Block Introduction. Some self study examples and proof of the theorems are left to the readers to check their progress.

Abstract Algebra The word Algebra is derived from the Arabic word al-jabr. Classically, algebra involves the study of equations and a number of problems that devoted out of the theory of equations. Then the term modern algebra is used as a tool to describe the information based on detailed investigations. Consequently a special branch of algebra namely abstract algebra, a generalization of modern algebra is introduced in which the algebraic systems (structures) are defined through axioms. The present course is divided into three blocks. The first block containing two units deal with the advanced study of group and ring theory. Then second block containing two units, introduces the concepts of vector spaces and linearly independence, basis and dimensions including the theory of linear transformations in consecutive units 3 and 4. Finally in block III the interesting properties of a special class of vector spaces called inner product spaces have been established and it is unit 5. After the study of this course the students will realize that abstract algebra allow us to deal with several simple algebraic systems by dealing with one representative system.

Block Introduction In previous classes the students have already studied elementary abstract algebra in which they have grabbed the elementary knowledge of two algebraic systems namely groups and rings including their properties. The main aim of this block is to deal with the further studies of these two systems. Ideally the goal is to achieve the information regarding groups and rings concerned with the prescribed course. Units 1This unit provides the special case of isomorphism of groups called automorphism and inner automorphism. Consequently automorphism group is described. Thereafter we also introduced conjugacy relation, class equation and counting of the conjugate elements corresponding to the elements of group. In the end of this unit, we shall discuss that a finite group G has a subgroup of every prime order dividing the order of group through Cauchy s & Sylow s theorems. Unit 2This unit presents the study of ring homomorphism and ideal (analogue of normal subgroup), quotient ring (analogue of quotient group), field of quotient group of an integral domain and a special class of integral domain called Euclidean ring. Then we introduced the polynomial rings over rational field and commutative rings. In the end of the unit, unique factorization domain has been discussed. Some self study examples and proof of the theorems are left to the readers to check their progress. 2

BLOCK 1 ABSTRACT ALGEBRA UNIT 1 GROUP THEORY Structure : 1.1Automorphism 1.2Inner Automorphism 1.3Automorphism groups 1.4Conjugacy relation and Centralizer 1.5Normaliser 1.6Counting principle and the class equation of a finite group 1.7 Cauchy s and Sylow s Theorems for finite abelian and non abelian groups. *Points for Discussion/ Clarification **References Before describing the main contents of unit, first we shall explain the fundaments concepts of group theory which are essential to understand the entire unit for convenience. Students may wish to review this quickly at first and then read the further required part of the prescribed course. 1.Function : The Notation f : A B to denote a function (or map) from A to B and the value of f at a is denoted by f(a). The set A is called the domain of f and B is called the codomain of f. The set f (A) = {bb b = f(a),for some aa} is called the range or image of A under f. For each subset C of B, the set f 1 (C) = { aa: f(a) C} Consisting of the elements of A mapping into C under f is called the pre image or inverse image of C under f. Let f : AB. Then (a) f is injective (or injection or one one) if whenever 3

a 1 a 2, then f(a 1 ) f(a 2 ) OR f (a 1 ) =f (a 2 ), then a 1 = a 2 (b) f is surjective (or surjection or onto) if for each bb there is some aa such each that f(a) = b. (c) f is bijective (or bijection) if f is both injective and surjective 2. Permutation : A permutation on a set A is simply a bijection from A to it self. 3. Binary relation : A binary relation on a set A is a subset R of A A and we write a b if (a, b) R. The relation on A is a) reflexive if a a, for all a A b) symmictric if a b b a, for all a, b A c) transitive if a b and b c a c, for all a, b, c A A relation is an equivalence if it is reflexive, symmetric and transitive. 4.Equivalence Class : If defines an equivalence relation on A, then equivalence class of a A is defined to be {x A x a}. It is denoted by the symbol [a]. i.e. [a] = {x A x a}. Elements of [a] are said to be equivalent to a. 5.Binary operation or composition : Let G be a non empty set. Then a binary on G is a mapping from G G in to G, defined by i.e. (a, b) ab a, b G ab G Example : Let A = { 1, 1} & B = {1, 2} operation Then the multiplication operation. is a binary operation on A but not B. Since for 2, 2 B 2.2 = 4 B. 6. Group: A non empty set G along with above binary operation is called a group if it satisfies the following axioms : G 1 G 2 G 3 For a, b G a b G (Closure axiom) For a, b, c G (a b) c = a (b c) (associative axiom) There exists an element e G called an identity element of G such that for all a G a e = a = e a G 4 For every a G, there exist a 1 G. Such that aa 1 = e = a 1 a. a 1 is called the inverse of a in G. The axiom G 1 is a super flous statement, i.e. It is a binary composition. G with axiom G 1 only is called a groupoid, with G 1 andg 2 is called semi group and with G 1, G 2 and G 3 only is called a monoid. 4

7.Commutative or abelian group : A group G is called an abelian or commutative group if for all a, b G ab = ba A group G is called finite if it is a finite set, other wise infinite. 8. Order of Group : The number of elements in a group G is called the order of G and It is denoted by 0(G). The infinite group is said to be of infinite order. The smallest group is denoted by {c} consisting only the identity element. It is clear that 0 ({e}) = 1. Example : (i) The set Z (or I) of integers forms an abelian an group w.r.t usual addition of integers but Z does not form a group w.r.t. multiplication since multiplicative inverse of every element of Z does not belong to Z. For example 2 Z but 2 1 = ½ Z / (ii) The set Q of rationals, R the set of reals are abelian groups w.r.t addition. (iii) Set of all 2 2 matrices over integers under matrix addition forms an abelian group. 9. Properties of group : In a group G : (i) The identity element is unique. (ii) The inverse of each a G is unique. (iii)(a 1 ) 1 = a, for all a G (iv) (ab) 1 = b 1 a 1 for all a, b G (Reversal law) (v) Left cancellation law ab = ac b = c (vi) Right cancellation law ba = ca b = c for all a, b, c G 10. Complex of a group : Every nonempty subset of a group G is called a Complex of G. 11. Sub Group : A non empty sub set H of a group G is a sub group of G if H is closedunder products and inverses. i.e. a, b H ab H and a 1 H The sub groups H = G and {e} are called trivial or improper sub groups of G and the sub groups H G and {e} are called nontrivial or proper sub groups of G. It can be easily seen that the identity of a sub group H is the same as the identity of group and the inverse of a in H is the same as the inverse of a in G. Example : The set {1, 1} is a sub group of the multiplicative group {1, 1, i, i} (ii) The set of even integers {0, 2, 4, } is a subgroup of the set of additive group Z = { 0, 1, 2,.} of integers. Criterian for a sub group : A sub set H of a group G is a sub group if and only if (i) H and (ii) For all a, b H a.b H and for all a H a 1 H 5

