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numbers the integers the rational numbers the real numbers the complex numbers A is a subset of B the empty set union of sets A and B intersection of sets A and B complement of the set A difference between sets A and B Cartesian product of sets A and B A × · · · × A (n times) identity mapping inverse of the function f a is congruent to b modulo n n factorial binomial coefficient n!/(k!(n − k)!) m divides n greatest common divisor of m and n power set of X 422 12 13 17 25 25 27 28 33 NOTATION Symbol Description 423 Page Zn lcm(m, n) U (n) Mn(R) det A GLn(R) Q8 C∗ |G| R∗ Q∗ SLn(R) Z(G) a |a| cis θ T Sn (a1, a2,..., ak) An Dn [G : H] LH RH d(x, y) dmin w(x) Mm×n(Z2) Null(H) δij G ∼= H Aut(G) ig Inn(G) ρg G/N ker φ G the integers modulo n least common multiple of m and n group of units in Zn the n × n matrices with entries in R determinant of A general linear group the group of quaternions the multiplicative group of complex numbers order of a group G the multiplicative group of real numbers the multiplicative group of rational numbers special linear group center of a group G cyclic subgroup generated by a order of an element a cos θ + i sin θ the circle group symmetric group on n letters cycle of length k alternating group on n letters dihedral group index of a subgroup H in a group G set of left cosets of H in a group G set of right cosets of H in a group G Hamming distance between x and y minimum distance of a code weight of x set of m by n matrices with entries in Z2 null space of a matrix H Kronecker delta G is isomorphic to H automorphism group of G ig(x) = gxg−1 inner automorphism group of G right regular representation factor group of G mod N kernel of φ commutator subgroup of G 37 34 44 45 45 45 46 46 46 49 49 |
49 55 60 60 65 67 77 79 83 85 96 96 96 121 121 121 127 127 131 144 156 156 156 157 160 171 168 424 Symbol Description NOTATION Page (aij) O(n) x SO(n) E(n) Ox Xg Gx XG N (H) H char R Z[i] Z(p) R[x] deg p(x) R[x1, x2,..., xn] φα Q(x) ν(a) F (x) F (x1,..., xn dim V U ⊕ V Hom(V, W ) V ∗ F (α1,..., αn) [E : F ] GF(pn) F ∗ G(E/F ) F{σi} matrix orthogonal group length of a vector x special orthogonal group Euclidean group orbit of x fixed point set of g isotropy subgroup of x set of fixed points in a G-set X normalizer of a subgroup H the ring of quaternions characteristic of a ring R the Gaussian integers ring of integers localized at p ring of polynomials over R degree of p(x) ring of polynomials in n variables evaluation homomorphism at α field of rational functions over Q Euclidean valuation of a field of rational functions in x field of rational functions in x1,..., xn a is less than b meet of a and b join of a and b largest element in a lattice smallest element in a lattice complement of a in a lattice dimension of a vector space V direct sum of vector spaces U and V set of all linear transformations from U to V dual of a vector space V smallest field containing F and α1,..., αn dimension of a field extension of E over F Galois field of order pn multiplicative group of a field F Galois group of E over F field fixed by automorphisms σi 180 183 184 187 187 215 215 215 217 233 245 249 248 265 269 269 272 272 292 297 303 303 307 309 309 311 311 311 329 332 332 332 337 340 361 361 377 382 NOTATION Symbol Description FG � |
�2 field fixed by automorphism group G discriminant of a polynomial 425 Page 382 398 Index G-equivalence classes, 227 G-equivalent, 215 G-set, 213 nth root of unity, 67, 390 Abel, Niels Henrik, 388 Abelian group, 43 Ackermann’s function, 35 Adleman, L., 107 Algebraic closure, 344 Algebraic extension, 337 Algebraic number, 337 Algorithm division, 273 Euclidean, 30 Artin, Emil, 304 Ascending chain condition, 295 Associate elements, 293 Atom, 315 Automorphism inner, 156, 177 of a group, 156 Basis of a lattice, 192 Bieberbach, L., 196 Binary operation, 42 Binary symmetric channel, 119 Boole, George, 320 Boolean algebra atom in a, 315 definition of, 312 finite, 314 isomorphism, 314 Boolean function, 224, 323 Boolean ring, 265 Burnside’s Counting Theorem, 220 Burnside, William, 48, 166, 227 Cancellation law for groups, 47 for integral domains, 248 Cardano, Gerolamo, 282 Carmichael numbers, 113 Cauchy’s Theorem, 231 Cauchy, Augustin-Louis, 85 Cayley table, 44 Cayley’s Theorem, 148 Cayley, Arthur, 149 Center of a group, 55 of a ring, 265 Centralizer, 55 of a subgroup, 217 of an element, 167 Characteristic of a ring, 249 Chinese Remainder Theorem for integers, 258 for rings, 266 Cipher, 103 Ciphertext, 103 Circuit parallel, 318 series, 317 series-parallel, 318 Class equation, 217 Code 426 INDEX BCH, 371 cyclic, 363 dual, 142 group, 124 Hamming definition of, 142 perfect, 143 shortened, 143 linear, 127 minimum distance of, 121 polynomial, 364 Commutative diagrams, 173 Commutative rings, 244 Composite integer, 30 Composition series, 206 Congruence modulo n, 17 Conjugacy classes, 217 Conjugate elements, 378 Conjugate fields, 397 Conjugate permutations, 101 Conjugate, complex, 64 Conjugation, 214 Constructible number, 349 Correspondence Theorem for groups, 174 for rings, 254 |
