* 8 Groups, with Appendix containing Rings and Fields.


 Antonia Gallagher
 4 years ago
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1 * 8 Groups, with Appendix containing Rings and Fields Binary Operations Definition We say that is a binary operation on a set S if, and only if, a, b, a b S Implicit in this definition is the idea that is closed, or as we often say, S is closed under The idea here is that we can give a value to a b from the set in which we started Example In Chapter 4 we saw that 8 is closed on {, [2] 8,, [6] 8 } but not on {[2] 8,, [6] 8 } So the set S is important in the definition above Examples Seen in the course: + on Z, Z m, on Z, Z m or Z m,, composition of permutations on S n, the set of permutations on {1, 2,, n} Example Not seen in the course: (a, b) (c, d) = (ac, bc + d) on Z Z Multiplication of n n matrices with real entries Example Subtraction,, is a binary operation on Z, since m, n Z, m n Z But is not a binary operation on N since 1, 2 N, yet 1 2 / N A binary operation might satisfy certain properties that we have seen before (PJE p18 for real numbers and p71 for sets) Definition (i) A binary operation is commutative if, a, b S, a b = b a, (ii) A binary operation is associative if, a, b, c S, (a b) c = a (b c) Example (S, ), where S is the set of all function Ω Ω, under composition, is not commutative, but is associative Example Subtraction,, on Z is neither commutative, (counterexample: ) nor associative (counterexample: 1 (2 3) (1 2) 3) 1
2 Identities and Inverses Definition Given a set S and binary operation we say that e S is an identity if, for all a S, e a = a and a e = a We have to check both e a and a e since we are not assuming that is commutative Example (i) 0 is the additive identity on R (or Z, N, C, Z m ) since 0 + x = x + 0 = x for all x R (or Z, N, C, Z m respectively) (ii) 1 is the multiplicative identity on R (or Z, N, C, Z m ) since 1 x = x 1 = x for all x R (or Z, N, C, Z m respectively) (iii) {,,,, } Here the identity is This last example is important, it shows that we get identities different to 1 and 0! Note the use of the word an in the definition But Lemma Suppose that is a binary operation on a set S and that (S, ) has an identity The identity is unique Proof Suppose that e and f are two identities on S Then e = e f since f is an identity (used here on the right), = f since e is an identity (used here on the left) So we could replace the word an by the in the definition If, in the multiplication table for (S, ), we can find an element whose row (and whose column) is identical to the heading row (respectively heading column), then we have found the identity 2
3 Definition Let S be a set with a binary operation and an identity element e S We say that an element a S is invertible if there exists b S such that a b = e and b a = e We say that b is the inverse of a, and normally write b as a 1 Example In (Z 6, ) the element [2] 6 has no inverse If it had an inverse, ie [b] 6 : [2] 6 [b] 6 = [1] 6, multiply both sides by [3] 6 to get [6] 6 [b] 6 = [3] 6, ie [0] 6 = [3] 6, a contradiction The problem here is that 6 = 2 3 is composite We have got round this in two ways in this course First we can look at (Z p, ) with p prime, when every nonzero element has an inverse The second way it to look at (Z m, ) where we simply throw away all the elements that don t have an inverse! If, for an i S we can look in its row in the multiplication table and find the identity in column j, say, and find in row j the identity in column i then i and j are inverse to each other If we can do this for every i S then every element will have an inverse Example {,,,, } Since the identity is we note = so [4] 1 20 =, = so [8] 1 20 =, = so [12] 1 20 = (The inverse of the identity is always itself!) 3
