n then ( N).( N) = (.(N )N) (By ssotve property of group G). = ( N) N (Sne N s norml therefore N = N ). = N = N for = whh s gn left oset. Conversely suppose tht g N. gn = g N (g ) - g N. gn (g ) - g N N g N g N N where g =(g ) - g g G. MCA-05 4

35 Ng N (g ) - N Ng (g ) - NN - =(g ) - N g G () (For sugroup N, NN - =N ) But g g N g (g ) - N g N = (g ) - N ( y theorem.7. ) () Now usng () n () we get Ng g N As orer of Ng n g N s sme, therefore, Ng = gn g G Hene N s norml sugroup of G...7 Theorem: Let N n M e the norml sugroups of group G suh tht N M = (e). Prove tht for ny n N, m M we hve nm =mn. Proof: Let us onser n element n - m - nm for n N n m M. Beuse N s norml n G, m - nm elongs to N. Sne n - lso elongs to N, we hve n - m - nm N. Smlrly, s M s lso norml, n - m - n M n hene n - m - nm elongs to M. Therefore, n - m - nm N M. But y gven onton N M = (e), therefore, n - m - nm = e mn n - m - nm = mn nm = mn. Hene the theorem s prove. MCA-05 5

36 ..8 Theorem: If we efne NM {nm / n N, m M}. Prove tht f N s norml sugroup of G then NM s lso sugroup of G. If oth N n M re norml sugroups of G then NM s lso norml sugroup of G. Proof: Let nm n rt e two elements of NM. Then (rt) - = t - r - s n G.we wll show tht nm (rt) - NM. Sne nm (rt) - = nm t - r - = n(m t - )r - (m t - ) - m t -. Now r - elongs to N, therefore, (m t - )r - (m t - ) - lso elongs to N (euse N s norml) n hene n(m t - )r - (m t - ) - elongs to N. Further mt - elongs to M mples tht n(m t - )r - (m t - ) - m t - = nm (rt) - elongs to NM. Hene NM s sugroup of G. Now we wll show tht g - nm g NM. As g - nm g =g - ngg - m g elongs to NM { euse g - ng N (N s norml n G) n g - m g M (M s norml n G)}. Hene NM s norml sugroup of G. Exmple: Show tht V 4 s norml n A 4 where V 4 s {I, ( )( 4),( )( 4), ( 4)( )} n A 4 s {I,, ( ), ( 4), ( 4), ( 4), ( ), ( 4 ), ( 4 ), ( 4 ), ( )( 4), ( )( 4), ( 4)( )}. MCA-05 6

37 Soluton: we wll show tht every left oset s equl to rght oset. Now V 4 s frst left oset whh s {I, ( )( 4), ( )( 4), ( 4)( )} n orresponng rght oset s {I, ( )( 4), ( )( 4), ( 4)( )}. More over ( )( 4) V 4 = ( )( 4) V 4 = ( 4)( ) V 4 = V 4 = V 4 ( )( 4) = V 4 ( )( 4) = V 4 ( 4)( ). (y use of theorem.7.) Now we lulte left oset of V 4 generte y ( ) ( ) V 4 = {( ) I, ( )( )( 4), ( ) ( )( 4), ( )( 4)( )} = {( ), ( 4 ), ( 4 ), ( 4)}. An orresponng rght oset s V 4 ( ) = {I ( ), ( )( 4) ( ), ( )( 4) ( ), ( 4)( ) ( )} = {( ), ( 4), ( 4 ), ( 4 )}. We get ( ) V 4 = V 4 ( ). ( 4 )V 4 = ( 4 ) V 4 =( 4) V 4 = ( ) V 4 = V 4 ( ). =V 4 ( 4) = V 4 ( 4 ) = V 4 ( 4 ). y use of theorem.7. For thr left oset we get ( 4) V 4 = {( 4) I, ( 4)( )( 4), ( 4) ( )( 4), ( 4)( 4)( )} = {( 4), ( 4), ( 4 ) ( )}. For thr rght oset we get V 4 ( 4) = {I ( 4), ( )( 4) ( 4), ( )( 4) ( 4), ( 4)( ) ( 4)} = {( 4), ( 4 ), ( ),( 4)}. We get tht ( 4) V 4 = V 4 ( 4). From ove we get tht eh left oset of V 4 n A 4 s equl to rght oset of V 4 n A 4. Hene V 4 s norml n A 4. MCA-05 7