For all a, b H ab 1 H It can also be check that the intersection of two subgroups is a subgroup but union of two subgroups is not necessarily a subgroup. 12 Order of an Element : The order of an element a G is the least positive integers n such that OR a n = e, where e is the identity of G. Example : Let G = {1, 1. i, i}. Then G is a multiplicative group. Now 1 1 = 1 0(1) = 1 ( 1) 1 = 1, ( 1) 2 = 1 0 ( 1) = 2, i 1 = I, i 2 = 1, i 3 = i, i 4 = 1 o(i) = 4 ( i) 1 = i, ( i) 2 = 1, ( i) 3 = i, ( i) 4 = 1 0( i) = 4 13 Cyclic group : If in a group G there exist an element a G such that every element x G is of the form a m, where m is some integer. Then G is a cyclic group and a is called the generator of G. Example G = {1, 1, i, i} = {i 4, i 2, i, i 3 }= {i, i 2, i 3, i 4 } Then G is a cyclic group generated by i. Note : There may be more than one generators of any cyclic group. Some useful result of cyclic group: (i) (ii) (iii) (iv) Every cyclic group is abelian. Every subgroup of a cyclic group is cyclic. If a is a generator of a cyclic group G. Then 0(G) = 0(a) Every group of prime order is cyclic. Prime No. : An integer p > 1 whose only divisors are 1 and p other numbers are called composite numbers. The first prime number is 2. Where 2 is the only even prime All other prime are odd. Composite No. : A number is said to be composite if = r, while B > 1, r >1 14 Cosets: Let H be any sub group of a group G and a G. Then the set Ha = {ha h H } is called a right coset of H is G generated by a and ah = {ah h H} is called a left coset of H in G generated by a. 6

Remark: (i) e G He = { he h H} = { h h H}= H Similarly, eh = H i.e. He = H = eh, means H it self is a right as well as left coset of H in G. 15 Properties of cosets: If H is a subgroup of G and (a) h H, then Hh = H = hh (b) If a, b are two distinct elements of G, Then (i) Ha = Hb ab 1 H and ah = bh b 1 a H (ii) a Hb Ha = Hb & a bh ah = bh (c) Two right (or left) cosets are either disjoint or identical. (d) If H is a subgroup of G. Then G is equal to the union of all right cosets (or left cosets) of H in G. i.e. G = H Ha Hb, where a, b, c are elements of G. i.e. cosets partition to G. 16 Lagrange s theorem: The order of each sub group of a finite group divides the order of group. 17. Index of Sub group: The number of distinct right (or left) cosets of subgroup H of a group G is called index of H in G. It is denoted by By Lagrange s theorem I G (H) or [G ; H] I G (H) = 0( G ) 0( H ) Corollary (i) If a be any element of a finite group G. Then 0(a) 0(G). (ii) If G is finite an a G then a 0(G) = e 18 Normal Subgroup : A subgroup H of a group G is normal in G if g G & h H ghg 1 H If gh g 1 = {ghg 1 G H}. Then H is normal in G if ghg 1 H Results on normal subgroups: (a) (b) (c) A subgroup H of G is normal in G iff g Hg 1 = H, g G A subroup H of G is normal in g iff each left coset of H in G is a right coset of H in G. A subgroup H of G is normal in G iff the product of any two right cosets of H in G is a right coset of H in G. 7

(d) Every subgroup of an abelian group is normal. 19 Quotient group : Let N be any normal subgroup of G. Then the collection G/N of all right cosets of N in G forms a group under the product of two right cosets. This group is called the quotient group or a factor group. Some results on quotient group : (a) If G is a finite group and N is a normal sub group of G. Then 0 G N 0( G ) 0( N ) (b) Every quotient group of an abelian group is abelian (c) Every quotient group of a cyclic group is cyclic. 20 Homomorphism of Groups : Let (G, * ) and (G 1, o) be any two groups. then a mapping f : G G 1 is called a homomorphism from G into (onto) G 1 if (i) if is into (onto) (ii) f (a * b) = f (a) o f (b), a, b G. When the group operations for G and H are not explicitly mentioned, then the homomorphism condition becomes simply. f (ab) = f (a) f (b), for all a, b G If homomorphism f is onto i.e. f (G) = G. Then G is called homomorphic image G. Example : Let R be the group of real numbers under addition and R be the multiplicative group of non zero real numbers and f : R R 0 defined by f (x) = 2 x, x R Then f is a homomorphism. (ii) If R be the additive group of real numbers and R + the multiplicative group of positive real numbers. Then the map f : R + R defined by f(x) = log 10 x is a homomorphism. Properties of homomorphism : If f : G G be a homomorphism of groups and e, e be respectively the identity of G and G then (v) f (e) = e (ii) f (a 1 ) = [f (a)] 1 for all a G. Kermel of Homomorphism : If f : G G be a homomorphism of a group G into G. Then Kernel of f denoted by k or k f or Kerf is the set of all those elements of G whose image under f is given by the identites e 1 of G 1. i.e. k = {x G f (x) = e 1 } 8

21 Isomorphism of groups : Let G and G 1 be two groups. Then a mapping f : G G 1 is said to be an isomorphism from G into (onto) G 1 if (vi) f is into (onto). (vii) f is one one. (viii)f is a homomorphism. i.e. the An isomorphism is a special case of homomorphism when it is also one one. If an isomorphism is onto then G and G 1 are said to be isomorphic groups symbolically it is denoted by G G. Some result on homomorphism and isomorphism (a) If f : G G is a homomorphism from a group G into G with Kernel K. They K is a normal sub group of G. (b) A homomorphism f of a group G into G is an isomorphism iff ker f = {e} (c) Fundamental theorem on homomorphism: If f is a homomorphism of a group G onto G with kernel k, then G/k G Every homomorphic image of a group G is isomorphic to some quotient group. OR 1.1Automorphism: Definition 1.1.1. An Isomorphism from a group G onto itself is called an automorphism of G. i.e. a mapping f : G G is an automorphism if (i) f is one one (ii) ) f is onto (iii) f is a homomorphism. The set of all automorphisms of group G is denoted by Aut (G). Examples: Example (1): An identity map I : G G defined by I(x) = x, x G is an automorphism. Solution: I is obviously a one one and onto map. Also for all x, y G x. y G I (x. y) = x. y I (x. y) = I(x). I(y) Example (2): Let f : Z Z, where Z is the group of integers with respect to addition, defined by f(x) = x, z Z is an automorphism. Solution: Let f : Z Z is defined by f (x) = x, z Z.(i) 9

f is one one : Let f(x 1 ) = f (x 2 ), for x 1, x 2 Z x 1 = x 2 x 1 = x 2 f is onto : For any x Z x Z f ( x) = ( x) = x. Thus f is onto f is a homomorphism : For Any x 1, x 2 Z we have f (x 1 + x 2 ) = (x 1 + x 2 ), for x 1 + x 2 Z = x 1 x 2 = ( x 1 ) + ( x 2 ) = f(x 1 ) + f(x 2 ) Hence f : Z Z is an automorphism. Example (3) : For any group G, f : G G is defined by f(x) = x 1, x G is an automorphism if and only if G is abelian. Solution : first assume that G is abelian. Let f (x) = f (y) for any x, y G x 1 = y 1 1 1 1 1 x y x = y thus f is one one Also f (xy) = (xy) 1 = y 1 x 1 = f (y) f(x) = f(x) f(y) Since f is onto. Hence f is an automorphism. Conversely assume that f : G G is an automorphism. To prove G is abelian, Let x and y be any two elements of G, then x, y G x. y G f (x y) = (x. y) 1 G = y.. 1 x 1 (By reversal Law) = f(y). f (x) = f (y. x) x. y = y. x ( f is 1 1) Example 4 : Let G be a finite abelian group of order n (n = odd > 1). Show that G has a nontrivial automorphism. Solution : Step I : Define f : G G by f(x) = x 1, x G and prove f is an automorphism as above Step II : To prove that f is a nontrivial automorphism we shall prove that f I 10