4 Semigroups Definition Let (S, ) be a nonempty set with a binary operation Then (S, ) is a semigroup if is associative Definition A semigroup (S, ) is called a commutative semigroup if is commutative Examples (Z, ), (N, +), (Z m, ), (Z m, ) are all commutative semigroups while S, be the set of all maps from X to X with the composition of functions is not Definition Suppose (S, ) is a semigroup and let a S Define iteratively the positive powers of a by a 1 = a and a n+1 = a a n for all n 1 Example This is exactly how we defined positive powers of a permutation Note For the theory of semigroups (and later, groups) we think of as a form of multiplication If the operation were additive, written say as, then the corresponding definition of powers would be (n + 1) a = a na Lemma Index Laws Let (S, ) be a semigroup and a S Then for all n, m N, (i) a m a n = a m+n, (ii) (a n ) m = a mn Proof (i) Let n 1 be given Use proof by induction for m 1 to show that m 1, a m a n = a m+n, (1) First consider m = 1 Then a 1 a n = a a n by definition of a 1 = a n+1 by definition of a n+1 Thus result holds for m = 1 Assume result holds for m = k, ie a k a n = a k+n Consider a k+1 a n = ( a a k) a n by definition of a k+1, = a ( a k a n) since is associative, = a a k+n by inductive hypothesis, = a (k+n)+1 by definition of power, = a (k+1)+n 4
5 Hence result holds for m = k + 1 Thus by induction, (1) holds Since n was arbitrary we have shown as required n 1, m 1, a m a n = a m+n (ii) Let n 1 be given Use proof by induction for m 1 to show that First consider m = 1 Then Thus result holds for m = 1 m 1, (a n ) m = a mn, (2) (a n ) 1 = a n by definition of first power = a 1 a Assume result holds for m = k, ie (a n ) k = a kn Consider (a n ) k+1 = a n (a n ) k by definition of k + 1th power, = a n a kn, by inductive hypothesis, = a n+kn by part (i), = a (k+1)n Hence result holds for m = k + 1 Thus by induction, (2) holds Since n was arbitrary we have shown n 1, m 1, (a n ) m = a mn as required Note this means that in particular, a n+1 = a n a (3) This could have been taken as the definition of a n+1 Remark If the semigroup (S, ) is commutative we also have a, b S, (a b) n = a n b n This need not happen in general Can you think of an example? One reason for introducing semigroups is the following 5
6 Lemma Suppose that (S, ) is a semigroup and a S has an inverse Then the inverse is unique Proof If an element a has two inverses, b and c say, then b = b e = b (a c) since c is an inverse of a, = (b a) c by associativity, = e c since b is an inverse of a, = c Groups Definition Given a set G and binary operation we say that (G, ) is a group if, and only if, G1 G is closed under, G2 is associative on G, G3 (G, ) has an identity element, ie e G : a G, e a = a e = a, G4 every element of (G, ) has an inverse, ie a G, a G : a a = a a = e Note that a group is a semigroup satisfying G3 and G4 Thus earlier results hold, namely that the identity element will be unique as will each inverse Also for m, n N and g G we have g m g n = g m+n and (g n ) m = g nm Definition For a group (G, ) we say it is an infinite group if G is an infinite set and we say that (G, ) is a finite group if G is a finite set In the latter case G denotes the number of elements in G and is called the order of (G, ) We say that (G, ) is a commutative or abelian group (after Niels Abel) if, and only if, it is a group and is commutative Example (Z, +) and (Z m, + m ) are additive groups, the first infinite and the second finite Example ({i, 1, i, 1}, ), where i 2 = 1, is a finite multiplicative group Example The set of permutations of a set of n elements, (S n, ), is a (nonabelian) group In fact, if you look back you can see we called it a symmetric 6
