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II] SEMESTER B.Sc., , MATHEMATICS PAPER — III, , , , GROUPS, , Dr.SHIVASHARANAPPA SIGARKANTI, , ASSOCIATE PROFESSOR AND HOD, DEPARTMENT OF MATHEMATICS, GFGC K.R.PURAM, BANGALORE -560036

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GFGC K.R.PURAM III SEM B.Sc PAPER — III ALGEBRA Ill GROUPS, , PAPER III ALGEBRA III GROUPS, RECAP, , Binary operation: An operation * defined on a set G is said to be a binary operation if, abE€G>a*beEG., , Group : Anon empty set G together with an operation * is said to be a group, , if the following axioms are satisfied, , 1) Closure law: for alla,b € Gimpliesa*be G, , 2) Associative law : for alla,b,c € G, a*(b*c)=(a*b)*c, , 3) Identity : there exists an element e inG such thata*e =exa=a forallaG, 4) Inverse : for every element ain G there exists anelement a+ inG such that, axa t=a'*a=e., , In addition if the commutative holds inG i.e.a*b=b+*a foralla,binG, , then the group is called abelian group., , If only the closure law and the associative law holds ,thenG is called a semi group., Every abelian group is a group and every group is a semigroup, , Notation : (G,*), , Examples are : (Z,+),(R, +), , Finite group : If G has finite number of elements then G is called finite group ,, otherwise it is an infinite group., , Order of a group:The order of a group G is defined as the number of elements in it, and is denoted by O(G)., , Properties :, , 1 )Identity element is unique, , 2) Inverse element is unique, , 3)(at)t=a, , 4) (axb)-+ = bt «at, , 5) Left cancellation law ab = ac implies b =c, Dr.SHIVASHARANAPPA SIGARKANTI HOD OF MATHEMATICS Page 1 of 15

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GFGC K.R.PURAM III SEM B.Sc PAPER — III ALGEBRA Ill GROUPS, , 6) Right cancellation law ac = bc implies a = b, , 7) If aand bare any two elements of G then the equations, , ax = band ya = b have unique solutions inG, , 8) If a1 =a for allainG ,thenG is abelian, , 9) If (ab)? = a?b? for alla, binG ,then G is abelian, , 10) Every group of order less than 6 is abelian, , 11)If Gis a group of even order then there exists an element a such that, a=a™'or equivalently a? =e, , 12) Anon empty subset H of a group (G,*) is said to be a subgroup of G, if H itself is a group w.r.t the operation *, , 13) G and {e} are subgroups of G which are called improper subgroup., Subgroup other than these two is called a proper subgroup of G., , 14) Anon empty subset H of a group G is a subgroup of G if and only if, DQabeH>a*b EH iijaeH>a' EH, , 14) A non empty subset H of a group G is a subgroup of G if and only if, abe€H>a*b' €H, , 15) The intersection of any two subgroups is againa subgroup but the union need not be., , 16) The identity and the inverse of subgroup are same as that of the group., , End of RECAP, , oh ok 2 ok oe 2 oe OK eK KK KK EK, Order of an element : The order of an element a of a group G is the least positive integer such, that a"=e., Ex.1.Find the order of eachelement of the group G, )G={1Lo,07?;*) ii)G={1,-1i,-i; +}, iii) G = {0,1,2,3,4; +5} iv) G = {1,2,3,4 ; Xs} v) G = {0,1,2,3,4,5,6 ; +7}, Soln : The order of identity is always 1, i) G = {1,0,w?; *) identity is1, w? =1,(w?)? =1 « O(w) = O(w*) =3, ii) G = {1,-1,i,-i;*} identityis1,, Dr.SHIVASHARANAPPA SIGARKANTI HOD OF MATHEMATICS Page 2 of 15

