Prepred y Dr. A.R.VIJAYALAKSHMI
Algeric systems Semi groups nd monoids Groups Sugroups nd homomorphms Cosets Lgrnge s theorem Ring & Fields (Definitions nd exmples
Stndrd Nottions. N :Set of ll nturl numers. Z :Set of,,3,... ll integers,...,,0,,,3,... 3. R :Set of ll rel numers 4. Z :Set of ll positive integers 5. Q : Set of ll rtionl numers p q, q 0 where p & q re integers 6. Z :Set of ll negtiveintegers,-,-3,... 3
Algeric system A system consting of non-empty set & one or more n-ry opertions on the set clled n Algeric system Note: A mpping f : A A A... A n - ry opertion nd n clled the order A or of An A clled the opertion. Note : For n, f : A A A clled inry opertion. 3.Binry opertion denoted y re, -,., x,, * nd the inry opertions 4
Properties of inry opertions:-.closure property : A inry opertion : G G G sid to e closed if for ll, G, * G.Associtivity property : A inry opertion : G G G sid to e ssocitive if ( c ( c,,c G 3.Extence of identity: eg such tht e e G The element e clled the identity element 4.Extence of inverse: For G, G such tht e The element clled the inverse of 5
5.Commuttivity: If,,G, then sid to e commuttiv e in G. 6.Dtriutive Property: (. c (.( c(left (. c ( c.( c(right 7.Cncelltion Property: ( c c(left dtriutive dtriutive cncelltion lw lw lw c c(right cncelltion lw 8.Idempotent Property: An element G sid to e idempotent if 6
Definition of Semigroup :A non empty set S together with inry Opertion * clled semigroup if the following conditions re stfied. i S, S ( Closure ii ( c = ( c,, c S ( Associtiv e Exmple : Let N,,3,... e the set of ll nturl numers. (N, +, (N, re semigroups under the inry Opertion ddition & sclr multipliction Exmple : (Z, - not semigroup,since ssocitive property not stfied. 7
Definition of monoid: A non empty set M together with inry opertion * clled monoid if the following conditions re stfied i M, M ( Closure ii ( c = ( c,,c M ( Associti ve iii e M such. tht e e M( Identity Exmple : Let N,,3,... e the set of ll nturl numers. (N, monoid with s identity Exmple :(N, + semigroup ut not monoid.the dditive identity zero ecuse +0 = 0 + = But 0 N (i.e. 0 not n element of N 8
Exmple3 : (Z, +, (Z, re semigroups s well s monoid Show tht the set N of nturl numers semigroup under the opertion x* y = mx(x,y. Is it monoid? Is it commuttive? Solution: Let N = {,,3, } e the set of nturl numers Let Let x x, y x, y, z (y z = N. N. x Now Now mx(y, x z y = mx mx(x, y x or y closed N. x, mx(y, z mx x, y, z mx(x, y, z mx x, y, z ( x y z = mx(x, y z mx ssocitiv e Hence (N, semigroup. 9
Now N & for ny x N x = mx(x, mx(,x x x x x x the identity element Hence (N,, monoid. Also for ny x,y N, x y = mx(x,y mx(y,x y x monoid commuttiv e 0
. If S denotes the set of positive integers 00, for x, y S,define x y =min{x,y}. Verify whether (S, monoid ssuming tht ssocitive. Solution: 00 the identity element in (S,, since x 00 = min{x,00} = x, since x 00 for ll x S. Hence (S, monoid.