291 of quotients, 291 prime, 303 splitting, 345 Finitely generated group, 201 Fior, Antonio, 282 First Isomorphism Theorem for groups, 172 for rings, 254 Fixed point set, 215 Freshman’s Dream, 359 Frobenius map, 373 Function bijective, 10 Boolean, 224, 323 composition of, 10 definition of, 8 domain of, 9 identity, 12 injective, 10 invertible, 13 one-to-one, 10 onto, 10 order-preserving, 322 range of, 9 surjective, 10 switching, 224, 323 Fundamental Theorem of Algebra, 345, 395 of Arithmetic, 30 of Finite Abelian Groups, 203 of Galois Theory, 385 G¨odel, Kurt, 320 Galois field, 361 Galois group, 377 Galois, ´Evariste, 48, 389 Gauss’s Lemma, 299 Gauss, Karl Friedrich, 301 Gaussian integers, 248 Generator of a cyclic subgroup, 60 Generators for a group, 201 Glide reflection, 188 Gorenstein, Daniel, 166 Greatest common divisor of elements in a UFD, 304 of two integers, 27 of two polynomials, 275 Greatest lower bound, 308 Greiss, R., 166 Grothendieck, A., 354 Group p-group, 202, 231 abelian, 43 action, 213 alternating, 83 INDEX 429 automorphism of, 156 center of, 92, 167, 217 circle, 67 commutative, 43 cyclic, 60 definition of, 42 dihedral, 85 Euclidean, 187 factor, 160 finite, 46 finitely generated, 201 Galois, 377 general linear, 45, 182 generators of, 201 Heisenberg, 53 homomorphism of, 169 infinite, 46 isomorphic, 144 isomorphism of, 144 nonabelian, 43 noncommutative, 43 of units, 44 order of, 46 orthogonal, 183 permutation, 77 point, 193 quaternion, 46 quotient, 160 simple, 162, 166 solvable, 209 space, 193 special linear, 50, 182 special orthogonal, 187 symmetric, 77 symmetry, 190 torsion, 210 Hamming distance, 121 Hamming, R., 124 Hellman, M., 107 Hilbert, David |
, 196, 256, 320, 354 Homomorphic image, 169 Homomorphism canonical, 172, 253 evaluation, 251, 272 kernel of a group, 171 kernel of a ring, 250 lattice, 322 natural, 172, 253 of groups, 169 ring, 250 Ideal definition of, 251 maximal, 255 one-sided, 253 prime, 255 principal, 252 trivial, 251 two-sided, 253 Idempotent, 266 Indeterminate, 269 Index of a subgroup, 96 Induction first principle of, 24 second principle of, 25 Infimum, 308 Inner product, 126 Integral domain, 244 Internal direct product, 153 International standard book number, 57 Irreducible element, 293 Irreducible polynomial, 277 Isometry, 188 Isomorphism of Boolean algebras, 314 of groups, 144 ring, 250 Join, 309 Jordan, C., 166 Jordan-H¨older Theorem, 207 Kernel of a group homomorphism, 171 of a linear transformation, 331 of a ring homomorphism, 250 Key definition of, 103 private, 104 430 public, 103 single, 104 Klein, Felix, 48, 179, 256 Kronecker delta, 131, 185 Kronecker, Leopold, 354 Kummer, Ernst, 354 Lagrange’s Theorem, 97 Lagrange, Joseph-Louis, 48, 85, 100 Laplace, Pierre-Simon, 85 Lattice completed, 311 definition of, 309 distributive, 311 homomorphism, 322 Lattice of points, 192 Lattices, Principle of Duality for, 309 Least upper bound, 308 Left regular representation, 149 Lie, Sophus, 48, 235 Linear combination, 327 Linear dependence, 327 Linear functionals, 332 Linear independence, 327 Linear map, 179 Linear transformation definition of, 11, 179, 331 kernel of, 331 null space of, 331 range of, 331 Lower bound, 308 Mapping, see Function Matrix distance-preserving, 185 generator, 128 inner product-preserving, 185 invertible, 181 length-preserving, 185 nonsingular, 181 null space of, 127 orthogonal, 183 parity-check, 128 similar, 16 unimodular, 193 INDEX Matrix, Vandermonde, 368 Maximal ideal, 255 Maximum-likelihood decoding, 119 Meet, 309 Metric, 141 Min |
imal generator polynomial, 366 Minimal polynomial, 338 Minkowski, Hermann, 354 Monic polynomial, 269 Mordell-Weil conjecture, 354 Multiplicative subset, 304 Multiplicity of a root, 381 Nilpotent element, 265 Noether, A. Emmy, 256 Noether, Max, 256 Normal extension, 384 Normal series of a group, 205 Normal subgroup, 159 Normalizer, 233 Null space of a linear transformation, 331 of a matrix, 127 Odd Order Theorem, 239 Orbit, 92, 215 Orthogonal group, 183 Orthogonal matrix, 183 Orthonormal set, 185 Partial order, 306 Partially ordered set, 307 Partitions, 16 Permutation conjugate, 101 definition of, 12, 76 even, 83 odd, 83 Permutation group, 77 Plaintext, 103 Polynomial code, 364 content of, 299 cyclotomic, 284 definition of, 269 INDEX 431 degree of, 269 error, 374 error-locator, 375 greatest common divisor of, 275 in n indeterminates, 272 irreducible, 277 leading coefficient of, 269 minimal, 338 minimal generator, 366 monic, 269 primitive, 299 root of, 274 separable, 381 zero of, 274 Polynomial separable, 359 Poset definition of, 307 largest element in, 311 smallest element in, 311 Power set, 33, 307 Prime element, 293 Prime field, 303 Prime ideal, 255 Prime integer, 30 Prime subfield, 303 Primitive nth root of unity, 68, 391 Primitive element, 381 Primitive Element Theorem, 381 Primitive polynomial, 299 Principal ideal, 252 Principal ideal domain (PID), 294 Principal series, 206 Pseudoprime, 113 Quaternions, 46, 246 Repeated squares, 68 Resolvent cubic equation, 287 Right regular representation, 157 Rigid motion, 40, 188 Ring Artinian, 304 Boolean, 265 center of, 265 characteristic of, 249 commutative, 244 definition of, 243 division, 244 factor, 253 finitely generated, 304 homomorphism, 250 isomorphism, 250 local, 305 Noetherian, 295 of integers localized at p, 265 of quotients, 305 quotient, 253 with |