38 Exmple: Show y n exmple tht we n fn three sugroups E, F n G of group suh tht E F G. E s norml n F n F s norml G ut E s not norml n G. Soluton: Let us tke S 4 (symmetr group of egree 4 over the set{,,, 4}). Tke E = {I, ( )}, F = V 4 n G =A 4. Frst we se tht E s suset of F. More over E s lso group (See prolem ) uner the sme operton s F s. Therefore, E s sugroup of F. Sne nex of E n F s two, therefore y Theorem..5, E s norml n F.By prevous exmple F s norml n G. We wll show tht E s not Norml n G. for t we wll show tht there exst n element of G for whh left oset of E s not equl to rght oset of E. Clulte left oset of E gven y ( ) ( )E = { ( )I, ( )( ) } = {( ), ( )}.() An rght oset of E gven y ( ) E ( ) = { I ( ), ( )( ) } = { ( ), ( ) } ().we see tht set gven n () s not equl to set gven n () so ( )E E ( ),therefore, E s not norml n G...7 Remrk: For fnte group G, (g) 0(G) = e g G. Ths result s ue to Lgrnge s theorem. MCA-05 8

39 . SEMI GROUP AND FREE SEMI GROUP.. Defnton: A set G s lle sem group f there s nry operton on G whh s ssotve on G..e. f t s nry operton then. (. ) = (. ).,, G. Exmple: Set of nturl numer s sem group uner ornry ton. Sne sum of two nturl numers s nturl numer n + ( + ) = ( + ) +,, N... Remrk: Every group s sem group ut onverse my not e true. Aove exmple shows ths se...4 Defnton: For gven set G = {,,, n } we efne,,,, or 4 s the sequene of elements of G. If G* s set of ll fnte sequenes n for α, β G* where α = k, β = then α.β = k. Uner ths relton whh s nry operton on G* eomes sem group s lle free semgroup generte y set G..4 APPLICATION OF ALGEBRAIC STRUCTURE IN MODULAR ARITHMATICS..4. In moulr rthmet: As we know tht set of ll postve ntegers less thn m n oprme to m forms group uner x multplton mo m, whh s enote y Z m. Now the numer of postve ntegers less thn m n o prme to m x x 0 re extly φ (m). Sne for Z m we hve ( Z ) m mo m. In ft we hve φ(m) = mo m for ll postve ntegers. For exmple tke m =, then Z = {, 5, 7, }, t s group MCA-05 9 x

40 uner multplton mo. As multplton mo s ssotve, ts entty element n nverse of s, 5 s 5, 7 s 7 n s tself mo. Now y remrk..7, φ() = mo, x Z But φ() = φ(4) φ() = φ( ) φ() = = 4. Hene 4 = mo, s 4 = mo, 5 4 = mo, 7 4 = mo n 4 = mo. In ft 5 7 mo. x.4. Defnton: Let Z, then orer of mo m s smllest m postve nteger t suh tht t mo m n t s enote y 0()..4. Note :In ove exmple 0 () =, 0(5) =, 0(7) =, 0() =. It s lso ler tht orer of every element lso ves orer of group x m Z x m x Z y orer of Z. m. So we hve n e out orer of.4.4 Exmple: Fn remner when 00 s ve y 7. Soluton: Tke group of nteger mo 7 we see tht {,,, 4, 5, 6} re the postve nteger less thn 7 whh forms group uner multplton mo 7. So φ(7) mo 7.e. 6 gves us remner when ve y 7. [Here 0 () = 6]. We wrte t s 6 mo 7 =. 6 = (7k + ) [Q f n nteger gves us remner when ve y 7 t s of the form 7k + ] MCA-05 40