Homomorphm: Let A mpping f : X Y (X, & (Y, Δ e lgeric systems. clled homomorphi sm ( or simply morphm if for ny x, x X x, f ( x x f ( x f ( Semi group Homomorphm: Let (S, & (T,Δ e semigroups, mpping f : S T sid toe semigroup homomorphi sm if f ( f ( f ( for, Monoid Homomorphm: Let (M,, e M & (T, Δ, e T e h : M N such tht for ny nd h(e M = e T S ny two monoids. A mpping, M, h( = h( Δ h( clled monoid homomorphi sm
Epimorphm: A homomorphm f : X Y sid to e epimorphm if f onto Monomorphm: A homomorphm f : X Y sid to e monomorphm if f one to one Isomorphm: A homomorphm f : X Y sid to e omorphm if f one to one & onto (or f ijective Endomorphm: Into homomorphm Automorphm: clled n endomorphm An omorphm onto itself clled n utomorphm 3
Prove tht monoid homomorphm preserves commuttivity & invertiility Let g :(M,, e M (N, Δ,eN such tht for ny, M, g( Let Thus, M monoid such tht g ( g( e monoid homomorphm = g( Δ g( & g(e M = e N = g( = g( homomorphm preserves = g( g( commuttivity Let e n invertile element in M,then M nd = e M g( = g( g( em g( g( = g( g( e g( ( g(. N Thus monoid homomorphm g preserves invertiility. 4
Cyclic Monoid : A monoid (M, sid to e cyclic if for some M, every element x M of the form n,where n n integer. (i.ex n. Wehere clled the genertor of the cyclic monoid Note : In the ove defn. replce monoid y semigroup we get cyclic semigroup Every Cyclic Monoid commuttive Definition: Let (M,,e e monoid nd M. Then if n element M suchtht e,then clled n inverse of (i.e sid to e invertile. 5
Theorem: Let (M,,e e monoid nd M. If invertile then its inverse unique Proof: Let nd c e elements of monoid M such tht = = e ( = e = ( c(using = ( c = c (i.e. = c Hence inverse unique. nd c = c = e ( 6
Definition: Let(S, & (T,Δ e semigroups.the direct productof (S, & (T,Δ the lgeric system (S T, in which the opertion on S T defined y (s,t (s,t (s s,tt (s,t,(s,t ST Definition: Let (S, e n lgeric system. Then zero of S n element S such tht x x x S Prove tht epimorphm preseves zeros Proof: Let f ( M,, e ( T,, e : M T e monoid epimomorphm Let x M e zero of M. Then x x x M 7
f ( z f ( z f ( z f ( z f ( f ( z f ( f ( z As f n epimorph m,every t T cn e written s t f( for some M.Thus f z the required zero of T Theorem: The composition of semigroup homomorphm lso semigroup homomorphm Proof: Let ( S,, ( T, nd ( V, e semigroup nd f : S T nd g S V e semigroup homomorphm. Let,S. f : S T semigroup homomorphm, f ( f ( f ( where f (, f ( ( T,. g T V e semigroup homomorphm. then for ny f (, f ( T, g[ f ( f ( ] g(f( g(f( (g f( (g f( 8
Now consider the composition function g f : S V (g f( g[ f ( ] g[ f( Δ( ] (g f( ( g f ( g f : S V semigroup homomorphm Definition: Su semigroup Let ( S, e semigroup. A suset T of S sid to e su semigroup ( T, if T closedunder theopertion (i.et itself semigroup 9
Definition: Su monoid Let ( M,, e e monoid. su monoid ( T,, e if T A suset of T of M closed under the sid to e opertion nd e T ( i. e T itself monoid under the sme opertion with the sme identity e Theorem: For ny commuttivemonoid ( M,, thesetof idempotent elements of M forms sumonoid Proof: Let (M, e commuttiv e monoid. Let S M/ e the set of idempotent element of M. Clerly e S, s e e e. Let, S, then & Now ( ( ( ( ( ( M, commuttiv e 0
( ( ssocitiv e ( ( ( i. e ( (. S. ( S, su monoid of ( M,