An equivalence relation in X is intimately connected with a partition of X, i.e. a decom position of X into disjoint subsets of X such that every element of X belongs to some subset. Examples of partitions of {I, 2, 3, 4, 5} are and {I}, {2,3}, {4,5} {I, 3, 4}, {2}, {5} On the other hand {I,2}, {2}, {3, 4, 5} is not a partition of {I, 2, 3, 4, 5}. If R is the equivalence relation (1.2) in {I,2,3,4}, then all the elements of {I,3,4} are R-related to 1, i.e., lRI, IR3, IR4 This suggests a means of getting a partition of a set X. In order to explain, we need some additional notation. Let R be an equivalence relation in a set X. If x E X, we define xR = {y lyE X and (x, y) E R}. xR is thus a certain subset of X. This subset xR is called the R-class of x, or the R-equivalence class of x, or the R-block of x. A subset of X will be called an R-class or R-block if it is the R-class or R-block of some element x E X. To illustrate these terms, consider the equivalence relation R given by (1.2). Here IR = {I, 3, 4}, 2R = {2}, and 3R = 4R = IR Thus the R-classes here are simply {I, 3, 4} and {2}. Notice that these R-classes constitute a partition of {I,2,3,4}. More generally, we have the following 10 SETS, MAPPINGS AND BINARY OPERATIONS [CHAP. 1 Theorem 1.2: Let X be a non-empty set and let R be an equivalence relation in X. Then if xRnx'Rr~, then xR=x'R, (i) (ii) x E xR for every x EX. Thus the R-classes constitute a partition of X, for (i) guarantees that dis tinct R-classes are disjoint, while (ii) shows that every element of X |
appears in at least one of the R-classes. Proof: First, we verify (i). Suppose xR n x'R r~. Then there is an element y E xR which lies also in x'R, i.e. (x, y) E R and (x', y) E R. As R is an equivalence, it follows from the symmetric property that (y, x') E R. But (x, y) E Rand (y, x') E R imply, by the transitive property, (x, x') E R. Now if z E x'R, then (x', z) E R; and hence by the transi (x, z) E R. This means, by the very tive property applied to (x, x') and (x', z), we find definition of xR, that z E xR. Since z was any element of x'R, we have proved x'R c: xR. The reverse inequality follows by a similar argument. Hence x'R = xR as required. The verification of (ii) is trivial since (x, x) E R means x E xR. This completes the proof of Theorem 1.2. Problems 1.13. (i) Prove that E = {(O,O), (1,1), (2,2), (3,3), (0,2), (1,3), (2,0), (3, I)} is an equivalence relation in 8 = {O, 1, 2, 3}. (ii) Find the E-equivalence blocks (a) OE, (b) IE, (c) 2E, (d) 3E. Solution: (i) (x, x) E E (y, z) = (2,0) or (2,2) and The reflexive property holds, i.e. for all x E 8, since (0,0), (1,1), (2,2), (3,3) E E. To show that E is symmetric, let us examine all pairs (x, y) where x ¥= y. There are only four, namely (0,2), (1,3), (2,0), (3,1). Clearly if (x, y) is anyone of the four, so is (y, x). When x = y, ( |
x, y) = (y, x). Thus (x, y) E E implies (y, x) E E. E is also transitive. Let (x, y) E E and (y, z) E E. Suppose x ¥= y. Then (x, y) can be (0,2), (1,3), (2,0) or (3,1). If (x, y) = (0,2), then (x, z) E E. Sim ilarly if (x, y) = (1,3), (2,0) or (3,1), it can be shown that (x, z) E E. When x = y, (y, z) E E means (x, z) E E. Therefore for any (x, z) E E and- E (x, y) E E and is transitive. (a) OE = {0,2}, (b) IE = {1,3}, (c) 2E = {2,0}, (d) 3E = {3, I}. Observe that OE = 2E and IE = 3E. (x, z) = (0,0) or (0,2) respectively. Hence (y, z) E E, we have (ii) 1.14. Let A = {p I p = (x, y) E Z2 with x - y divisible by 3}. Prove that A is an equivalence relation in Z and find the R-equivalence classes. Solution: It was shown in Problem 1.1l(iii), page 7, that A satisfies the three conditions of an equivalence relation. The R-equivalance classes are: (1) OA = {3q I q E Z}; for if (0, x) E A, 0 - x = -x is divisible by 3. Also, (0,3q) EA. (2) 1A = {I - 3q I q E Z}; for if (1, x) E A, 1 - x = 3q and hence x = 1 - 3q. If x = 1 - 3q, 1 - x is divisible by 3; hence (1, x) E 1.4.. (3) 2A = {2 - 3q I q E Z}; for if (2, x) E A, |