41 Now 00 = (6 + ) = ( 6 ).9 = (7k + ). 9 Hene 00 mo 7 = 9 [Q (7k + ) gves us remner when ve y 7 usng the result tht f mo m then t mo m] Now 9 mo 7 = Hene 00 mo 7 =..5 SOME DEFINITION AND RESULTS ABOUT CODE WORDS AND CODES. Coe s me of oe wors. Weght of oe wor s numer of non-zero entres n tht oe wor. For exmple 000 s oe wor of weght three.. When we tlk out lner oe of length n, t men oe wors re n tuples n set of ll oe wors forms sugroup of group of ll n tuples [Here 5 tuple men sequene ontns extly 5 elements.e. 000 s 5 tuple]. If we tke n tuple over the set {0, } then B n s tht set whh ontns ll n tuple n s lle set of nry n tuple. For exmple elements of B re {(0,0,0), (0,0,), (0,,0), (,0,0), (,,0), (,0,), (0,,), (,,)}.e. eght elements. The set B n forms group uner omponent wse ton mo.e. (,, ) + (', ', ) = ( +, +, + ) (,, 0) + (0,, 0) = (+0, +, 0+0) = (,0,0). Now we tke suset of B = {(000), (0), (0), (0)} t s sugroup of group B uner omponent-wse ton mo. then ths s oe of tuples. Dstne etween two oe wors s the numers of poston t whh they ffer wth eh other for exmple stne etween (0) n (0) s. MCA-05 4

42 4. In ths se we tke reeve wor n tke ts stne wth oe wors. We hoose tht oe wor whh hs mnmum stne wth reeve wor n ssume tht the ove oe wor ws sent..6 COSET LEADER DECODING Now we efne oset leer eong. As we know tht f H s sugroup of B n, then B n n e wrtten s unon of stnt osets of H n B n. Sne reeve wor s n n tuple les n B n n hene les n oset lso we hoose the wor of mnmum weght n oset s oset leer n we sutrt tht wor from reeve wor n otn the requre oe wor whh ws sent. So we use osets to otn orret oe wor. For exmple n B we see tht H = {(000), (0), (0), (0)} s sugroup. Then we get followng osets. (0,0,0) + H={(0,0,0), (,0,), (,,0), (0,,)} (,0,0) + H={(,0,0), (0,0,), (0,,0), (,,)} re two osets. Here (0,0,0) s oset leer for frst oset. An (,0,0) s oset leer for seon oset. Now f we reeve () s wor we see tht t les n seon oset wth oset leer (,00), hene we eoe t to () (,00) = (0) s oe wor whh ws sent..7 LANGUAGES We onsere the set S* onsstng of ll fnte strngs of elements from the set S. There re mny possle nterprettons of the elements of S*, epenng on the nture of S. If we thnk of S s set of wors, then S* my e regre s the olleton of ll possle sentenes forme from wors n S. Of ourse, suh sentenes o not hve to e menngful or even sensly MCA-05 4