7 group before we knew what group meant This will be one of our main examples when illustrating the properties of groups in general It is a group because: G1 the composition of bijections is a bijection, G2 composition of functions is always associative, G3 the permutation that fixes every element is the identity and G4 every bijection has an inverse Question But why do we call (S n, ) the symmetric group? Consider, as an example, n = 4 Think of a square in the plane, center at the origin, with vertices labelled clockwise, 1,2,3 and 4 What symmetries does the square have? It has rotational symmetries about the origin If we rotate by π/2 in the clockwise direction we see that corners map 1 2, 2 3, 3 4 and 4 1 So this rotation can be represented by the cycle (1, 2, 3, 4) In the other direction what would (1, 2) (3, 4) represent? It would be a reflection in a line through the origin For Student: What are the permutations that represent the other symmetries of the square? In this way we see that S 4 contains the symmetries of the square Hence the use of the word symmetry in the name of (S n, ) Definition Let (G, ) be a group and g G Define the nonpositive powers of g by g 0 = e and g n = ( g 1) n for n 1 Example This is exactly how we defined negative powers of a permutation (So the earlier, iterative definition should be used for the positive power h n where h = g 1 ) 7
8 Elementary consequences of the axioms: Theorem Let (G, ) be a group (i) For a, x, y G if x a = y a then x = y (a cancellation result), (ii) For a, x, y G if a x = a y then x = y (a cancellation result), (iii) e 1 = e, where e is the identity, (iv) For all x G, (x 1 ) 1 = x, (v) For all x, y G, (x y) 1 = y 1 x 1, (vi) For x 1, x 2,, x n G, (x 1 x 2 x n ) 1 = x 1 n x 1 2 x 1 1, (vii) For m, n Z and g G, g m g n = g m+n and (g n ) m = g nm Proof not given in course Note We have already seen examples of (i) and (ii) for (Z m, ) and permutation groups (S n, ) Result (vii) was known previously in semigroups only for m, n N End of course in 2008 Definition Let (G, ) be a group with identity e and let g G If there exists n such that g n = e then the order of g is the least positive d : g d = e If no such n exists we say g has infinite order Example This generalises the definition of order for permutations given in Chapter 6 Example (i) (Z 5, ) : [2] 2 5 = [4] 5, [2]3 5 = [3] 5, [2]4 5 = [1] 5 So the order of [2] 5 is 4 For students, show that the orders of [3] 5 and [4] 5 are also 4 (ii) In ({i, 1, i, 1}, ) the order of 1 is 2, of i and i are 4 (iii) (Z 6, +) is an additive group, identity [0] 6 To find orders we add until we get zero So [2] 6 + [2] 6 = [4] 6, [2] 6 + [4] 6 = [6] 6 = [0] 6, [3] 6 + [3] 6 = [6] 6 = [0] 6 Thus the order of [2] 6 is 3, and the order of [3] 6 is 2 Something for future years: for finite groups the order of an element divides the order of the group! 8
9 Multiplication Tables for Finite Groups Suppose is a closed binary operation on a finite set S = {a 1,, a n } Then the multiplication table for (S, ) is a 1 a 2 a j a n a 1 a 1 a 1 a 1 a 2 a 2 a 2 a 1 a i a i a j a n Examples of these tables have been seen throughout this course, for example for (S 3, ), (Z 5, +), (Z 8, ), (Z 5, ) and (Z 8, ) If we can fill in every entry of the table with an element of S then the operation is closed In a group (G, ) we have an identity so, as noted before, in the multiplication table we will find an element whose row (and whose column) is identical to the heading row (respectively heading column) Also, in a group every element has an inverse, hence for any i S we can look in its row in the multiplication table and find the identity in column j, say, and find in row j the identity in column i Another property of the table of a group is Lemma If (G, ) is a group then every row (and every column) contains each element of G once Proof If in row a two different columns, b and c say, contained the same element then a b = a c The cancellation rule gives b = c which contradicts them being different Similarly, If in column α two different rows, β and γ say, contained the same element then β α = γ α The cancellation rule gives β = γ which contradicts them being different 9