2 - x = 3q and so x = 2 - 3q. If x = 2 - 3q, 2 - x is divisible by 3; hence (2, x) EA. OA, lA, 2A are the only R-blocks, for any integer can be written as 3q, 1 - 3q, or 2 - 3q. Consequently OA u lA u 2A = Z. 1.15. Let Z* be the set of nonzero integers and let 8 = Z X Z*. (Recall that Z is the set of all integers.) Let E = {p I p = «r, s), (t, u» E 8 2 with ru = st}. Prove that E is an equivalence relation in 8. Sec. 1.3] MAPPINGS 11 Solution: implies (r, s) E 8 E is reflexive, for ((r, s), (r, s)) E 8 2; and since rs = sr, ((r, s), (1', s)) E E. The symmetric property of E is established by noticing ((r, s), (t, w)) E E means (r, s) and (t, w) E 8 2, and rw = st. But rw = st can be rewritten as ts = wr. Hence ((t, w), (r, s» E E. To show E is transitive, let ((r, s), (t, u) E E and ((t, u), (v, w») E E. Then ru = st and tw = vu. Since u oF 0, r = st and rw = stw = ~tw = ~vu = sv. Thus rw = sv and so u u transitive. Therefore E is an equivalence relation. ((r, s), (v, w)) E E. Hence E is u u 1.16. Prove that 8 X 8 is an equivalence relation in 8. Solution: 8 X 8 is reflexive since (x, x) E 8 X 8 and, by definition of 8 X 8, (y, z) E 8 an equivalence relation on 8. imply x, y and z E 8. But then (y, x) E 8 X 8. Hence 8 X 8 for all x E 8. If (x, y) E 8 X 8, then x and y E 8 and is symmetric. Now ( |
x, z) E 8 X 8 and 8 X 8 is transitive. Thus 8 X 8 is (x, y) E 8 1.17. Prove that a set X is infinite if and only if there are infinitely many equivalences in X. (Hard.) Solution: Assume there are an infinite number of equivalences in X. If X is finite, then there are at most a finite number of distinct subsets of X2. Therefore when X is finite there are at most a finite number of equivalences in X, which contradicts our hypothesis. Hence X must be infinite. Con versely, assume X is infinite. We exhibit an infinite number of equivalences in X as follows: For each pair a, b E R, a oF b, we define R(a,bl = {p I p = (x, y) E X2, where either x = y, (x, y) = (a, b) or (x, y) = (b, a)} Now R(a,bJ = R(c,dJ if and only if {a,b} = {c,d}. Therefore since X is infinite, we can find an infinite number of different pairs a, b E X each of which gives a distinct set R(a,bJ' Furthermore, each R(a,bl is an equivalence. To prove that R(a,bJ satisfies the three conditions of an equivalence (x, x) E R(a,bJ for all x E X, by the very definition of R(a,bJ' Secondly, relation, we first notice (y, x) = (x, y) = (a, b) or (b, a), R(a,bJ is symmetric since (b, a) or (a, b) respectively, or x = y and then (x, y) = (y, x). Thirdly, if (x, y) and (y, z) E R(a,bJ, (x,y) = (a,b) then (x,z) E R(a,bJ' To see this, notice that (x,y) can only be (a,b), (b,a) or (x,x). (x, z) = (a, a) or (a, b), which are both elements of R(a,bJ' implies (x |
, z) = Similarly, (y,z) E R(a.b)' (y, z) = (b, a) or (b, b); hence (x, y) = (x, x) means (x, z) E R(a, bJ' Finally, (x, y) = (b, a) (x, y) E R(a,bJ means in which case implies d. The division notation We find it useful to introduce a notation for the R-classes of an equivalence relation R in a set X, namely X/R. Problems 1.18. What is 8/E in Problem 1.13? Solution: 8/E = rOE, IE}. 1.19. What is Z/A in Problem 1.14? Solution: Z/A = {OA, lA, 2A}. 1.3 MAPPINGS a. Definition of mapping Assign to each even integer the value 1, and to each odd integer the value -1. Let us give the name a to this assignment; thus a assigns to each even element in Z, the set of all integers, the unique element +1 in the set {I, -I} and to each odd element in Z the unique element -1 in {I, -I}. In less detailed terms a assigns to each element in Z a unique ele ment in {I, -I}. Such an assignment is termed a mapping from Z into {I, -I} or a map 12 SETS, MAPPINGS AND BINARY OPERATIONS [CHAP. 1 from Z into {I, -I}. More generally, if Sand T are any two non-empty sets, a mapping or a map from S into T is an assignment of a unique element of T to each element of S. For the most part we shall denote mappings by lower case Greek letters such as a, {3, y. If a is a mapping from S into T, we shall express this fact more briefly by writing a: S ~ T; this is read "a is a mapping from S into T". We call S the domain and T the codomain of a. We find it useful to provide further notation and definitions. Suppose that a: S ~ T. If a assigns to S in S the element t in T, we write a: S ~ t and read this as "a sends S into t". We call t the image of s ( |
under a) if a: S ~ t. It is convenient to have a number of notations for the image of an element s in S under a mapping a: S ~ T; thus we shall write Sa or sa, or even a(S) for the image of sunder a. For the most part we use the first notation. If t E T and Sa = t, we call s a preimage of t. By Sa we mean {Sa Is E S}. We call Sa the range of a. Problems 1.20. is defined by Suppose a: P --> P (i) a: n --> n 2 for all n E P (ii) a: n --> n + 1 for all n E P (iii) a: n --> 2n for all n E P (iv) a: n --> 1 for all n E P (v) a: 1 --> 2, n --> 1 for all n E P, n > 1 Note: In (i)-(v) above, the mapping a is defined by describing its "action" on every element of P. For example, in (i), 1a = 12 = 1, 2a = 4, 3a = 9,.... Note that each element of P has a unique element assigned to it. (e) Is every element In each case determine: of P an image of some element of P in (i)-(v)? How many preimages (under a) does 2 have in (v)? How many preimages does 1 have in (iv)? in (v)? (b) a preimage (under a) of 2, 5, 6, 27. (a) 2a, 5a, 6a; Solution: (i) (ii) (iii) (iv) (v) (a) 2a = 4; 5a = 25; 6a = 36. (b) There is no preimage of 2, 5, 6 or 27. (e) (a) (b) (e) (a) (b) Not every element of P is an image, e.g. 2 is not an image. 2a = 3; 5a = 6; 6a = 7. 1a = 2; 4a = 5; 5a = 6; 26<1' = 27. Hence 1,4,5,26 are the required preimages. The only element of P which has no preimage is 1. 2a = 4; 5 |