43 onstrute. We my thnk of lnguge s omplete spefton, t lest n prnple, of three thngs. Frst, there must e set S onsstng of ll wors tht re to e regre s eng prt of the lnguge. Seon, suset of S* must e esgnte s the set of properly onstrute sentenes n the lnguge. The menng of ths term wll epen very muh on the lnguge eng onstrute. So f S s ny set, then suset of free sem group generte y S s lle lnguges on S..8 FINITE-STATE MACHINES We thnk of mhne s system tht n ept nput, possly proue out put, n hve some sort of nternl memory tht n keep trk of ertn nformton out prevous nputs. The omplete nternl onton of the mhne n ll of ts memory, t ny prtulr tme, s s to onsttute the stte of the mhne t tht tme. The stte n whh mhne fns tself t ny nstnt summrzes ts memory of pst nputs n etermnes how t wll ret to susequent nput. When more nput rrve the gven stte of the mhne etermne (wth the nput) the next stte to e oupe, n ny output tht my e proue. If the numer of sttes s fnte, the mhne s fnte-stte mhne. Suppose tht we hve fnte set S = {s 0, s,, s n }, fnte set I, n for eh x, funton f x : S S. Let F = f x x I. the trple (S, I, F) s lle fnte-stte mhne, S s lle the stte set of the mhne, n the elements of S re lle sttes. The set I s lle the nput set of the mhne. For ny nput x I, the funton f x esres the effet tht ths nput hs on the sttes of the mhne n s lle stte trnston funton. Thus, f the mhne s n stte s n nput x ours, the next stte of the mhne wll e f x (s). Sne the next stte f x (s) s unquely etermne y the pr (s, x), there s funton F: S I S gven y MCA-05 4

44 F (s, x) = f x (s). The nvul funtons f x n ll e reovere from knowlege of F. Mny uthors wll use funton F : S I S, nste of set f x x I}, to efne fnte-stte mhne. The efntons re ompletely equvlent. Exmple. Let S = {s 0, s } n I = {0, }. Defne f 0 n f s follows: F 0 (s 0 ) = s 0, f (s 0 ) = s, F 0 (s ) = s, f (s ) = s 0. Ths fnte-stte mhne hs two sttes s 0 n s n epts two possle nputs, 0 n. The nput 0 leves eh stte fxe, n the nput reverses sttes. We n thnk of ths mhne s moel for rut (or logl) eve n vsulze suh eve s n Fg... The output sgnls wll, t ny gven tme, onsst of two voltges, one hgher thn the other. Ether lne wll e t the hgher voltge n lne t the lower, or the reverse. The frst set of output ontons wll e enote s 0 n the seon wll e enote s. An nput pulse, represente y the symol, wll reverse output voltges. The symol 0 represents the sene of n nput pulse n so results n no hnge of output. Ths eve s often lle T flp-flop n s onrete relzton of the mhne n ths exmple. We summrze ths mhne n Fg... The tle shown there lsts the sttes own the se n nputs ross the top. The olumn uner eh nput gves the vlues of the funton orresponng to tht nput t eh stte shown on the left. The rrngement llustrte n Fg.. for summrzng the effet of nputs on sttes s lle the stte trnston tle of the fnte-stte mhne. It n e use wth ny mhne of resonle sze n n onvenent metho of spefyng the mhne. MCA-05 44

45 Fg.. Fg.. Exmple : Conser the stte trnston tle shown n Fg... Here n re the possle nputs, n there re three sttes, s 0, s n s. The tle shows us tht f (s 0 ) = s 0, f (s ) = s, f (s ) = s n f (s 0 ) = s 0, f (s ) = s, f (s ) = s B s 0 s 0 s 0 s s s s s s fg.. If M s fnte-stte mhne wth sttes S, nputs I, n stte trnston funtons {fx x I}, we n etermne relton R M on S n nturl wy. If s, s j S, we sy tht s R M s j f there s n nput x so tht f x (s) = s j. Thus s R M s j mens tht f the mhne s n stte s, there s some nput x I tht, f reeve next, wll put the mhne n stte s j. The relton R M permts us to esre the mhne M s lelle grph of the relton R M on S, where eh ege s lelle y the set of ll nputs tht use the mhne to hnge sttes s nte y tht ege. We see tht output n fnte stte mhne s n element of permutton group of nput symols. MCA-05 45