10 Example If we look back at the table for (Z 8, ) we see that it is not a group [2] 8 [6] 8 [2] 8 [6] 8 [2] 8 [2] 8 [6] 8 [2] 8 [6] 8 [6] 8 [2] 8 [2] 8 [6] 8 [6] 8 [6] 8 [2] 8 [6] 8 [2] 8 [6] 8 [2] 8 For example, the row for does not contain every element As discussed above, if we throw away the noninvertible elements we get a group, (Z 8, ), Unfortunately there is no way to look at the table of a closed binary operation on a finite set to see if the operation is associative Finally, a group is abelian if, and only if, the table is symmetric about its leading diagonal 10
11 Appendix Contents Elementary consequences of the Group axioms: Maps between Groups Rings Fields More examples of binary operations and groups Elementary consequences of the Group axioms: Theorem Let (G, ) be a group Proof (i) (ii) (i) For a, x, y G if x a = y a then x = y (a cancellation result), (ii) For a, x, y G if a x = a y then x = y (a cancellation result), (iii) e 1 = e, where e is the identity, (iv) For all x G, (x 1 ) 1 = x, (v) For all x, y G, (x y) 1 = y 1 x 1, (vi) For x 1, x 2,, x n G, (x 1 x 2 x n ) 1 = x 1 n x 1 2 x 1 1, (vii) For m, n Z and g G, g m g n = g m+n and (g n ) m = g nm x a = y a (x a) a 1 = (y a) a 1 x ( a a 1) = y ( a a 1) by associativity, x e = y e definition of inverses, x = y a x = a y a 1 (a x) = a 1 (a y) ( a 1 a ) x = ( a 1 a ) y by associativity, e x = e y definition of inverses, x = y 11
12 (iii) e 1 = e 1 e since e is the identity, = e since e 1 is the inverse of e (iv) By definition (x 1 ) 1 is the inverse of x 1 Yet x 1 x = x x 1 = e means that x is also the inverse of x 1 We know that the inverse of an element is unique, hence (x 1 ) 1 = x as required (v) By definition (x y) 1 is the inverse of x y Yet (x y) ( y 1 x 1) = ( (x y) y 1) x 1 by associativity, = ( x ( y y 1)) x 1 again by associativity, = (x e) x 1 = x x 1 = e Similarly, (y 1 x 1 ) (x y) = e So y 1 x 1 is also an inverse of x y We know that the inverse of an element is unique, hence (x y) 1 = y 1 x 1 as required (vi) Use induction based on having used part (v) (x 1 x 2 x n ) 1 = ((x 1 x 2 x n 1 ) x n ) 1 = x 1 n (x 1 x n 1 ) 1, (vi) Proof of g m g n = g m+n : If m 1, n 1 the result has been seen earlier in the Lemma for semigroups If either m = 0 or n = 0, use the fact that g 0 = e, the identity If m 1, n 1 write m = r, n = t when, by the definition for negative powers, g m g n = ( g 1) r ( g 1 ) s = ( g 1) r+s by the result for semigroups, = g (r+s) = g m+n If either m 1 and n 1 or m 1 and n 1 we have g m g n equalling 12
13 either (g 1 ) r g n or g m (g 1 ) s Consider ( g 1 ) r g n = = = = ( (g ) 1 r 1 g 1) ( g g n 1) by definition of n th power and by (3) above, (( (g ) ) 1 r 1 g 1) g g n 1 associativity, ( (g ) 1 r 1 ( g 1 g )) g n 1 associativity ( (g ) ) 1 r 1 e g n 1 = ( g 1) r 1 g n 1 Continue, to get (g 1 ) r n, if r n, or g n r otherwise In both cases we get g n r = g n+m since m = r The similar result follows from g m (g 1 ) s, namely it equals g m s = g m+n since n = s In all cases we have g m g n = g m+n Proof of (g n ) m = g mn : If m 1, n 1 the result has been seen earlier in the Lemma for semigroups If either m = 0 or n = 0, both sides are equal to the identity If n 1 and m 1 write m = r Then (g n ) m = (g n ) r = ( (g n ) 1) r by definition of negative exponent, = (( g 1) n) r, by part (vi) of this Theorem, = ( g 1) nr by this result for positive exponents, = g nr by definition of negative exponent, = g nm If n 1 and m 1 write n = s Then (g n ) m = ( g s) m = (( g 1 ) s) m by definition of negative exponent, = ( g 1) sm by this result for positive exponents, = g sm by definition of negative exponent, = g nm 13