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GFGC K.R.PURAM III SEM B.Sc PAPER — III ALGEBRA Ill GROUPS, , (-1)? = 1,i* =1,(-)* =1 « O(-1) = 2,07) =4,0(-) = 4, iii) G = {0,1,2,3,4; +5} identity is 0,, , 18 = 14,5 1+51+51+51 = 0,, 25 = 245 2+52+s52+52 = 0,, , 35 = 345 34+53+53+53 = 0,, , 45 = 4454454454454 = 0,, 0(1) = 0(2) = 0(3) = 0(4) =5, , iv) G = {1,2,3,4 ; Xs} 1is the identity, , 24*=16=1,, 3*=81=1,, V=16=1, , 0(2) = 4,0(3) = 4,0(4) = 2, v) G = {0,1,2,3,4,5,6; +7} identity is 0,, q7, , 14+, 1+71+71+71+71+71 = 0,, 27 = 247 24+72+72+72+72+72 = 0,, 37 = 347 34+73+73+73+73+73 = 0,, , 4? = 447 4474474474474474 = 0,, 57 = 5 +7 5+75+75+75+75+75 = 0,, 67 = 6 +7 6+76+76+76+76+76 = 0,, , 0(1) = 0(2) = 0(3) = O(4) = 0(5) = 0(6) =7, , Note : If O(a)=n,, then a"=e conversely if a“=e, then O(a) < m, for example in the above, example 114 = 1 +, 1+71+71+71+71+71+71+71+71+71+71+71+71 = 0 but 0(1) =, 7<14, , Theorem 1: The order of each element of a finite group if finite., , Proof : Let G be a finite group and let O(G)= n. Let a be any element of G. Consider the set, K= { a,a?,a3,...,}. , a is an element of G therefore any power of a is also an element of G, which, implies K is a subset of G. If a is of infinite order then K is an infinite set ,which is a contradiction, to the fact that K is a subset of G. Hence a must be of finite order., , Theorem 2: The order of any element of a group G is equal to the order of its inverse, i.e. If ais an element of a group G, then O(a)=O(a")., , Dr.SHIVASHARANAPPA SIGARKANTI HOD OF MATHEMATICS Page 3 of 15

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GFGC K.R.PURAM III SEM B.Sc PAPER — III ALGEBRA Ill GROUPS, , Proof : LetO(a)=m and O(a)=n, , O(a)=m >a™ =e, consider (a“1)™= (a™)-1=e71 =e, , > O(a") <mi.en<m-—-—(1), , Again O(a!)=n >(a1)" =e, consider a” = (a1) ((a“1)") =e 1=e, => O(a) <ni.eam<n-—--(2), , from (1) and (2) we getm=n -. O(a) = O(a“), , Theorem 3 : Let a be an element of a group G of order n. Then for an integer m, a=e if and only, if n divides m., , Proof : Given O(a) = nimplies a” =e., , Let a™ = e by division algorithm m = nq +r, , where q is the quotient and r is the remainder when m is divided by n,, clearly OS r<n> a™*" =e >a™a" =e >(a")4a" =e >e41a" =e ea" =e, => a" =e since nis the least positive integer suchthaa" =eand0<r<n, =>r=0 «m=nqi.e.ndividesm, , Conversely let n divides m.Thenm = nq for some integer q, , consider a™ = a™4 = (a")4 =e1=e, , Theorem 4: Let a and b are any two elements of a group G. Then O(a)=O(b*ab)., Proof : Let O(a) = mand O(b"'ab) =n., , Thena™ =e and (b~tab)" =e, , Consider (b~tab)™ = (b~1ab).(b~1ab). (b~*ab), ...m times, , = b-+(a(bb).a(bb~). a(bb-4), ....m times)b, , =b(a™)b = b-'eb = bb =e = O(b“'ab) <mi.e.n<m———(1), Again O(b~tab) =n => (b-1ab)" =e, , => (b-tab).(b~tab).(b~1ab), ...n times = e, , => b-l(a(bb7').a(bb).a(bb~*), ....n times)b = e, , > b (a )b=e, , =a" =beb™, , > a" =e > O(a) <ni.e.m<n-—-—-—(2), , from (1) and (2) m =n i.e. O(a) = O(b~*ab), Dr.SHIVASHARANAPPA SIGARKANTI HOD OF MATHEMATICS Page 4 of 15