a = 10; 6a = 12. 1 is a pre image of 2; 5 has no preimage; 3 is the preimage of 6; 27 has no preimage. 1 has no preimage, so not every element of P is an image. (c) (a) 2a = 1; 5a = 1; 6a = 1. (b) 2, 5, 6 and 27 have no preimages. (c) 1 has every element of P as a preimage and no other element of P has a preimage. (a) 2a = 1; 5a = 1; 6a = 1. (b) 1 is a preimage of 2; 5, 6 and 7 have no preimages. (c) 2 has one preimage, namely 1. 1 has an infinite number of preimages. In fact any ele ment of P not equal to 1 is mapped onto 1. 1 and 2 are the only elements of P which have pre images. 1.21. Let S = {I, 2, 3}, T = {I, 4, 5}. (i) Write down a mapping of S into T. (ii) Let a: S --> T be defined by 1a = 4, 2a = 5, 3a = 4. element in S under a? Give preimages of 1 and 4. Is every element of T an image of some Is every element of T the image of more than one element in S? Solution: (i) For example, a: S --> T defined by 1a = 1, 2a = 4, 3a = 5. (ii) Not every element of T is an image, for 1 has no preimage. Only 4 is an image of more than one element in S. 1 has no preimage, and 4 has preimages 1 and 3. Sec. 1.3] MAPPINGS 13 1.22. Suppose a: p-.c, C is defined by a: X -.c, x2 (i) (ii) a: X -.c, 2x + 27 (iii) a: X ~ z where z E C is such that Z2 == x. (iv) a: X -.c, z where z E C is such that z3=x-l. (v) (vi) a: X -.c, ix + 1 ix + 1 0:: X --> ix -1 (vii) a: |
X -.c, 10giO X (a) Do all of these descriptions really define mappings of Pinto C? (b) Is every element in C a preimage of some element of P in (i), (ii), (v), (vi), (vii)? Solution: (a) (i) and (ii) define mappings since every X E P has a unique image. (iii) does not define a map ping; for example, either 2 or -2 could be taken as 4a. Similarly (iv) is not a mapping, since there are three complex cube roots of x -1, so that each x has three different images. (v), (vi) and (vii) define mappings. (b) In (i), (ii), (v) and (vii), i has no preimage: for (i) i = x 2 implies X = i -227 It P; (v) ix + 1 = i gives X = 1 + i It P; (vii) if 10giO x = i, and thus x It P. In (vi) 1 has no preimage because ix + 1 = 1 implies 1 = -l. ix -1 implies X = VI It P; (ii) 2x + 27 = i then 10i = x 1.23. What is Pa in each of the cases in Problem 1.20? Solution: (i) Pa is the set of squares {1, 4, 9, 16,... }. (ii) P a = {2, 3, 4, 5,... } (iii) Pa = {2, 4, 6,8,... }, i.e. all the even integers. (iv) Pa = {1} (v) Pa = {1, 2} h. Formal definition of mapping The reader may ask whether our definition of mapping is precise. After all, it depends upon an English word, assignment, a word that is used in many different ways. A comparison with Section 1.2b is valuable. In Section 1.2b we introduced the concept of equivalence relation in X, but as we felt uneasy about it, we redefined it in terms of a subset of X2. Here too we feel uneasy about our definition of mapping and so we shall redefine it in terms of sets. A subset a of S X T is called a mapping of S into T if (s, t |
l) and (s, t 2) E a occurs only if tl = t 2, and for each s E S there exists an element (s, t) Ea. S is called the domain and T the codomain of a. If a is a mapping of S into T (written briefly as a: S ~ T) and (s, t) E a, we call t the image of s under a and write a: S ~ t. We also write t = Sa. lt is easy to see the relationship between the old definition and the new. In the old definition the elements of S were assigned unique elements of T. Consider the subset of S X T consisting of the pairs (s, t) where t is assigned to s. The two conditions of the new definition are satisfied by this subset. In the sequel we will use the definition of Section 1.3a, being confident that if necessary we could justify our arguments using the definition of a mapping in terms of a subset. c. Types of mappings We have talked of mappings without defining what is meant by the equality of two mappings. We will now remedy this. Suppose a: S ~ T and [3: S' ~ T'. Then we define a = [3 if and only if S = S', T = T', and, for every element s E S, Sa = Sa'. In other words, two mappings are equal if and only if they have the same domain, the same codomain, and the same "action" on each element of S. For example, let S = {I, 2, 3, 4}, T = {4, 5, 6}. Let a: S ~ T be defined by Ia = 4, 2a = 5, 3a = 6, 4a = 4; let [3: S ~ T be defined by 1[3 = 4, 2[3 = 5, 3[3 = 6, 4[3 = 5. Then a oF [3 since 4a oF 4[3. 14 SETS, MAPPINGS AND BINARY OPERATIONS [CHAP. 1 It is important to distinguish certain types of mappings. Thus suppose a: S ~ T. Then we say a is a mapping from S onto T (notice that into has now been replaced by onto) if every element in T has at least one preimage in S, i.e. if for every t E T there is at |