Assume if possible f = I f(x) = I(x) = x x 1 = x xx 1 = xx e = x 2 0 (x) = 2 0(x) 2, x G 0(x)= 1 or 0(x) = 2 If 0(x) = 1, then x = e, which is not true and if 0(x) = 2 0(x) 0(G) 2 0(G) 0(G) is even Which is again a contradiction, Hence our assumption that f = I is wrong. f I CHECK YOUR PROGRESS: (i) If G is non abelian group then the mapping f : G G defined by f(x) = x 1 is not an automorphism. (ii) Let G be the multiplicative group of all positive real numbers. Then a mapping f : G G defined by f(x) = x 2 is an automorphism. (iii) Let C be the multiplicative group of all non zero complex numbers. Then the mapping f : C C defined by f(x+iy) = (x iy) is an automorphism. Theorem 1.1.2 : Let G be a group and an automorphism of G. If a G is of order o(a) > 0, then 0( (a)) = o(a). Proof : Suppose that is an automorphism of a group G, and suppose that a G be such that o(a) = n (i.e. a n = e But no lower positive power). Then (a) n = (a) (a) (a)..n times) = (a, a..n times) = (a n ) = (e) = e Suppose, if possible, (a) m = e for some 0 < m 0 < n, then (a m ) = (a, a..m times = (a) (a) (a)..m times) = (a) m = (e) = (e) ( is homomorphism) a m = e ( is 1-1) This is a contradiction. Thus we conclude that o( (a)) = n. Hence o( (a)) = o (a) 11

1.2Inner automorphism : Definition 1.2.1. Let G be a group then for a fixed element g G, the mapping T g : G G defined by T g (x) = gxg 1, x G (i) is an automorphism of G and is called an inner automorphism of G. Proof : T g is one one : For x 1 and x 2 be any two elements of G such that T g (x 1 ) = T g (x 2 ) gx 1 g 1 = gx 2 g 1, x G (by (i)) g 1 (gx 1 g 1 )g = g 1 (gx 2 g 1 )g (g 1 g) x 1 (g 1 g) = (g 1 g) x 2 (g 1 g) x 1 = x 2 T g is one one T g is onto : For any y G, an element g 1 y G such that T g (g 1 yg) = g (g 1 yg) g 1 = (gg 1 ) y (gg 1 ) = eye = y Thus T g is a mapping of G onto G itself. T g is a homomorphism : Let x, y G. Then T g (xy) = g (xy)g 1 = g(xg 1 gy) g 1 = (g xg 1 ) (gyg 1 ) = Tg (x) Tg(y) Thus T g is a homomorphism. But T g is also one one and onto. Hence T g is an automorphism of G. The set of all inner automorphisms of G is denoted by I (G) i.e. I(G) = { Tg Aut (G) g G } = { g xg 1 / x G} Remark (i)since T g is an automorphism of G defined for a particular element of G. Therefore T g is a special kind of an automorphism called an inner automorphism. Definition 1.2.2 : Self Conjugate element : Let G be a group. An element a G is said to be self conjugate iff a = x 1 ax, x G i.e. iff x a = ax, x G Thus an element a of a group G is self conjugate iff a commutes with every element of G. 12

Definition 1.2.3 : Centre of a group : The set of all self conjugate elements of a group G is called the centre of G and is denoted by Z (G) or simply by z. i.e. Z = { z G zx = xz, x G} Lemma 1.2.4 The Centre of a group Z is a normal sub group of G. Proof: We have centre of G i.e. z is given by Z = { z G zx = xz x G} First we shall prove that Z is a sub group of G. Let z 1, z 2 G. They z 1 x = xz 1 & z 2 x = x z 2 x G Now z 2 x = x z 2, x G 1 1 z ( z x )z = 2 2 2 z 1 2 ( xz 2 )z 1 2, x G. (i) x z 1 1 z x, x G (ii) z 2 1 2 2 Z, z 2 Z 1 To prove Z is a sub group we shall prove that z 1 z2 Z, z 1, z 2 Z We have ( z z 1 1 2 )x z ( z 1 1 2 x ) 1 = z ( xz ) = 1 1 2 ( z x )z 1 2.(by ii) 1 = ( xz ) (by i) 1 z 2 1 = x ( z z ), x Z 1 2 1 z 1 z 1 Z, z 1, z 2 Z Hence Z is a sub group of G. To prove Z is a normal sub group. Let x be any element of G and z Z. Then zx = xz x G. Now xzx 1 = (xz)x 1 = (zx)x 1 ( z x = xz) = z (xx 1 ) = z But z Z x z x 1 Z, x G, z Z Hence Z is a normal sub group of G. 1.3. Automorphism group: Theorem 1.3.1 : The set of all automorphism of G i.e. Aut(G) forms a group under the usual composition of mappings. 13

Proof : Let Aut (G) be the set of all automorphisms of G. Since I is a trivial automorphism of G. therefore I Aut (G). Thus Aut(G) is a nonempty set. Now we shall prove that Aut(G) forms a group under the composition of mapping first of all note that Aut(G) is closed under the composition. Let T 1, T 2 Aut(G) we shall show that T 1 T 2 Aut(G). Since T 1 & T 2 are bijective only. Therefore T 1 T 2 is also a bijective map. Also for any x, y G, we have (T 1 T 2 ) (xy) = T 1 (T 2 (xy)) = T 1 (T 2 (x) T 2 (y)) = T 1 (T 2 (x)) T 1 (T 2 (y) = (T 1 T 2 ) (x) (T 1 T 2 ) (y) Thus T 1 T 2 is an automorphism of G. T 1 T 2 Aut (G), for all T 1, T 2 Aut (G) Since the composition of mapping is associative, therefore Aut(G) also holds associativity. Again I, the identities mapping is an automorphism of G. Therefore I Aut(G) is an identity of Aut(G). We have only to show that if TAut (G) then T 1 Aut (G), Now T(T 1 (x)t 1 (y) = (T(T 1 (x) (T (T 1 )(y)) = ((TT 1 )(x) (TT 1 )(y) = I (x). I (y) = xy This means that T 1 (xy) = T 1 (x). T 1 (y) Hence Aut(G) is a group w.r.t. the composition of mappings as composition. Discussion :We have just established that Aut(G), the set of all automorphisms of group G forms a group. It is now a pertinent question whether Aut(G) is a nontrivial group or whether it consists of trivial automorphism I only. In other words we want to confirm that do there exist nontrivial automorphisms. The confirmation is left to the student in the following exercise. CHECK YOUR PROGRESS: (I) If G is a group having only two elements, then Aut(G) consists only of I(trivial - automorphism). (II) If G has more than two elements than Aut(G) always has more than one elements i.e G has a non trivial automorphism. (Hint: See the self study example 2) 14