14 If n 1 and m 1 then (g n ) m = ( g s) ( r ((g ) ) = 1 s) 1 r by definition of negative exponent, ( ((g = ) 1) ) 1 r by part (vi) of this Theorem, = (g s ) r by part (iv) of this Theorem, = g sr by this result for positive exponents, = g mn since mn = rs Hence in all cases (g n ) m = g mn Maps between groups At the end of Chapter 4 there was a vague comment that the multiplication tables for (Z 8, ) and (Z 5, ) are in some way different We then ask, what of the tables for (Z 5, ) and (Z 4, +), written as [1] 5 [2] 5 [3] 5 [4] 5 [1] 5 [1] 5 [2] 5 [3] 5 [4] 5 [2] 5 [2] 5 [4] 5 [1] 5 [3] 5 [3] 5 [3] 5 [1] 5 [4] 5 [2] 5 [4] 5 [4] 5 [3] 5 [2] 5 [1] 5 and + [0] 4 [1] 4 [3] 4 [2] 4 [0] 4 [0] 4 [1] 4 [3] 4 [2] 4 [1] 4 [1] 4 [2] 4 [0] 4 [3] 4 [3] 4 [3] 4 [0] 4 [2] 4 [1] 4 [2] 4 [2] 4 [3] 4 [1] 4 [0] 4 (4) Do they not have the same form? They do because there is a bijection θ between the tables given by θ ([1] 5 ) = [0] 4, θ ([2] 5 ) = [1] 4, θ ([3] 5 ) = [3] 4 and θ ([4] 5 ) = [2] 4 This has the property that θ ([a] 5 [b] 5 ) = θ ([a] 5 ) + θ ([b] 5 ), for all a, b {1, 2, 3, 4} So not only are the sets mapped to each other, but the tables are mapped to each other In general, let (G, ) and (H, ) be two groups (so, may be different sets and different binary operations) If they are the same in the way above we have a bijection θ : G H which maps the multiplication table for G onto that for H So (G, ) g 2 g 1 g 1 g 2 (H, ) h 2 = θ (g 2 ) h 1 = θ (g 1 ) h 1 h 2 θ, 14
15 thus g 1 g 2 must map onto h 1 h 2 This means that θ (g 1 g 2 ) = h 1 h 2 = θ (g 1 ) θ (g 2 ) Something for future years: Study maps θ : G H satisfying θ (g 1 g 2 ) = θ (g 1 ) θ (g 2 ), for all g 1, g 2 in a group Such maps are called isomorphisms, and if there exists an isomorphism between two groups, they are called isomorphic It can be hard to find isomorphisms Note that in (4) we wrote the table for (Z 4, +) in the nonstandard order of [0] 4, [1] 4, [3] 4 and [2] 4 to see the similarity with (Z 5, ) Imagine how hard it might be to see this similarity if the tables are far larger, or the groups are infinite! Sometimes it can be easier to show that two groups are not isomorphic It can be shown that if (G, ) and (H, ) are isomorphic with isomorphism θ then if h = θ (g), h and g have the same orders So the orders of elements in (G, ) must match the orders of elements in (H, ) Example In (Z 8, ) every element, other than has order 2 Above we saw that in (Z 5, ), the element [2] 5 has order 4 Hence these groups are not isomorphic, a result alluded to at the end of Chapter 4 Further study would be to find how many nonisomorphic groups there are of each order The first few results are Rings n = 6 n = 8, n = 9, n = 10, n = 12, 2 groups, 5 groups, 2 groups, 2 groups, 5 groups Definition A ring is a nonempty set R along with two binary operations on R, addition + and multiplication, such that (i) (R, +) is an abelian group, (ii) R is closed under multiplication, (iii) multiplication is associative on R (iv) For all a, b, c R we have a (b + c) = a b + a c (b + c) a = b a + c a 15
16 These are called the Distributive laws, we are distributing the a throughout the terms of the bracket These laws are important in that they combine both operations, + and Note that we don t demand that (R, ) is a group Nonzero elements in R may fail to have inverses Even more basic, there may not be a multiplicative identity! And we also don t demand that multiplication is commutative Examples of rings: (i) (Z, +, ) is the first example of a ring From that we can go to a polynomial ring (Z [x], +, ) and a finite ring (Z m, +, ) (ii) The set of n n matrices with real coefficients, (M n (R), +, ), is a ring We have a multiplicative identity, the identity matrix, but not every matrix has an inverse Also this ring is noncommutative The interest in rings lies in how the two operations interact with each other We saw before, in the section on primes, that if we ask additive questions about multiplicative objects, primes, we get questions, such as Goldbach s Conjecture, that have withstood hundreds of years In the case of a general ring we have the additive identity 0 What would we expect of the multiplication 0 a for a R? Unsurprisingly Example For all a R we have 0 a = 0 Solution Let a R be given Since 0 is the additive identity, we have = 0 Then (0 + 0) a = 0 a 0 a + 0 a = 0 a, by distributive law, 0 a = 0, on adding (0 a) to both sides Since the distributive law is the only axiom to contain both operations, + and it is no surprise we have to use it in the proof above Again, 1 is the additive inverse of 1 as is a for any element of R But are a and 1 a the same? Unsurprisingly Example For all a R we have 1 a = a Solution Let a R be given Start from 1 + ( 1) = 0 Then (1 + ( 1)) a = 0 a = 0 by result above, 1 a + ( 1) a = 0 by distributive law, a + ( 1) a = 0 since 1 is the multiplicative identity, a + (a + ( 1) a) = a + 0 adding a to both side, ( a + a) + ( 1) a = a, by associativity on LHS and 0 identity on RHS, 0 + ( 1) a = a, ( 1) a = a, 16