least one element S E S for which So' = t; in this case we call a an onto mapping. On the other hand we say a is one-to-one if So' = S' a implies s = s', i.e. distinct elements of S have distinct images in T (under a). Finally, we say a is a matching of Sand T or that a matches S with T or a is a bijection if a is both onto and one-to-one. Two sets are termed equipotent or of the same cardinality if there exists a matching of the one with the other. If S is finite and a matches T, then we say Sand T have the same number of elements. We denote the number of elements in a set S by lSI. If S is infinite this definition no longer makes sense unless one takes from all the sets which match S a single fixed set which we then term lSI, the cardinality of St. A set which matches P is called enumerable, countable, or countably infinite. Important results are that the set of rational numbers is enumerable and the set of real numbers is nott. We require one further definition. Suppose a: S ~ T and suppose S' ~ S. Then we define a mapping from S' into T by simply restricting the domain of a to S'; this mapping is denoted by a ls'(read a restricted to S') and is called the restriction of a to S'. To be quite explicit, a lS': S' ~ T is defined by a lS': S ~ So' for all s E S' Problems 1.24. Which of the mappings defined in Problems 1.20-1.22 are (a) onto, (b) one-to-one, (c) matchings? Solution: (a) None of the mappings defined in Problems 1.20 and 1.22 is onto by the solution already given to these problems. The mapping defined in Problem 1.21(ii) is hot onto, since 1 has no preimage. (b) Problems 1.20(i), (ii), and (iii) define one-to-one mappings: for in (i), if na = ma, then n 2 = m 2 and so n = m, since there is only one positive square root of an element in P; in (ii), na = ma gives n + 1 = m + 1 or n = m; and |
in (iii), na = ma implies 2n = 2m or n = m. Clearly Problems 1.20(iv) and (v) do not define one-to-one mappings. The mapping a in Problem 1.21(ii) is not one-to-one since la = 30'. All the mappings defined in Problem 1.22 are one-to-one: for in implies 2x + 27 = (i), Xa = x'a means x 2 = X,2 and, since x, x' E P, x = x'; in (ii), Xa = X'a implies ix + 1 = ix' + 1 or x = x'; (vi) Xa = X'a means 2X' + 27 or x = x'; in (v), Xa = X'a ix + 1 _ ix' + 1 oglO x = oglo X = Y an 1 d ix _ 1 hence lOY = x == x'. (Note that (iii) and (iv) are not mappings.) 1X or x - x,In VII, Xa - X a gIves ix' -1 or 2' - 2' I I,'( " ) ~x - 1 • I I - - (c) None of the mappings defined in Problems 1.20-1.22 is a matching, since none of them is onto. 1.25. Which mappings in Problems 1.20-1.22 are equal? Solution: None. Problem 1.20 features a: P -> P; Problem 1.21, a: S -> T; and Problem 1.22, a: P -> C. Hence we need only compare the mappings in each exercise. 1.26. Let a: P -> Z be defined by na == -n for all n E P. Is a onto? One-to-one? A matching? Solution: a is neither onto nor a matching, since 1 has no pre image. a is one-to-one, for if na = n'a, then -n == -n' and hence n == n'. tFor more details see, for example, G. Birkhoff and S, MacLane, A Survey of Modern Algebra, Macmillan, 1953. Sec. 1.3] MAPPINGS 15 1.27. Suppose m is a fixed positive integer. Every integer n can |
be written uniquely in the form n = qm -I- r where the remainder l' is an element of the set {O, 1,..., m I}. For example, if m = 4, forms: 4k,4k+I,4k+2,4k+3. Let a:Z---{O,I,...,m-I} be defined by na = 1', if n=qm+1' where 1'E{O,I,...,m-I}. Prove: integer can be expressed in precisely one of then every following the (a) a is onto {O, 1,..., m - I}. (b) If n1' n2 are any two integers, then (n 1n2)a = (njan2a)a. Solution: (a) a is onto because 0, 1,..., m - 1 have pre images 0, 1,..., m - 1 respectively. (b) Let n1 = q1m + r1 and n 2 = q2m -I- 1'2' Then 1.28. Check which of the following mappings are onto, one-to-one, bijections: a: C.... R defined by a: a -I- ib --> a2 + b2 (i) (ii) a: Z --> P defined by a: n --> n 2 + 1 (iii) a: P.... Q defined by a: n --> 2n + 1. n Solution: (i) a is neither onto nor one-to-one, because a2 + b2 "" 0, for any a, b E R, and -Ia = Ia = 1. Hence a is not a bijection. (ii) As 3 has no preimage, a is neither onto nor a bijection. Also a is not one-to-one, since Ia = 2 and -Ia = 2. (iii) 1 has no preimage, since 2n ~ 1 = 1 implies n,= -1 fl P. Therefore a is not onto and hence n = 2n n + n; ence n = n'. no a 1Jec 10n. na = n a means 2n -I- 1 = 2n' + 1 or Thus each image has a unique preimage and a is one-to-one. 2'+ n n b·· " t' n n h |