Definition 1.3.2Permutation: Let G be a finite group of order n. Then a one one mapping of G onto itself is called a permutation of degree n. Theorem 1.3.3: Let G be a group. Let A(G) be the group of all permutations on G. Then Aut(G), the set of all automorphisms of G is a sub group of A(G). Proof : Since every member of Aut(G) is a one one homomorphism mapping of a group G auto G itself i.e. every number of Aut(G) is a permutation on G. Therefore T Aut (G) T A (G) Aut(G) A(G) The remaining proof is left to the students. Hints : Step (1) Take T 1 T 2 Aut (G) show that T 1 T 2 Aut (G) by proving T 1 T 2 is also one one, onto & homomorphism. Step (2) Take T Aut (G) & show that T 1 Aut (G). Hence Aut(G) is a sub group of A(G). 1.4 Conjugacy relation and Centralizer : Definitation 1.4.1 : Conjugate element : Let G be a group. An element a G is called conjugate to an element bg if there exists an element xg such that a = x 1 bx, for some xg If a = x 1 bx, then some times we also say that a is transform of b by x. Remark : Here the element x is not unique. Definition 1.4.2 Conjugacy Relation: If a is conjugate to b then we write a b and this relation is known as relation of conjugacy. Lemma 1.4.3 : Conjugacy is an equivalence relation on G. Proof To prove that the relation of conjugacy is an equivalence relation on G. We shall prove that this relation is reflexive, symmetric and transitive. (i) Reflexivity : Since a = e 1 ae, for each ag. Where e is the identity of G. a a, for each a G Thus the relation is reflexive. (ii) Symmetry : Let a,, bg be such that 15

16 a b a = x 1 b x, for some x G x 1 G such that xax 1 = x (x 1 bx) x 1 xax 1 = (xx 1 )bxx 1 xax 1 =b (x 1) 1 ax 1 = b y 1 ay = b, for y = x 1 G b a. Thus the relation is symmetric. (iii) Transitivity : Let a, b, c G be such that a b and b c x, y G such that a = x 1 bx and b = y 1 cy a = x 1 (y 1 cy)x a = (x 1 y 1 ) c (yx) a = (yx) 1 c (yx) (by reversal law) a = z 1 cz, for z = yx G a c. Thus the relation is transitive. Hence the relation is an equivalence relation on G. Examples 1 Let S 3 be a symmetric group on three symbols 1, 2, 3, then (i) (1, 2, 3) is conjugate of (1, 3, 2) (ii) (1, 3) is conjugate of (2, 3) Solution : (2 3) 1 (1 32) (2 3) = = 1 2 3 3 2 1 1 1 1 2 2 3 3 1 2 3 3 2 1 1 = 3 2 2 3 1 1 1 2 2 3 3 1 2 3 3 2 1 1 = 3 2 2 3 1 1 2 3 1 2 3 1 3 2 1 1 2 3 = 3 2 1 3 2 1 = 1 3 3 2 2 1 = (1 2 3) Hence (1 2 3) = ( 2 3) 1 (1 3 2) (2 3)

(2) Solution is Left to the students. Theorem 1.4.5 : Any two elements of an abelian group G are conjugate if and only if they are equal. Proof : Let G be an abelian group. First assume that a, b G are conjugate i.e. a b a = x 1 bx, for some x G a = (x 1 b) x a = (bx 1 )x ( G is abelian) a = b (x 1 x ) = bc a = b Conversely assume that a = b, where a, b G a = be a = b (x 1 x) a = (bx 1 ) x a = (x 1 b) x, (G is abelian) a = x 1 bx a b Hence a and b are conjugate. Definition1.4.6 Conjugate class: Since an equivalence relation defined on a set decomposes (partition) the set into mutually disjoint equivalence classes. Hence the relation of conjugacy on a group G partitions G into mutually disjoint equivalence classes known as classes of conjugate elements. For an element a G, the conjugate class of a is denoted by C(a) defined as the set of all those elements of G which are conjugate to a. i.e. C(a) = set of all conjugates of a = { x 1 ax x G} i.e. C(a) consists of the set of all distinct elements of the form x 1 ax as x ranges over G. If G is finite, then the number of distinct elements in C(a) will be finite and denoted by c a i.e. o C(a) = c a Note : (i) For the identity e of G. C(e) = {y G ye } = { y G y = x 1 ex, for some x G} = {x 1 ex x G} = {e} (ii) If G is an abetian group, then C(a) = { x 1 ax x G} = {ax 1 x x G} = {a}, a G Theorem 1.4.7: Let the conjugacy relation defined on G, then for a, b G 17

(i) a C(a) (ii) b C(a) C(a) = C(b) (iii) Either C(a) C(b) = or C(a) = C(b) i.e. two conjugate classes are either disjoint or identical. Proof : (i) Since the conjugacy relation reflexive on G. a a, for each a G a C(a), for each a G. (ii) Let b C(a) b a (i) Let x C(a) x a (ii) From (i) and (ii) x a and b a x a and a b ( is symmetric) x b ( is transitive) x C(b), x C(a) C(a) C(b).. (iii) Now let y C(b) y b. (iv) From (i) and (iv), y b and b a y a ( is transitive) y C(a), C(b) C(b) C(a) (v) From (iii) and (v) C(a) = C(b) Conversely assume that C(a) = C(b) Since every element belongs to its conjugate class, therefore b C(b) b C(a) (since C(a) = C(b)) Proof (iii) Suppose C(a) C(b) =. Then the theorem is proved. So let C(a) C(b) an element say x such that x C(a) C(b) x C(a) & x C(b) x a and x b a x & x b ( is symmetric) a b ( is transitive) a C(b) C(a) = C(b) (by part (ii) ) Definition 1.4.8: Self conjugate element : An element of a group G is called self conjugate iff a is the only element conjugate to a. 18

i.e. C(a) = {a} x 1 ax = a, x G ax = xa, x G i.e. a is self conjugate iff a commutes with every element of the group. Definition 1.4.9 Centralizer : For any non empty subset A of a group G, the Centralizer C(A) of A is defined as C(A) = { x G ax = xa,, a A } Definition 1.4.10 : Centralizer of Sub group : The centralizer C(H) of a sub group H of a group G is defined as C(H) = {x G hx = xh, h H } Example: Let G = { 1, i, j, k}. Then G defines a group under usual multiplication together with i 2 = j 2 = k 2 = 1 and ij = ji = k, jk = kj= i & ki= ki = j.g is called quarterion group. If H = { 1, i} then H is a sub group of G and C(H) = { 1, i} Theorem 1.4.11: C(H) is sub group of G. Proof: We have C(H) the centralizer of a sub group H is defined as C(H) = { x G hx = xh h H } Since eh = he, h H e C(H) C(H) Let x and y be any two elements of C(H) then x C(H) hx = xh, h H and y C(H) hy = yh, h H..(i) First we shall prove that y 1 C(H), for any y C(H) hy = yh, h H y 1 (hy) y 1 = y 1 (yh)y 1 y 1 h = hy 1, h H..(ii) y 1 C(H) Finally we shall prove that xy 1 C(H), x, y C(H) Now for any h H (xy 1 ) h = x (y 1 h) = x (hy 1 ) by (ii) = ((xh) y 1 ) = (hx) y 1 by (i) = h (xy 1 ) 19