Group A non empty set G together with inry opertion * clled group if the following conditions re stfied.. i G, G ( Closure ii ( c = ( c,,c G ( Associtiv e iii e G such tht e = e = G ( Identity iv G such tht = = e G ( Inverse. Aelin group In group, if =, G then ( G, clled elin group
Order of group: The numer of elements in group G when G finite sid to e order of group nd denoted y O(G. Order of n element: Let G e group nd e e the identity element of G. Let G. If there lest positive integer m such tht m = e then m clled the order(or period of G,. which denoted y O(. Sugroup: A suset H of group G clled sugroup of G if H itself forms group under the opertion of G. 3
Properties of group The identity element of group G unique Proof: Let e nd e e two identity element of G. Let e e the identity element nd e ct s some element of G Then e e = e e = e ( Now, let e e the identity element nd e ct s some element of Then e e = e e = e ( G. From (, e = e e = e [ using ( ] Hence identity element of G unique. 4
The inverse of n element in group G unique. Proof: ' '' Let nd e the two inverses of. Then ' = ' = e nd '' = '' = e ' = ' e = ' ( '' = ( ' '' = e '' = '' Hence inverse of unique For ny, G, ( = 5
Prolem. If (G, n elin group, prove tht ( = for ll, G. Solution: Given tht G elin group. =,G. Then ( = ( ( = ( = ( = ( (.Check whether {0,,4,6,8} under ddition mod 0 group. Solution: + 0 0 4 6 8 0 4 6 8 0 4 6 8 4 6 8 0 4 6 8 0 6 8 0 4 8 0 4 6 = Set closed under ddition mod 0. 0 the identity element. Inverse of 0,,4,6,8 re 0,8,6,4, respectively Hence {0,,4,6,8} group under ddition mod 0 6
3.Prove tht the only idempotent element of group its identity element Solution: If n idempotent element then we hve Now, = e = = ( = ( = e =. Hence the only idempotent element of G its identity element. 4.Show tht the intersection of sugroups of group sugroup. Solution: Let H nd K re sugroups of G. Then tlest the identity element e H nd ek. e H K Thus H K non empty suset of G. Let,HK,H nd,k H nd K H K 7
5.Prove tht the set of ll non-singulr mtrices forms n elin group with respect to mtrix multipliction. Solution: 0, with Let G e the set of ll mtrices of the form G c d d c Let, G d c c d c d d c c d d c Now (,. 0 ( ( d c with G closed.. 0 0 Let G. 0 0 0 0 element. identity the G 0 0 Also mtrix multipliction ssocitive 8
c d d c Let of inverse the e 0 0 c d d c Now 0 0 ( d c c d c d d c 0 d c nd cd Compring we get,, d c get we solving, By of inverse The G group 9
d c c d c d d c c d d c Now ( Commuttive property stfied. Hence G n elin group. d c c d c d d c c d d c ( nd 30
Cyclic group : A group ( G, sid to e cyclic, if there exts n element G such tht ech nd every element in G cn e expressedin the form of n where n n integer. Here the element ' ' sid to e the genertor of the cyclic group. 6.Prove tht every cyclic group elin. Solution: Let (G, e cyclic group generted y n element G. For ny two elements x, y G we hve x = m nd y = n where m, n re integers. Now, x y = m n mn nm = n m = y x G elin 3
7.Show tht G = {,,i,-i} cyclic group. Solution: Let = i. Every element of G cn e expressed s integrl powers of i. i 4, - i, i i, - i i 3. Hence i the genertor of G nd therefore G cyclic. 8.Prove tht every group of order 3 cyclic ( so elin. Hw e c e c c e c e 3