46 .9 KEYWORDS AND SUMMARY In ths hpter we hve shown the pplton of Alger struture. Keywors re, semgroup oes n fnte stte mhne..0 SELF ASSESSMENT QUESTION. Fn the remner when 00 s ve y.. Prove tht group s lwys semgroup n onverse my not e true.. Defne oset leer eong on B 4 tkng ts sugroup. 4. Defne outputs of fnte stte mhne. 5 Show tht E ={I, ( ) s group.. SUGGESTED READINGS () I.N. Hersten, Tops n Alger, Wley estern Lt., New Delh, 975. () Surjeet Sngh n Quz Zmeerun., Moern Alger. () Seymour Lepshutz, Fnte Mthemts (Interntonl eton 98), MGrw-Hll Book Compny, New York. MCA-05 46

47 MCA-05: Mthemts II (Dsrete Mthemtl Strutures) Lesson No: Wrtten y Pnkj Kumr Lesson: Grph theory - I Vette y Prof. Kulp Sngh STRUCTURE.0 OBJECTIVE. INTRODUCTION. GRAPH. PATHS AND CIRCUITS.4 SOME DEFINITIONS WITH EXAMPLES.5 CONNECTED AND DISCONNECTED GRAPHS..6 MATRIX REPRESENTATION OF GRAPHS.7 KEYWORDS AND SUMMARY.8 SELF ASSESSMENT QUESTIONS.9 SUGGESTED READINGS.0 OBJECTIVE: Ojetve of ths hpter s to gn some knowlege out grphs, whh hs we pplton n omputer net workng, ruts et.. INTRODUCTION: In ths hpter we hve efne grph whh s ptorl representton of reltons on sets. We hve efne rete grph, unrete grphs, pths, ruts n mtrx ssote wth grphs.. GRAPH: A pr of set {V, E}, V φ, onsttute grph. Elements of set V re lle vertes whle elements of set E re lle eges or lnes or urves. Generlly lnes n ponts of plne represent the eges n vertes of the grph. Note:. If V s fnte set then we sy tht grph s fnte grph. MCA-05 47

48 . Eh ege s represente y pr of vertes sy u n v, these vertes re lle en pont of eges we wll enote t y E (u, v)... Drete grphs If we put u n v s n orere pr then ege s lle rete from u to v n suh grph n whh eh ege s rete s lle rete grph. For exmple, fgures gven elow re rete grphs. > > > > Fg.. Fg.. Another exmple s grph of relton s lwys rete grph... Unrete grph If grph s not rete s lle unrete grph n suh grphs eges re gven s E (u, v). Fgure. n.4 re unrete grph MCA-05 48

49 Fgure. Fg..4 Note: () Ege E (u, u) s lle self loop. A grph wth no self loop n no prllel ege s lle smple grph otherwse t s lle non-smple. For exmple grph of relton whh s nether reflexve nor symmetr s smple n tht of reflexve relton s not smple. Fgure.5 n.6 re grph whh re non-smple n smple respetvely. Fg..5 Fg..6 () Eges e n e re prllel eges f they hve sme vertes. Here e n e re prllel eges. e e. PATH AND CIRCUIT Fg..7 MCA-05 49

50 In the followng fgure, we hve e, ), e, ), e, ) e 4, ), ( ( ( 4 ( 5 e ( 5, 6 ) re the eges so tht we move from to 6 long these eges wthout usng n ege more thn one. Fg..8 Now we efne the followng: MCA-05 50

51 .. Wlk. Let G e grph. Then sequene of vertes v 0, v, v,. v t eh jent to the next n there s lwys n ege etween v, n v +, s lle wlk. The vertex v 0 s lle the ntl vertex n the vertex v t s lle termnte vertex of the pth. Numer of eges n wlk s lle ts length. A wlk s lle open wlk f t hs fferent egnnng n en ponts n s lle lose wlk f t s egnnng n en ponts re sme... Defnton: A Trl s wlk hvng ll stnt eges. A Pth s wlk n whh ll vertes re stnt. A lose trl s lle Crut. A rut n whh vertes (exept the frst n lst) o not repet s lle Cyle... Note: A pth s lwys trl ut trl nee not e pth. Smlrly yle s lwys rut ut rut s not yle lwys. In fgure gven elow for the grph, one s rut whle other s not rut. Fg..9 Fg..0 s not rut s rut n yle In Fgure., MCA-05 5