x y 1 C(H), x, y C(H) Hence C(H) is a sub group of G. 1.5 Normalizer: Definition 1.5.1 Normalizer of an element: The normalizer of an element a G is the set of all those elements of G which commute with a and is denoted by N(a). i.e. N(a) = {x G ax = xa} Note : (1) Since ex = xe, x G x N(e), x G G N(e) But N(e) G N(e) = G (ii) If G is abelian, then ax = xa, x G G N(a) But N(a) G N(a) = G Theorem 1.5.2 : The normalizer N(a) of a G is a sub group of G. Proof : We have N(a) = {x G ax = xa} Let x 1, x 2 be any two elements of N(a). Then x 1, x 2 N(a) ax 1 = x 1 a.. (i) and ax 2 = x 2 a...(ii) First we shall show that We have ax 2 = x 2 a Now we shall prove that x 1 We have a (x 1 x N(a) 1 2 1 2 x x x 1 2 x (ax 2 ) 1 2 1 2 1 2 N(a). x ) = (ax 1 ) = (x 1 a) x = 1 2 x ( x 2 a ) x 1 2 1 2 1 a = a x (iii) N(a) x 1 2 1 2 1 = x 1 (a x ) 2 x (by (i) ) 2 1 = x 1 ( x a) (by (iii)) = (x 1 x 1 x 2 x 1 2 1 2 ) a N (a), x 1, x 2 N(a) Hence N(a) is a sub group of G. Remark : N(a) is not necessarily a normal sub group of G. 20

Definition 1.5.3 Normalizer of a sub group: For any sub group H of a group g, the normalizer N(H) of H is defined as N(H ) = { x G xh = Hx } Theorem 1.5.4 N(H) is a sub group of H. The proof of this theorem is left to the students. Theorem 1.5.5 C(H) N(H). But the converse is not true. Proof : We have for any sub group H of a group G, the centralizer C(H) and the normalizer N(H) of a sub group H are defined as C(H) = { x G hx = xh, h H } & N(H) = { x H xh = xh} For any x C(H) hx = xh, h H Hx = xh x N(H) C(H) N(H) How ever C(H) need not be equal to N(H). As considered the quarterian group G = { 1, i, j, k} and H = { 1, i} Then N(H) = G and C(H) = { 1, i} showing that C(H) N(H) Theorem 1.5.6: Let a be any element of G. Then two elements x, y G give rise to the same conjugate to a if and only if they belong to the same right coset of the normalizer of a in G. Proof : Let x, y G give rise to the same conjugate of a in G, then x 1 ax = y 1 ay x(x 1 ax) y 1 = x (y 1 ay) y 1 axy 1 = xy 1 a xy 1 N(a) But N(a) is a sub group of G. xy 1 N(a) N(a) x = N(a) y ( a Hb Ha = Hb) x N(a) x = N(a) y ( a Ha) and y N(a) y = N(a) x x, y N(a)x = N(a)y Hence x, y belongs to the same right coset of the normalizer of a in G. Conversely assume that x, y G belongs to the same right cosets of a in G, Let x, y N(a) z, for some z G N(a) z = N(a) x = N(a) y N(a) x = N(a) y xy 1 N(a) (Ha = Hb ab 1 H) 21

axy 1 = xy 1 a x 1 (axy 1 )y = x 1 (xy 1 a)y x 1 ax = y 1 ay Hence x, y G give rise to the same conjugate of a in G. 1.6 Counting Principle and the class equation of group Theorem 1.6.1 : Let G be a finite group and a G. Let N(a) denotes the normalizer of a 0( G ) G and C(a), the conjugate class of a G. Then 0(C(a)) = i.e., number of distinct 0( N( a )) elements conjugate to a in G is the index of the normalizer of a in G. Proof. Let G be a finite group and a G. To show o(c(a)) = 0( G ) 0( N( a )) By definition, we have N(a)= {x G: ax = xa}and C(a) = { x 1 ax : x G } Let M be the set of right cosets of N(a) in G. i.e. M = { N(a)x : x G} Let us Define a map f : M C(a) such that f [N(a).x]=x l ax xg. Obviously, f is well defined. To show f is one-one and unto. (i) f is one-one : Consider f [N(a) x]=f [N(a) y], x, y G (ii) f is onto : x 1 ax = y _1 ay x (x 1 a x)y 1 = x (y 1 ay) y 1 (xx 1 ) (axy 1 ) = (xy 1 a) (yy 1 ) e(axy 1 ) = (xy 1 a)e axy 1 =xy 1, a a(xy 1 ) = (xy 1 )a xy _1 N(a) a(xy 1 ) = (xy 1 )a xy 1 N(a) N(a)x = N(a)y f is one-one. For given any x 1 axc(a), N(a) xm such that 22

Hence f (N(a)x) = x 1 ax f is onto. o (C(a)) = o (M) = number of distinct right cosets of N(a) in G. = index of N(a) in G. 0( G ) =, (by definition of index.) 0( N( a )) Theorem 1.6.2 : If G is a finite group, then o(g)= element a in each conjugate class. 0( G ) where this sum runs over one 0( N( a )) Proof. Since, we know that, the relation of conjugacy is an equivalence relation on the set G. There fore relation partitions the G into disjoint equivalence classes. Let C(a),C(b), C(c) etc. respectively denote the conjugate classes of elements a, b, c G. Also let 0(C(a))=c a,o(c(b))=c b, 0(C(c)) = c c. Then G = C(a)) C(b) C(c)... o (G) = o(c(a))+o (C(b))+o (C(c))+... ( C(a), C(b), C(c) are mutually disjoint) But o(c(a)) = 0(G) = = o(c(a)) where the sum runs over each elements of conjugate class. 0( G ) 0( N( a )) 0( G ) 0( N( a )) Remark : The above equation (I) is known as class equation of G. [By theorem 1.6.1]..(1) Theorem 1.6.3. Let Z denote the centre of a group G. Then show that a Z iff N (a) = G. Also show that if G is finite, then a Z iff 0(N(a)) = 0(G) Proof. Let Z denotes the centre of group G. Let us first suppose a Z. To show N(a) = G. Since, we know that N(a) is a subgroup of G and N(a)={x G : a x = xa} and Z = {x G : z x = x z, xg.} Let y G and a Z, then ay = ya (by definition of Z) y N{a) G N(a) But N(a) G G = N(a). Conversely, let G = N(a), where a G. 23