t, 1.29. Let S be the set of open intervals (a, b) on the real line and let T be the set of closed intervals [a, b]. Define a: S.... T by (a, b) a = [a, b]. Is a one-to-one? Onto? Solution: The mapping is one-to-one and onto. For, if (a, b) a = (a', b') a then [a, b] = [a', b'l. But this equality holds iff a = a' and b = b'. a is therefore one-to-one. a is also onto since a closed interval [a, b] has a preimage (a, b). 1.30. How many mappings are there from {I,2} into itself? From {I, 2, 3} into itself? In each case, how many of these mappings are one-to-one? Onto? Solution: There are four mappings of {I,2} into itself, namely: a1 defined by Ia1 = 1 and 2al = 2; a2 defined by Ia2 = 1 and 2a2 = 1; a3 defined by Ia3 = 2 and 2a3 = 1; a4 defined by 1a4 = 2 and 2a4 = 2. Only a1 and a3 are one-to-one and onto. To find the number of mappings of {I, 2, 3} into itself, we proceed as follows: 1 may have any of three images under such a mapping, i.e. 1 --> 1 or 1.... 2 or 1 --> 3. Also, 2 may have any of 3 images, either 1, 2 or 3. So we have in all 3 X 3 possibilities for the actions of mappings on 1 and 2. Then 3 can be sent into 1, 2 or 3, giving 3 X 3 X 3 = 27 possible mappings of {I, 2, 3} into itself. There are 3 X 2 X 1 = 6 possible one-to-one and onto mappings; for when we once choose an image for 1 there are only two possible images for 2, and then the image of 3 is uniquely determined. 1.31. Let S = {I, 2, 3}, T = {3, 4, 5}, U = {4, 5, 6}, and let a: S |
.... T be defined by a: 1 --> 3, 2 -> 3,3.... 5. Let {3 and y be the mappings from T into U given by {3: 3 --> 4, 4 --> 6, 5 -> 4, y: 3.... 4, 4.... 4, 5.... 4. Compute (Ia){3, (2a){3, (3a){3, (Ia)y, (2a)y, (3a)y. Solution: (Ia){3 = 3{3 = 4; (2a){3 = 3{3 = 4; (3a){3 = 5{3 = 4; (la)y = 3y = 4; (2a)y = 3y = 4; (3a)y = 5y = 4. 16 1.32. SETS, MAPPINGS AND BINARY OPERATIONS [CHAP. 1 Let ", f3, y be the mappings of Q into Q defined by,,: x ---. 2"' f3: x ---. x + 1, y: x -> x - 1. Prove that x if x is any element of Q, then «xy)a)f3 x+1 2 What is,,!p? a lZ? Is alP = a? lZ Solution: «xy)a)f3 = «(x - 1)a)f3 = -2- f3 = x-1 into Q. a lZ is a mapping of Z into domain of a lZ • x-1+2 2 x+1 2 alP is a mapping of P since the domain of a!p is not the same as the 1.33. If S is a non-empty set, prove that S is infinite if and only if there are infinitely many mappings of S into S. Solution: First we show that if there are infinitely many mappings of S into S, then S is infinite. Assume, lSI = n. Now each of the n elements can be mapped onto at on the contrary, that S is finite and most n images. Hence there are at most nn different mappings of S into S. This contradicts our assumption that there is an infinite number of such mappings. Hence S is infinite. Conversely, let S be infinite. We define, for each S E S, the mapping as: S ---. S by as: x -> S for all |
xES. {as I S E S} is an infinite set since as = as, if and only if S = s', and we assumed S to be infinite, Therefore we have found an infinite number of mappings of S into S. 1.34. If S is any non-empty set, verify that S matches S. Solution: Define the mapping a: S -> S by a: S ---. S for each S E S. a is clearly one-to-one and onto. Hence a is a matching, 1.35. If S matches T, prove that T matches S, Solution: If S matches T, then there is a mapping a: S -7 T which is one-to-one and onto. Define a: T -7 S as follows. Let t E T. Then there is an S E S such that Sa = t. The image of t under a is defined to be s. t E T, We now show a is a matching. In the first place, a is a mapping. For if ta = sand ta = s', then by definition of a, Sa = t and s' a = t. But a is one-to-one, so that Sa = s' a for some implies S = s'. Thus the image of an element under a is unique. Secondly, a is one-to-one, because ta = t' a = s t = tf. Thirdly, if s E S, then Sa = t for some t E T. By definition of a, tii = s. Hence every element of S has a preimage under a and a is onto. implies Sa = t and Sa = tf, which in turn implies, since a is a mapping, 1.36. If S matches T and if T matches U, prove that S matches U. (Hard.) Solution: Let a: S -> T and f3: T -7 U be matchings. Then a: S -> U, defined by Sa = (sa)f3, is a matching. a is a mapping of S into U; for Sa E U, and if Sa = Ul and Sa = u2 then (Sa)f3 = Ul and (Sa)f3 = U2, which implies Ul = u 2 since Sa has a unique image under f3. a is one-to-one, for Sa = s'a implies (sa)f3 = (S'a)f3 |
. But a andf3 are one-to-one, so that Sa = sfa and s = Sf. a is also onto, for if U E U then there is atE T such that tf3 = u. Now t has a pre image S E Sunder a. Thus Sa = (sa)f3 = tf3 = u. 1.37. Let a be the mapping of S = {1, 2, 3,4,5,6,7,8,9, 10} into itself given by a: 1 --> 3, 2 -7 4, 3 -7 5, 4 -7 7, 5 --> 9, 6 --> 10, 7 --> 1, 8 --> 3, 9 -> 4, 10 --> 5. Then there is a useful alternative way of describing a: we list the elements of S on one line (in any order) and on the following line we place under each element of S its image under a, enclosing the entire description in parentheses as follows. Describe the following mappings of S into S, using this notation 10 1 8 3 9 10) 4 5 Sec. 1.4] COMPOSITION OF MAPPINGS 17 (i) f3: 1.... 2, 2.... 2,3.... 2, 4.... 2, 5.... 2, 6.... 3, 7.... 4,8.... 4, 9.... 4,10.... 5. (ii) y: 1.... 6, 2.... 7, 3 -> 8, 4.... 9, 5.... 10, 6.... 1, 7 -> 1, 8 -> 1, 9.... 1, 10.... 1. Use these descriptions to decide whether (iii) f3 = y, (iv) f3 is one-to-one, (v) y is onto. Solution: (i) (ii) 2 G 2 (~ 10 15 ) 9 4 9 ~O) 1 (iii) f3#y since 1f3 = 2 and 1y = 6. It is only necessary to compare the bottom rows. (iv) f3 is not one-to-one, e.g. 1f3 = 2f3 = 2. It is only necessary to find a repetition on the bottom row. (v) y is not onto, e.g. 2 has no preimage. It is only necessary to check whether all the integers 1,2,...,10 appear in the bottom row. 1.4 COMPOSITION OF MAPPINGS Definition |