Prolem: Let If where G e G of n finite group order n then the lest n generted positive e nd G y integer n, for element, 3,.., which G. n n e e Solution: Given O(G = n m tht e, m n. Suppose SinceG cyclic,ny element of By divion lgorithm, Clim : n k = mq e G cne writtens + r, where 0 k r < m for somek Z. k mq r Hence mq r ( m q every element some r, 0 r < m. r e k q r e. r G cn e expressed s r r for 33
G {,, 3,.., m,e} (i.e. O(G m n, which contrdiction.[since O(G n ] Hence n e. Also,, 3,.., n re ll dtinct where n e For th, suppose i = j for i < j n i -j j -j i- j = e where i. - j n, which the contrdiction. Hence i j for i - j n. G {,, 3,.., m,e} 34
Theorem: The necessry non empty suset H of H & sufficient condition G e sugroup for, H Proof: Necessrycondition: Assume H sugroup of G. Since H sugroup of G, it must e closed under multipliction., H H Since H sugroup of G, ech element of H must posses inverse. H H, H,hence for ny H H ( y closure property of H. 35
Sufficient condition: Assume tht H non empty suset of group G nd, H H. Clim : H sugroup of G. i Replcing y in the given condition H, we get H e H. Thus identity exts in H. ii Replcing y e in the given condition H, we get e H H. Thus inverse exts in H. iii Let, H, H [ y (ii ] ( H H. Thus closure lw true in H. iv Since ssocitive property holds in G, it holds in H. H sugroup of G. 36
Group homomorphm G, nd H, Δ Let two groups. A mpping f : G H clled group homomorphi sm if for ny, G, f( = f( f( nd f(e G e H, f( [f(] Note.A group homomorphm f clled omorphm, if f & onto. A homomorphm of semigroup into itself clled semigroup endomorphm 37
Theorem: If f:g G' group homomorphm then i Let e e the identity of (G, then f(e = e' where e' e the identity of G, (or Group homomorphi sm preserves identity. ii For ny G, f( = [f(] (or Group homomorphm preserves inverse. Proof: For ny, G, we hve f( = f( f(. i Since e the identity of G, we hve e = f( e = f( f( f(e = f( f( f(e = f( [since G, f( G' f(e = e' f( e' = f( ] 38
- =e f ( =f ( e - - f ( f ( =f ( e =e ' - nd =e f ( - =f ( e - f ( f ( =f ( e =e ' - - f ( f ( =f ( f ( = e '. H e nc ef ( - =[ f ( - ] f or l l G. ii For ny G, G. f( = f(e f( f( = f(e = e' nd = e f( - = f(e f( - ( = f(e = e' f( f( = f( f( =e'. Hence f( = [f(] for ll G. iii f(g sugroup of G' (or Group homomorphm. preserves sugroup 39
iii Let f(g = { f(x / x G} then f(g non empty suset of G' Let ', ' f(g. We hve f( =', f( = ' for, G. ' ( ' = f( [f(] = f( f( = f( - f(g. For ny ',' f(g ' (' f(g. Hence f(g sugroup of G' Kernel of homomorphm Let f : G G' e grouphomomorphm then thekernelof f (or Ker(f defined s Ker(f = { x G / f(x = e', e' the identity of G' 40}.
Left coset Let (G, e group nd i For ny G, the set clled the left coset element G Cosets (H, e sugroup of (G, H H h/h H of H G determined y the Right coset ii For ny G, the set H H h /h H clled the right coset of H G determined y the element G Note: ithe right (or left coset of H in G non empty ii H, H re lso susets of G. iiiif G elin, then H H Any two right (left cosets of H in G re either djoint or identicl 4
Prolems Find ll the cosets of the sugroup H = {, } in G = {,, i, -i} with the opertion multipliction Solution: Right coset of H in G Left coset of H in G H.( = {, } H.(= {,} H.( i = {i, -i}.h = {, } (.H = {, } (i.h = {i, -i} H.(-i = {-i, i} (-i. H = {-i, i} Find the left cosets of {[0],[3]} in the ddition modulo group (Z6, +6. Solution: Let Z 6 = {[0],[],[],[3],[4],[5] } e group H = {[0],[3]} e sugroup of (Z6, + 6. 4