To show that a G. N(a) = G G N (a) y G y N(a) ay = ya ay = ya, yg a Z. Now, let G is finite. To show that a Z 0(G) = 0(N(a)) We have, az N(a) = G. Also, G is finite 0(N(a)) = 0(G) Theorem 1.6.4 : If G is a finite group and Z is its centre, then class equation of G is expressible as 0( G) o(g) = o(z) +, a Z 0( N( a)) where the summation runs over one element a in each conjugate class. Proof : Let a be any element of a group G. Then class equation of G is given by We know that or 0(G) = a Z 0(N(a)) = 0(G) 0( G ) a Z = 1 0( N( a )) 0( G ) a Z 0( N( a )) 0( G ) 0( N( a )) 0( G ) = 1 0( N( a )) (I) = 0(Z).(2) 0( G ) Now (1) can be written as o(g) = 0( G) + a Z 0( N( a )) a Z 0( N( a)) 0( G) o(g) = o (Z) + (By (2) ) a Z 0( N( a)) Theorem 1.6.5 : If G is group of order p n, where p is a prime number then Z(G) (e) OR A group of prime order must always have a non-trivial center. Proof : If a G, since N (a) is a subgroup of G, by Lagrange s theorem we have o(n(a)) o(g) 24

o (N(a)) p n, p being prime o (N(a)) = p n, where n a n. Let Z(G) denote the center of G. Then a Z(G) o(n(a)) = o(g) p n a = p n n a = n Let o(z(g)) = z, writing the class equation of G, we have p n o( G ) = o(g) = n a no( N( a )) o( G ) o( G ) = + = n a no( N( a )) n a no( N( a )) n o( G ) p = + n a Z( G ) o( N( a )) n n p a n p = o(z(g) + n p a na n n p = z + n p a na n a..(1) Since p is a divisor of p n, since n a < n for each term in the Σ of the right side, we must have p n p n a = n na p p, n a < n Now p p n p and p p na n n na p p p na n n na p (p n n p na ) p na n p o(z(g)) Since xe = ex x G e Z (G) and since z 0, it must be a positive integer divisible by the prime p. But 2 is the least prime number, and so z > 1. This situation demands that there must be an element, besides e, in Z (G). Hence Z(G) (e) Theorem 1.6.6. A group of order p 2 is abelian, where p is prime number. 25

Proof. Let G be a group of order p 2. Then we have to show that G is abelian. For this, we shall show that Z = G. (Then G will be abelian, since the centre Z is abelian). Since p is prime, then Z (e). Therefore, o(z) > 1. Again since Z is a subgroup of G. Then, by Lagrange's theorem o (Z) o(g) i.e., o(z ) p 2 If o(z) = p 2 either o(z) = p or o(z) - p 2. o(z) = o(g) Z = G, where Z is abelian G is also abelian and the result follows. Now, suppose o(z). = p, so o(z) < o(g). Then there must exist an element which is in G but is not in Z. Let a G and az. Since a commutes with itself, a N(a). Also, N(a) is a subgroup of G. Let xz so xa = ax x commutes with a x N(a). Thus Now x Z x N(a) Z N(a) a Z a N(a), Z N(a) o(n(a)) > 0 (Z) o(n(a)) > p But o(n(a)) o(g) = p 2 (By Lagrange s theorem) o(n(a)) = p 2 (o(n(a))/p 2 ) N(a) = G a Z which is a contradiction. Thus o(z) = p is not possible o(z)=p 2. Hence Z = G G is abelian. Remark : It can be seen that a group of order 9 is abelian. For, p = 3 p 2 = 3 2 = 9 i.e. o (G) = p 2, where p = 3 is a prime o (G)=3 2 = 9 G is abelian. (By above theorem) 1.7 Cauchy s Theorem and Sylow s Theorems for finite abelian groups and Non-Abelian Groups : Thorem 1.7.1(Cauchy s Theorem for abelian Groups) : Let G be a finite abelian group, and let p be a prime. If p o(g), then there is an element a e G such that a p = e. 26

Proof. We shall prove this theorem by induction on o(g). In other words, we assume that the theorem is true for all abelian groups of order less than that of o(g). From this we shall prove that the result is also true for G. We note that the theorem is obiviously true for group G with o(g) = 1; i.e., G = {e}. If G has only improper subgroups; i.e., G has no subgroups H (e), G, then G must be of prime order since every group of composite order possesses proper subgroups. Now p is a prime number and p o(g). So, o(g)= p. Further, we know that every group of prime order is cyclic. Therefore, each element a e of G will be a generator of G. Hence excluding e, the remaining p - 1 elements a ( e) of G are such that a 0(G) = e i.e., a p = e. So suppose that G has proper subgroups N, then N (e), G. Now there arise two possibilities : either p o(n) or p o(n). If p o(n), by our induction hypothesis, since o(n) < o(g) and N is abelian, b N, b e such that b p = e. Since b N G, So we have shown that an element G, b e"such that b p = e Now Let p o(n). Since G is abelian, N is a normal subgroup of G, so G / N is a group. Moreover, 0 (G/ N) = Since, p o(g) and p 0( G ) 0( N ) 0( G ) o(g) = 0(N) o(g) = 0 (G/ N) 0(N) (Since N is normal) 0( N ) P o(g\n) = o(n), it follows from (1) that 0( G ) < 0 (G) [o(n) > 1] 0( N ) Also, since G is abelian, G/N is abelian. Thus by our induction hypothesis, an element Nb G/N satisfying (Nb) n = N, the identity element of G/N, Nb N; in the quotient group G/N we have o(nb) = P. We also have (Nb) p = N, Nb N Nb P =N, Nb N b p N, b N, b e Now using Lagrange's theorem, we have (b p ) 0(N) = e b 0(N)p = e (b 0(N) ) p = e Let b 0{N) = c. Then certainly c p = e. b 27

In order to show that c is the required element that satisfies the conclusion of the we must finally show that c e. However, if c = e then we have b 0(N) = c, c = e b 0(N) = e Nb 0(N) = Ne (Nb) 0(N) = N.(2) Also in the quotient group G/N, o(nb) =p; i.e., (Nb) p = N, p o(n), p is a prime number, find from relation (2) that Nb = N theorem bn, a contradiction to the fact that bn Thus c e, c p = e and we have completed the induction. Hence the theorem. we Theorem 1.7.2 : (Cauchy's Theorem for Non-abelian Groups): If G is a finite group, p is a prime number and p o(g), then G has an element of order p. Proof. Let G be a finite group, p be a prime number and p o(g). We seek an element a eg satisfying a p = e. To prove its existence we apply the method of induction on o(g); that is, we assume that the theorem is true for all groups, of order less than o(g) and finally we shall show that it is true for G. If o(g) = 1; i.e., G = {e} then the theorem is obviously true, we start the induction. Suppose H is a proper subgroup of G. Now two easels arise (i) p 0(H), and (ii) p o(h). Suppose there exists a subgroups H G of G such that p / 0(H), then o(h) < o(g). Therefore by induction the theorem is true for H. So an element ah such that o(a) = p. But a H, H G A G Therefore, in element a e G such that o (a) = p Now we assume that p 0(H) t where H is any proper subgroup of G. Suppose Z(G) is the center of G. We now write the class equation for G : o(g) = o(z(g)) + 0( G ) a Z( G ) 0( N( a )) 0( g ) i.e. o(g) = o(z(g)) + N ( a ) ( G ) 0( N( a )) We know that N(a) is a subgroup of G. If a Z(G), then clearly N(a) G. Therefore, by our assumption, p o(n/(a)). farther, if N(a) G..(1) Then c a = 0( G ) 0( N( a )) 28