52 Fg.. () s trl s no ege repets () e, n e re pths () e s rut ut not yle (v), n e re yles..4 SOME DEFINITIONS WITH EXAMPLES.4. Degree of vertex: In non-rete grph G, the egree of vertex v s etermne y ountng eh loop on v twe n eh other ege one. It s enote y (v). Exmple: (V ) =, (V ) =, (V ) = re the egrees of V, V n V n followng fgure. Fgure. MCA-05 5

53 .4. Theorem: The sum of (v ) for eh v of unrete grph G (V, E) s twe the numer of eges n G Proof: Sne G s unrete grph, eh ege of G s nent wth two vertes, therefore, ontrutes to the sum of egree of ll the vertes of the unrete grph. Therefore, the sum of egrees of ll the vertes n G s twe the numer of eges n G. Exmple: Drw smple grph wth three vertes.e. rw grph wth no self loop n no prllel eges. Soluton: Fgure shown elow s smple grph..4. REMARK If we see followng grph Fg.. Fg..4 MCA-05 5

54 There re fve eges n ths grph n whh ege e 4 n e 5 hs sme vertes (, ). Therefore these eges re prllel eges. Further sum of egrees of the vertes s +++=0 =.5= tmes the numer of eges..4.4 Defntons. Isolte vertex: A vertex on whh no ege nent s lle solte vertex,.e. vertex v suh tht (v) = 0.. Null grph: A grph G = (V, E) suh V φ n E = φ s lle null grph, therefore null grph n whh every vertex s solte..4.5 Theorem: In non-rete grph, the numer of vertes of o egree s lwys even. Proof: Let the numer of vertes n grph G e n. Wlog suppose tht the egree of frst k vertes sy v 0, v, v,. v k e even n remnng n-k vertes e o.e. the vertes wth o egree. n k Now ( v ) = ( v ) + ( v ) = = = k+ n... () But we know y Theorem (.4.) tht L.H.S. of () s even. As (v ) n frst term of R.H.S. s even, therefore, ( v ) k = s lso even. It gves us tht n = k+ ( v ) must e even. But eh (v ) n tht ( v ) n = k+ s o. Moreover we know tht sum of o numer s even f they re tken even numer of tmes. So here n-k must e even..e. o numer vertes n the grph must e even. Hene the proof s over. MCA-05 54

55 .5 CONNECTED AND DISCONNECTED GRAPHS..5. Defnton: If n grph we n move from ny vertex to the ny other vertex of the grph then suh grphs re lle onnete grphs otherwse t s lle sonnete grph. Smple we n sy tht f there exsts pth etween every pr of vertes the grph s lle onnete. For exmple, Grph n Fg..5 s onnete grph whle n Fg..6 t s sonnete. Fg..5 Fg Defnton. Let G e onnete grph. The ege onnetvty of G s the mnmum numer of lnes (Eges) whose removl results n sonnete or trvl grph. It s generlly enote y δ(g).. Vertex onnetvty of grph G s the mnmum numer of vertes whose removl results n sonnete or trvl grph s lle the vertex onnetvty of G. It s generlly enote y k(g)..5. Theorem. The ege onnetvty of onnete grph G nnot exee the mnmum egree of G,.e. λ(g) δ(g). Proof: Let G e onnete grph n v e vertex of mnmum egree n G. Then the removl of eges nent wth the vertex v sonnets the vertex v from the grph G. Thus the set of ll eges nent wth the vertex v forms ut set of G. But from the efnton, ege onnetvty s the ege onnetvty of G nnot exee the mnmum egree of v,.e. λ (G) δ (G). MCA-05 55