[5] 6 H {[5],[]} The left cosets of H re [0] 6 H {[0], [3]} [] 6 H [] 6 H {[], [4]} {[], [5]} [3] 6 H [4] 6 H {[3], {[4], [0]} []} [5] 6 H {[5],[]} Here [0] 6 H [3] 6 H, [] 6 H [4] 6 H, [] 6 H [5] 6 re the three dtinct left cosets of H in (Z 6, + 6. H 43
Lgrnge s theorem If G finite group nd H sugroup of G, then O(H divides O(G. Proof. Since G finite group, O(G = n. Let H e the sugroup of G, therefore H lso finite. (i.e. O(H = m (m n n m We must prove tht m the divor of n. (i.e. k where k n integer. Let H = {h, h,..h m } re the m memers of H. For G, H the right coset of H in G. H { h, h,..h m}. Since there one to one correspondence etween H nd H, we hve O(H = O(H = m Since we know tht ny two right cosets of H in G re either djoint (or identicl,the numer of dtinct right cosets of H in G will e finite, sy k. 44
The union of these k dtinct right cosets of H in G equl to G. Hence if H, H, H k re the dtinct right cosets of H in G, then G = H H H3 Hk O(G O(H O(H.. O(H n = m + m +..+ m (k times n = km n O( G k ( i. e k m O( H Hence O(H the divor of O(G. k Find sugroup of order two of the group (Z 8, + 8. Sol. Z 8 = {[0],[],[],[3],[4],[5],[6],[7]} If H sugroup of z 8, then O(H divides 8. Hence O(H =,, 4 or 8. 45
Deductions The order of every element of finite group G divor of the Order of the group. Proof. Since G finite, O(G = n nd O( = m. Let O(H = H = {[0],[x]} nd [x]=[0] x = 4. H = {[0],[4]}. H p = {, p Z } e cyclic sugroup generted y. Since O( = m, then H will hve m elements. By Lgrnge s theorem, O(H divides O(G O(H O( m (i.e. O( divides O(G. If G finitegroupnd G then O(G e. Proof. Let O(G = n nd O( = m then m = e. Since the order of n element divides the order of the group, we get n = km. Now, O(G = n = km = ( m k = e k = e. 46
Norml sugroup e to sid (G, group of (N, sugroup A nd G g every for if G, of sugroup norml N. gng N, n every for e (N, nd group elin n e (G, Let G. of sugroup N n let nd G g Let N = n = ne - = n(gg - = (ngg = (gng gng N. gng N, n nd G g - sugroup. norml (N, Hence 47
Theorem: Kernel of (G, homomorphm f from group to group (G', norml Proof: Ker(f = {x G / f(x = e', e' the identity of Clerly f(e = e' [since e G] e Ker(f Hence Ker(f non empty suset of (G, G' }. Let, Ker(f. f( = e' nd f( = e' Now,f( = f( f( = f( [f(] = e' (e' = e' e' e' Ker(f. Hence Ker(f sugroup of (G, 48.
f(g f(k = f(g g k f(g Now, [f(g] f(g = ' e ' e - [f(g] f(g = Ker(f. g k g. (G, of sugroup norml Ker(f Hence 49
(G, Stte nd prove Fundmentl theorem on homomorphm of groups. If f then Proof: homomorphm of G K Given omorphic f :G to G'. G onto G' with kernel K, G ( i. e G ' K G' homomorphm with kernel K, so tht K norml sugroup of G. G G To prove G' let us define mp g :G K K g( = K, G. Now we define mp h : such tht h(k = f(. such tht G K G' 50
To prove h n omorphm, first we shll show tht the mpping h well defined (i.e. for ny,g, K = K h(k = h(k. We hve K = K f( K =e' f( f( f( [f(] = e' f(= e' f( K K f( e' = e' f( f( = f( h(k = h(k h well defined. K = K = K = K K 5
Clim : h homomorphm Let K, K Now, h(k G K K = h[k( ] = f( = f( f( = h(k h(k h homomorphm. 5
Clim : h one to one Let h(k = h(k f( = f( f( f( = f( f( f( = f( = f(e = e' f( = e' K K = K. h one to one. 53
Clim : h onto Let y e n element of G'. Since f onto, there exts n element G such tht f( = y. By definition, h(k = f( h(k = y. Thus h onto Hence h n omorphm of G ( i. e G ' K G K onto G'. 54