o(g) = c a o(n(a))...(2) Since p o(g) and p o(n(a)), it follows from (2) that 0( G ) p c a ; i.e. p ; if N(a) G and so 0 ( N( a )) p 0( G ) N ( a )G 0(( N( a )) 0( G ) P 0(G), and p N ( a )G 0(( N( a )) 0( G ) p 0 ( G ) N ( a )G 0(( N( a )) p o(z(g)) [by(l)] Thus, Z(G), is a proper subgroup of G such that p o(z(g)). But, by our assumption, p is not a divisor of any proper subgroup of G, and so Z(G) cannot be a proper subgroup of G. This situation forced us to accept the only possibility left us, namely, that Z(G) = G. But then G is abelian. We now invoke the Cauchy's theorem for abelian groups which ensures us. the existence of an element A e G such that a p = e. But p is a prime number and so o(a) = p. this proves the theorem. Definition 1.7.3 Sylow s p-sub Group : Definition 1 (p-group) Let p be a fixed prime number. Then a group G is said to be p-group if every element of G has a order, a power of p. Definition 2 (p-sub group) Let p be a prime number. Then a subgroup H of a group G is said to be p-sub group if every element of H is a power of p. Definition Sylow p-sub Group: Let G be a group and p be a prime number such that p n divide order of G and p n+1 does not divide it. Then, a subgroup H of G such that o(h) = p n is known a Sylow p-sub group of G or p-sylow subgroup of G. 29

Theorem 1.7.4. (Sylow's first Theorem) Let G be a group and p be any prime number and m be a positive integer such that p m divides o(g). then, there exists a subgroup H of G such that o(h) = p m. Proof. We shall prove this theorem by induction. When o(g) = 1, then theorem is obvious. Further, assume that result is true for all groups with order less than o(g) let p m 0(G). If K is a subgroup of G such that K G and p m o(k), then, by induction H K such that o(h) = p m. H K H G. Therefore, result holds in this case. Now, assume p m does not divide order of any proper subgroup of G. Consider class equation of G. 0( G ) 0(G) = 0 [Z(G)] + a Z( G ) 0( N( a )) But we have a Z (G) N(a) G p m / o (N(a)) P m / o (G) p m 0( G ). 0 (N(a)) 0N( a ) p p p 0( G ), for all a Z(G) as p m / o(n(a)) 0N( a ) 0( G ) a Z ( G ) 0( N( a )) 0( G ) 0 ( G ) = 0(Z(G)) a Z ( G ) 0( N( a )) x Z(G) such that o (x) = p Let K = { x} Z(G) K is normal in G Now 0(G / K ) < 0(G) and p m / o(g) = 0 (G) / K). 0 (K) P m / o(k) and therefore p m 1 p m o(g K) By induction hypothesis there exists a subgroup H / K of G / K such that O (H / K) p m 1 Therefore o(h) = p m Also, H / K G / K H G. Thus, result is true in this case also Hence, by mathematical induction theorem is proved. Remark : If p is prime such that p n 0(G) and p n+1 o(g), then a p-sylow subgroup of G. 30

Theorem 1.7.5 (Sylow s Second Theorem) : Any two sylow p-subgroup of a finite group G are conjugate in G. Proof : Let G be a group and P, Q be Sylow s p-sub groups of G such that O(P) = p n = o (Q), where p n+1 O(G) Let us suppose P and Q are not conjugate in G, i.e. P g Q g 1 for any g G. It is known that o( P )o( Q ) O (P Q) = 1 o( P xqx ) Now, since P ( x Q x 1 ) O (P (x Q x 1 )) = p m, m n If m = n, then we have P (x Q x 1 )) = p P (x Q x 1 )) P = (x Q x 1 )) [0 (xqx 1 ) = 0(Q) = 0(P)] Which is a contradiction Therefore, we get m < n and hence 0(P Q) = p 2n m, m < n, xg which implies 0(P Q) = p n+1 (P n m+1 ) = multiple of p n+1 Therefore o(g) = P Q x p n+1 o = multiple of p n+1 RHS p n+1 o(g) Which is a contradiction. Hence, P = g Qg 1, for some g G Remark : Let P be a Sylow p-subgroup of G. Then the number of Sylow p-subgroup of G o( G ) is equal to o( N( p )) Theorem 1.7.6 (Sylow s third Theorem) : The number of Sylow p-sub groups of G is of the form 1 + kp where (1+kp) o(g), k being a non-negative integer. Proof : Let G be a group and P be a Sylow p-subgroup G. Let o(p) = p n Now G = x P P xn( P ) P P xn( P ) P P x N(P) P x = xp 31

Thus P x = P x P [PP=P] xn( P ) P P xn( P ) P N (P) Since P N(P) and union of disjoint right sets equal the set, therefore x N(P) P x = x P xpx 1 P o(p xpx 1 ) = p m, m < n [By Sylow s second theorem] o(p xo) = p 2n m, m < n Therefore, o(g) = o (N(P) + o P P Thus = (N(P) + xn( P ) x N( P ) p 2n m 2nm n1 o( G ) p p.t 1 1 where t is an integer o( N( P )) o( N( P )) o( N( P )) Further, since LHS = integer p n 1.t = r, an integer o( N( P )) Therefore p n+1. t = r. o(n(p)) Also P N(P) o(p) o(n(p)) p n o(n(p)) o(n(p)) = p n.u Thus p n+1.t = r.o(n(p)) pt = r.u p ru If p u, then p n+1 o(n(p)) o(g) p n+1 o(g), which is a contradiction r Therefore, p r = integer p Thus t r = integer k = p p n1 o( G ) p.t 1 1 o( N( P )) o( N( P )) p.t u = 1 + kp Then by above remark, we get o( G ) number of Sylow p-subgroup of g. o( N( P )) 32

Thus the number of Sylow p-subgroup is of the from 1 + kp = Hence, ( 1 + kp) o(g) o( G ) o( N( P )) Remarks : (i) If o(g) = p n.q. (p q) = 1 then the number of Sylow p-subgroup is 1+kp where 1+kp p n q which implies (1+kp q as (1+kp, p n ) = 1 (ii) Every p-subgroup of a finite group G is contained in some Sylow s p- subgroup of G. (iii)if G is a finite group and P is a p-subgroup of G, then P is Sylow p-subgroup of G if and only if no p-subgroup of g properly contains P. POINTS FOR DISCUSSION/ CLARIFICATION: After going through the block you may like to seek discussion/ clarification on some points. If so mention these points below: (I) Points for Discussion: (II) Points for Clarification: 33