56 .5.4 Theorem: The vertex onnetvty of grph G s lwys less then or equl to the ege onnetvty of G,.e., k (G) λ (G). Proof: If grph G s sonnete or trvl the k (G) = λ (G) = 0. If G s onnete n hs rge e, then λ =. In ths se K =, sne ether G hs ut pont nent wth e or G s K.( k (G) λ (G) when λ (G) = 0 or ). Fnlly let us suppose tht λ (G). The G hs λ lnes whose removl sonnets G. Clerly the λ- of these eges proues grph wth rge e = {u, v}. For eh of these λ- eges selet n nent pont whh s fferent from u or v. The removl of these ponts (vertes) lso removes λ- eges n f the resultng grph s sonnete then k λ- < λ. If not the ege e = {u, v} s rge n hene the removl of u n v wll result n ether sonnete or trvl grph. Hene k λ n eh se n ths ompletes the proof of the theorem. Thus, the vertex onnetvty of grph oes not exee the ege onnetvty n ege onnetvty of grph nnot exee the mnmum egree of G. Hene the theorem gven elow..5.5 Corollry :For ny grph G, k(g) λ (G) δ (G) s sonnete..5.6 Theorem :A grph s sonnete f V n e wrtten s unon of two non-empty, sjont susets V n V suh tht there exst no element of E whose one vertex n V n other n V. Proof: Let us suppose tht G e onnete grph. Tke ny vertex v n G. Let V e the olleton of ll these vertes, whh re jone y pths to v. Sne G s not onnete V V [f V = V then G wll e onnete]. So tke MCA-05 56

57 v set hvng ll vertes of G whh re not n ll vertes of V whh re not n V. Therefore V n V re requre susets of V. Conversely, suppose tht V = V V, v Φ, v Φ, V V = φ, then f we tke v V n v V then there exst no ege etween v n v.e. grph s sonnete..5.7 Component of grph mens mxml onnete sugrph of grph G (V, E). For exmple n Fg..8. Fg..8 {v,v,v }, {v 4,v 5 }, {v 6,v 7,v 8,v 9 }, re omponents. Further t s ler tht grph s onnete f n only f t hs extly one omponent. MCA-05 57

58 MCA Theorem: A smple grph wth m vertes n r omponents n hve t most (m-r) (m r + )/ eges. Proof: Let G (V, E) e grph wth m vertes n r omponents let m 0, m, m r e the numer of vertes n eh omponents of G (V, E). Then we hve Σm = m n m () Now from () we get ( ) = = r r m m () Squrng () on oth ses we get ( ) mr r m m r + = = But ( ) ( ) ( )( ) = = + = + = r j r mr r m m m m m mr r m r m m r ) ( + + = ( 0 ) ( 0 ) ( j m n m Q ) = = + r r r mr r m m m = + + r m r mr r m m We lso know tht n smple grph wth m vertes hve t most m (m -)/. Thus the mxmum numer of eges n G s ( ) ( )( ) [ ] m r m r m m m m m r r r = = = = = < ( )( ) + r m r m

59 Ths ompletes the proof..5.9 Defnton: Two vertes u n v n grph re s to e mutully rehle f G ontns oth rete u-v wlk n rete v-u wlk. A grph s s to e strongly onnete f every two of ts vertes re mutully rehle. Exmple: Dgrph shown n the fgure.0 s strongly onnete..6 MATRIX REPRESENTATION OF GRAPHS Sne we know tht t s very esy to mnpulte mtres. We tke the mtrx ssote wth fferent grphs. There re two wys of representng grph- () nene mtrx; () Ajeny mtrx..6. Inene mtrx: Let v, v,.. v n e n vertes n e, e,, e m e m eges of grph G. Then n n m mtrx I = [ j ] whose n rows orrespon to n vertes n m olumns orrespons to m eges where j s s 0 j = f v s not n en pont of ege e j f v s not n en pont of ege e j f e j s self loop on v ths mtrx I = [ j ] s lle nene mtrx. Exmple: Wrte the nene mtrx of the grph n Fgure s MCA-05 59