Permuttions Let S e nonempty set. A ijection function S clled permuttion of S. Acycle of length clled trnspositions S Even permuttion - product of n even numer of trnspositions. 5 3 7 4 8 5 6 6 7 4 8 3 f = ( 5 6 (3 7 4 8 = ( ( 5 ( 6 (3 7 (3 4 (3 8 = product of n even numer of trnspositions Hence f even permuttion. 55
Odd permuttion - product of n odd numer of trnspositions. g 4 3 3 4 = ( 4 3 = ( 4 ( ( 3 = product of n odd numer of trnspositions Hence g odd permuttion. 56
3 4 5 5 4 3 nd 3 5 4 5 4 3 h f If } find = {,,3,4,5 set A the on ns permuttio re n g on A such tht permuttio f g g f Solution:. f h f g 3 5 4 5 4 3 f 4 5 3 5 4 3 f h 3 4 5 5 4 3 f h f g odd permuttions. n! even permuttions nd n! There re. n }e finite set with n elements,,..,, = { Let A n 57
Stte nd prove Cyley s representtion theorem on permuttion groups. Sttement: Every finite group of order n omorphic to sugroup of the permuttion group of degree n. Proof: Let G e the given group nd A(G e the group of ll permuttions of the set G For ny G, define mp f : G G' such f (x = x. tht f well defined nd one-to-one : Let x = y x = y f(x = f(y Thus f well defined nd one - to - one. 58
f onto: We hve x Thus f y Hence f f (x= x (i.e. onto y G for y = x ny y permuttion. (i.e. f G. A(G. Let K e the set of ll such permuttions. We cn show tht K sugroup of A(G. Since e G, f e K. Thus K non - empty. 59
Let f, f K Then (f f (x = f (f(x = f (x x f f = f K. Now, ( f f ( x f ( f ( x = f ( = = = f (x x x = ex = fe(x Thus the inverse of f f Hence K sugroup of A(G. K. Next we will show tht G omorphic to K 60
Define mp h :G K such tht h( = f. h well defined nd one-to-one : For ny, G, let = x = x f (x = f (x f = f h( = h( Thus h well defined nd one-to-one. Clerly h onto. (Since the permuttion ijective. 6
h homomorphm: Let h( = f = f f = h( h( Thus h homomorphm. G omorphic to K. 6
Ring An lgeric system (R, +,. sid to e ring if + nd. re inry opertions on R stfying the following properties: i (R, + n lgeric group. ii (R,. semigroup. iii The opertion. dtriutive over +, (i.e. for,,c R,.( + c =. +.c nd ( +.c =.c +.c Exmple : The set of ll integers Z, the set of ll rel numers R re rings under the usul ddition nd usul multipliction. 63
Zero divor An element '' in ring (R, +,. sid to e zero divor if 0 nd 0 in R such tht. = 0 Exmple : (Z 5, + 5,. 5 ring without zero divors. Integrl domin: A commuttive ring with identity nd without nonzero zero divors clled n integrl domin. Exmple : The ring Z of ll integers n integrl domin. Field: If (R, +,. commuttive ring with tlest two elements nd if (R-{0},. group, then the ring R clled field. Exmple : i The ring R of rel numers field. ii The ring (Z 7, + 7,. 7 field. 64
Commuttive ring. The ring (R, +,. clled commuttive ring, if for ll, R. Exmple: The set of even integers under usul ddition nd multipliction.. =. Show tht the set of ll integers form ring under usul ddition nd multipliction It ovious tht + stfies closure nd ssocitive property For the inry opertion +, the identity element 0 nd 0 z For the inry opertion +, the inverse of - nd - Z. (Z, + group. 65
Now consider (Z, It ovious tht ' ' stfies closure nd ssocitiv e property. (Z, semigroup. Moreover for,, c Z,.( + c =. +.c nd ( +.c =.c +.c Thus dtriutive over +. Hence (Z, +, ring. 66