17 What we can see here is that because we have only a few axioms defining a ring, the proof of something quite familiar, is surprisingly long We increase the number of axioms in the next section Fields Definition A field (F, +, ), is a nonempty set F along with two binary operations on F, addition + and multiplication, such that satisfies (i) (F, +, ) is a ring (ii) multiplication is commutative on F, (iii) There is a multiplicative identity in F, (iv) Every nonzero element of F has a multiplicative inverse The interest in fields comes from the fact that many of a our familiar arithmetic structures are fields Examples of fields: (i) (R, +, ), (C, +, ) and (Q, +, ) are infinite fields, (ii) (Z p, + p, p ) for p a prime, is an example of a finite field But note that (Z m, + m, m ) is not a field if m not prime Something for future years: (R, +, ), (C, +, ) and (Q, +, ) satisfy the axioms of a field and so, to this extent, are the same But what further properties do they satisfy that show they are different? On (R, +, ) and (Q, +, ) we can define an order relation a < b with properties such that if a < b then a + c < b + c and if a < b and b < c then a < c It can be shown that no such relation exists on (C, +, ) Hence (C, +, ) is different to both (R, +, ) and (Q, +, ) But are (R, +, ) and (Q, +, ) different? From the first half of the course you know they are different, R is uncountable while Q is countable But here I want to mention another difference Consider the sequence of rational numbers 1, 14, 141, 1414, 14142, ,, which gets arbitrarily close (converges to) the limit 2 Unfortunately, 2 / Q So, in (Q, +, ) it is not true that all convergent sequences converge But it can be shown that in (R, +, ) all convergent sequences converge These ideas take us away from algebra and into the areas of analysis More examples of binary operations and groups Example of binary operations not seen in course: S = M n (R), the set of n n matrices with real entries, with either matrix addition or matrix multiplication 17
18 Let X be any nonempty set, S = P (X), the power set of X, with either or Let Ω be a nonempty set and S the set of all functions Ω Ω along with, composition of functions Example Not a binary operation: If S = N then subtraction is not a binary operation since 1 2 / N If S = Q then a b = a is not a binary operation since 1 0 has no b meaning Example of binary operations not seen in course: Want to take the time to define sets of polynomials Definition For a set F, a polynomial over F with variable x is of the form a n x n + a n 1 x n 1 + a n 2 x n a 1 x + a 0, where a n, a n 1,, a 1, a 0 F The a i, 0 i n are the coefficients of the polynomial If x n is the largest power of x appearing in the polynomial then n is the degree of the polynomial, a n x n is the leading term and a n is the leading coefficient The collection of all polynomials with one variable x and with coefficients from F will be denoted by F [x] (Note the square brackets) Note that 0 F [x], being + 0x 2 + 0x + 0, but it is not said to have a degree, though some books give it degree 1 or even Examples 3x 2 + 5x 1 Z [x], x 2 π R [x], 3 7 x x2 + x Q [x], 5x 4 + x + 2 Z 7 [x] If we can add and multiply numbers in the set F then we can add and multiply the polynomials in F [x] Examples (i) In Z [x] the sum of 3x 2 + 5x 1 and 5x 3 3x 2 + 2x + 1 is ( 3x 2 + 5x 1 ) + ( 5x 3 3x 2 + 2x + 1 ) = 5x 3 + 7x (ii) Addition in Z 3 [x] The sum of 2x 3 + 2x 2 + x + 1 and x 3 + 2x is 2x 3 + 2x 2 + x ( x 3 + 2x ) = x 2 + x, 18