REFERENCES: 1. Topics in Algebra, I.N.Herstien, 2. Algebra (Graduate Texts in Mathematics), Thomas W. Hungerfold, Springer Verlag. 3. Abstract Algebra, David S. Dummit and Richard M. Foote, John Wiley & Sons, Inc.. New York. 4. Basic Algebra Vol. I, N. Jacbson, Hindustan Publishing Corporation, New Delhi 5. A First Course in Algebra, John B. Fraleigh, Narosa Publishing House, New Delhi BLOCK 1 ABSTRACT ALGEBRA UNIT 2 RING THEORY Structure : 2.1 Ring Homomorphism 2.2 Ideals and Quotients Rings 2.3 Field of Quotients of an Integral Domain 2.4 Euclidean Rings 2.5 Polynomial Rings 2.6 Polynomials Over the Rational Field 2.7 Polynomial Over the Commutative Rings 2.8 Unique Factorization Domain Ring Theory: In general Ring is an algebraic structure with two binary operations namely addition and multiplication. In fact ring is an abelian group with some extra axioms i.e., it is a generalized form of a group, therefore many of the concepts and results of group theory can be generalized to ring. Thus Before describing the main contents of the unit, first we shall explain the fundamental concepts and elementary results of Ring theory: 1. Ring : An algebraic structure (R, +, ) is said to be ring if (i) (R, +) is an abelian group (ii) (R, ) is a semi group (iii) Both the distributive laws are satisfied in R. 2. Ring with unity: If there exists multiplicative identity in ring R then it is called ring with unity. 3. Commutative Ring: R is said to be commutative ring if a b = b a, a, b R. 4. Ring with zero divisors: If there exists a, b R such that a 0, b 0, but ab = 0, then R is said to be ring with zero divisors and a and b are called zero divisors of R. 34

5. Ring without zero divisors: If there exists a, b R such that ab = 0 either a = 0 or b = 0, i.e., the product of no two non-zero elements of R is zero, then R is said to be ring without zero divisors. 6. Elementary properties of ring: For all a, b, c R, (i) a0 = 0a = 0 (ii) a(-b) = (-a) b = -(ab) (iii) (-a) (-b) = ab (iv) a(b - c) = ab - ac (v) (b - c) a = ba - ca (vi) For a 0, ab = ac b = c and ba = ca b = c (Cancellation law). 7. Integral Domain: A commutative ring with unit element and without zero divisors is called an integral domain. 8. Skew field or division ring: A ring R with unit element is said to be skew field or division ring if every non-zero element in R has its multiplicative inverse in R. 9. Field: A commutative ring F with unit element is said to be field if every non-zero element in F has its multiplicative inverse in F. "A field is necessarily is an Integral domain. However only a finite Integral domain is a field". 10. Subring: A non-empty subset S of a ring R is called a subring of R if S itself is a ring under the operations as defined in R. "The necessary and sufficient conditions for S to be a subring are that (i) a b S, a, b S and (ii) a b S, a, b S ". "The intersection of two subrings is a subring. However the union of two subrings is a subring only when one is contained in the other". 11. Subfield : A non-empty subset K of a field F is called a subfield of F if K itself is a field under the operations as defined in F. "The necessary and sufficient conditions for K to be a subfield are that (i) a - b K, a, b K and (ii) a b -1 K, a, b K ". 2.1. Ring homomorphism. Definition2.1.1: Homomorphism of Rings. Let R and R` be two rings with two binary operations additions and multiplication, then a mapping from the ring R into the ring R` is said to be a homomorphism if (i) (a + b) = (a) + (b) (ii) (ab) = (a)(b), for all a, b R. Remark Here both rings have same binary operations, in case if R and R have respectively the operations +,. and *, o then above conditions (i) and (ii) can be written as (i) (a + b) = (a)*(b) (ii) (ab) = (a)o(b) THINGS TO REMEMBER: (1) If is surjective, i.e., (R) = R`, then R` is said to be a homomorphism image of R. (2) If is injective, it is known as isomorphism. If is bijective, then R and R` are isomorphic and is denoted by the symbol of R R. 35

Definition 2.1.2 Isomorphism of rings: A homomorphism from a ring R into a ring R is said to be isomorphism if (i) is one to one (ii) is onto i.e, An isomorphism of rings is a special case of homomorphism if it is one-one also 2.1.3 Some useful Results on Homomorphism: Theorem 1 If is a homomorphism of R into R`, then (i) (0) = 0`, 0 R, 0` R`, where o and o` are respectively the zero elements of R and R`. (ii) (-a) = (a), a R. Proof: (i) Since 0 R and a be any arbitrary element of R, then a R a + 0 = a (a + 0) = (a) (a) + (0) = (a) ( is homomorphism) Also, 0 + a = a (0 + a) = (a) (0) + (a) = (a). (a) + (0) = (a) = (0) + (a), (a) R`. This show that (0) is the zero element of R`, If 0` is the zero element of R`, then (0) = 0`. (ii) Let a be an arbitrary element of R, then a R therefore, a + (a) = 0 [a + (-a)] = (0) (a) + (-a) = (0) Also, (-a) + a = 0 (-a) + (a) = (0) Thus (a) + (-a) = (0) = (-a) + (a) (a) R`. Hence (-a) = -(a). Theorem 2 Every homomorphism image of a commutative ring is a commutative ring. Proof: Let be a homomorphism from a commutative ring R onto a ring R`, then we shall have to show that R` is a commutative ring. Let a`, b` be any elements of R`. Since is onto, then there exists two elements a and b of R such that a` = (a) and b` = (b). Now, a`b` = (a)(b) = (ab) ( is homomorphism) = (ba) ( R is commutative) = (b)(a) ( is homomorphism) = b`,a`. Hence, R` is a commutative ring. Theorem 3 If is a homomorphism of a ring R into a ring R`, then (R) is a subring of R`. Proof: Let (a) and (b) be any arbitrary elements of (R) for some a, b R. Since the ring R is itself a subring so that a b R and a b R, therefore, (a) - (b) = (a) + (-b) [ (-b) = -(b)] = [a + (-b)] (R) ( is homomorphism and a b R) 36

Now (a)(b) = (ab) (R) ( ab R) Thus, (a) - (b) (R) and (a) (b) (R). Hence, (R) is a subring of R`. Definition 2.1.4 Kernel of a ring homomorphism: Let R and R` be two rings and be a homomorphism from a ring R into a ring R` then the set of all those elements of R` which are mapped onto the zero elements of R`, is said to be kernel of and is usually denoted by ker() or K. i.e., If 0` is the zero element of R`, then K= {a R : (a) = 0`} Since (0) = 0` 0 K K 2.2 Ideals and Quotient Rings 2.2.1 Ideals Definition (Left ideal). An additive subgroup S of a ring R is said to be a left ideal of R if a S, r R r a S. Definition (Right ideal). An additive subgroup S of a ring R is said to be a right ideal of R if a S, r R ar S. Definition (Ideal). A non empty subset S of a ring R is said to be an ideal of R if it is both left and right ideal. That is, an additive subgroup S of R is an ideal of R if a S, r R ra S, ar S. THINGS TO REMEMBER: In particular, the subset {0} consisting of 0 alone and the ring R itself are ideals in R. These two ideals {0} and R are called the improper ideals of R. All other ideals of R are called proper. 2.2.2 Useful result on Ideals Theorem 1 If S is an ideal of a ring R, then S is a subring of R Proof: Since S is an ideal of a ring R, therefore S is an additive subgroup of R and hence a S, b S a b S And a S, r R ra S and ar S Also, a S, b S a S, br [since S R] ab S, S being ideal of R. Theorem 2: The intersection of any two ideals of a ring R is again an ideal of R. Proof: Let S 1 and S 2 be any two ideals of a ring R, then we have to show that S 1 an ideal of R. S 2 is also 37