19 (iii) In Z [x] the product of x 2 + 2x + 3 and x is ( x 2 + 2x + 3 ) ( x ) = 3x x 3 + 8x +x 4 + 4x 2 = x 4 + 2x 3 + 7x 2 + 8x + 12 (iv) Multiplication in Z 2 [x] The product of x 3 + x + 1 and x 2 + x + 1 is ( x 3 + x + 1 ) ( x 2 + x + 1 ) = x 5 + x 4 + x 3 + x 3 + x 2 + x + x 2 + x + 1 = x 5 + x using 2 0 mod 2 Thus + and are binary operations on Z [x] and Z m [x] Aside The results of Chapters 13, on arithmetic, congruencies and congruence classes can be given for either Z [x], Z m [x] in place of Z This is because we can talk of one polynomial dividing another Definition If f, g F [x], we say that g divides f if there exists h F [x] such that f = gh Example In Z [x], x 1 divides x 3 2x since x 3 2x = (x 1) ( x 2 x 1 ) In Z 2 [x], x + 1 divides x since x = (x + 1) ( x 2 + x + 1 ) We could then talk of greatest common divisors (greatest in terms of degree) and linear combinations Or we could talk about congruencies, saying f (x) g (x) mod h (x) iff h (x) divides f (x) g (x) and congruence classes We could then construct new algebraic structures by defining addition and multiplication on these congruence classes of polynomials This is something for future years End of aside Example of identity In (P (X), ) the identity is X since X C = C X = C for all C P (X) Question Do we always have identities? 19
20 Examples Let 2Z be the set of even integers The product of two even integers is even so is a binary operation on 2Z Yet there is no identity in (2Z, ) (because 1 is not even) In (Z, ) we have a right identity, n 0 = n, but no left identity (the left identity won t be 0 since 0 n = n n when n 0 and no other possible value for the left identity will work) Question Do we always have inverses? Example In (M 2 (R), ) not every nonzero matrix here has an inverse, for example ( ) Examples of semigroups (X is a nonempty set) (Z, ), (Z n, +), (Z n, ) (N, +), (2Z, ), (P (X), ), (P (X), ), (S n, ), (Z [x], +), (Z [x], ), (Z n [x], +) and (Z n [x], ) Definition An important subset of (M 2 (R), ) is the collection of matrices that have an inverse, ie are invertible Such matrices have a nonzero determinant {( ) } a b GL 2 (R) = : ad bc 0 c d Here GL stands for General Linear So again we have thrown away the elements with no inverse Example In (M 2 (R), ) let ( 1 1 a = 0 1 Then But a 2 b 2 = (ab) 2 = ( ( ) ( 1 0, b = 1 1 ) 2 = ) ( ( ) = So we don t necessarily have a 2 b 2 = (a b) 2 Example (Z, +) is an additive group ) ) ( Verification If m, n Z then m + n Z and so G1 holds If m, n, p Z then (m + n) + p = m + (n + p) and so G2 holds 20 )
21 We have 0 Z and for all n Z, n + 0 = 0 + n = n and so G3 holds For any n Z we have n Z and n + ( n) = ( n) + n = 0 and so G4 holds Thus (Z, +) is an additive group with identity 0 and the inverse of n is n Examples: (Z m, + m ) is an additive group Verification We know (Z m, +) is a semigroup So only need note that [0] m is the identity and, for [a] m Z m, the inverse is [ a] m = [m a] m Example (M n (R), +) (Z [x], +) and (Z m [x], +) are further examples of additive groups In such groups the identity is normally denoted by 0 Example ({i, 1, i, 1}, ), where i 2 = 1, is a multiplicative group Verification From the table 1 i i i i 1 i i 1 1 i i i 1 1 i 1 1 i i 1 we see that G1, G3 (with e = 1) and G4 (with 1 1 = 1, i 1 = i, ( 1) 1 = 1 and ( i) 1 = i) are all satisfied G2 holds since multiplication of complex numbers is associative Examples Other multiplicative groups are (C \ {0}, ), (Q \ {0}, ), and ( Z p, p ) Question for students Why are (M n (R), ) (Z [x], ) and (Z m [x], ) not multiplicative groups? Note In the theory of groups a general group is normally written with a multiplicative operation In multiplicative groups the identity is normally denoted by 1, id or I Note that I say normally Example ({2, 4, 6, 8}, 10 ) From the table we see that the identity is 6 Also 2 1 = 8, 4 1 = 4, 6 1 = 6 and 8 1 = 2 21
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