Let a: S.... T, (3: T.... U. Because Sa E T, we can compute (Sa)(3. This suggests "composing the mappings a and (3", i.e. defining a mapping of S into U by performing a and (3 in succession on each of the elements in S. More precisely we define a 0 (3, the com position of a with (3 (in this order) as a mapping of S into U defined by S(a 0 (3) = (Sa)(3, for all S in S (Some authors use exactly the opposite order, so that their a 0 (3 is our (30 a.) For example, let S = {1,2}, T = {3,4,5}, U = {6,7} and let a: S.... T, (3: T.... U be defined by Then a: 1.... 3, 2.... 5, (3: 3.... 6, 4.... 7, 5.... 6 l(a o(3) 2(a o (3) (la)(3 (2a)(3 3(3 5(3 6 6 Hence a 0 (3: {I, 2}.... {6,7} is defined by a 0 (3: 1.... 6, 2.... 6 This notion of the composition of two mappings is of tremendous importance; hence we give the following drill problems. Problems 1.38. Let a: p.... C be defined by na = in + 1 and let i 2 =-1. Whatdo"of3,(aof3)oa,andao(f3oa) map nEP to? Why is aOf3#f3 o a? f3: C.... P be defined by f3: a + ib.... b2, where Solution: Let n E P. Then n(a ° f3) = (na)f3 = (in + 1)f3 = n 2 • Now n((a 0 f3) ° a) = (n(a ° f3))a = n 2a = (a ° 13) ° a = a ° (13 ° a). in2 + 1 and n(a ° (f3 ° a)) = (na)(f3 ° a) = ((in + 1)f3)a = n 2a = in2 + 1. Hence f3 ° a: C -'> C. Hence a ° 13 # f3 |
° a. a ° 13 : P -'> P while 1.39. Let,,: Q -'> Q be defined by a: a -'> a2 + 2 and let f3: Q -'> Q be defined by pute a ° f3, f3 ° a. Are these mappings equal? Compute (a ° f3) ° a, a ° (13 ° a). f3: a -'> ta - 2. Com 18 SETS, MAPPINGS AND BINARY OPERATIONS [CHAP. 1 Solution: Let a E Q. qa - 2)2 + 2. (~a2-1)2+2 l§a 2 -1)2 + 2. a(a ° (3) = (aa)(3 = (a2 + 2)(3 = ~(a2 + 2) - 2 = ta 2 - 1. a((3 ° a) = (a{3)a = (~a - 2)a = Clearly a ° (3 oF (30 a. Furthermore, a((a ° (3) 0 a) = and a(a ° ((3 0 a)) = (aa)((3 ° a) = (a2 + 2)(3 ° a = ((a 2 + 2){3)a = ((t(a2 + 2) - 2)a = (a(a ° (3))a = (ta2 - l)a Note then that (a ° (3) 0 a = a ° ((3 0 a). 1.40. Employing the notation of Problem 1.37, compute the following: (i) 3 1 5 G 2 3 4!) °C 2 3 4!) 2 3 4!)oG ii) (~ (iii) G 2 3 4 4 5 3 ~) G 2 3 4!) ° iv) Solution: (ii) G 2 3 4 4 1 3 (i) (~ 2 3 4!) Let a: S -> T, (3 : T -> U, 'I: U -> V. Prove that (a ° (3) 0'1 = a ° ((3 0 'I)' CHard.) iii)G ~) ~) 2 4 3 ~) ° G 2 3 4!) ~) (iv) G 2 3 4 5 3 4 ~) Solution: If 8 E S, then 8((a ° (3) 0 'I) = (8(a ° ( |
3))'1 Consequently (a ° (3) 0 'I = a ° ((3 0 'I)' Prove that if a : S -> T and (3: T -> U and a ° (3 is onto, then (3 is onto. ((8a)(3)y and 8(a ° ((3 0 'I)) (8a)((3 ° 'I) ((8a)(3)y. Is a onto? (Hard.) 1.41. 1.42. Solution: Let u E U. As a ° (3 is onto, we can find a preimage of u under a ° (3. Let 8 E S be a preimage of u under a ° (3, i.e. 8(a ° (3) = u. Thus 8(a ° (3) = (8a)(3 = u and 8a is a preimage of u under (3. Hence (3 is onto. a need not be onto, e.g. let S = {I}, T = {1,2}, U = {1}. Define a: S -> T by 1a = 1, and (3: T -> U by 1(3 = 2(3 = 1. a ° (3 is onto but a is not. 1.43. Prove that if a: S -> T, (Hard.) (3: T -> U and a ° (3 is one-to-one, then a is one-to-one. Is (3 one-to-one? Solution: Let 81> 82 E Sand 81a = 82a. 81a = 82a implies 81a ° f3 = 82a ° f3. But a ° f3 is one-to-one, so that 81 = 82' Hence a is one-to-one. and U = {I}. Define a: S -> T one-to-one, e.g. f3 : T -> U by 1f3 = 1 and 2f3 = 1. a ° f3 is one-to-one but f3 is not one-to-one. (81a)(3 = (82a)(3 and, by definition of a ° {l, f3 is not necessarily 1a = 1, and let S = {I}, T = {1,2} by 1.44. 1.45. Prove that f3: T -> U (T oF 0 |
) is one-to-one if and only if for every set S and every pair of map pings a: S -> T and a': S -> T, a ° f3 = a' ° f3 Solution: implies a = a'. (Hard.) First assume that for every set S and every pair of mappings a: S -> T and a': S -> T, 0'0 f3 = a' ° f3 implies a = a'. Under this hypothesis suppose f3 is not one-to-one. Then we can find (t oF t') such that tf3 = t' (3 = u E U. Let S = {1,2}, let a: S -> T be defined by 10' = t t, t' E T and 2a = t', and let a': S -> T be defined by since 1a ° f3 = (la)f3 = tf3 = U, 20' ° f3 = (2a)f3 = t' f3 = U, la' ° f3 = (la')f3 = t' f3 = U, and 2a' ° f3 = (2a')f3 = la' = t' and 20" = t. Now 0'0 f3 = a' ° f3, tf3 = u. But a oF a'. Hence the assumption that f3 is not one-to-one is false and f3 must be one-to-one. To prove the converse, let f3 be one-to-one. Say we can find a set S and a pair of mappings a and a' of S into T such that a ° f3 = a' ° f3 and a oF a'. So there exists 8 E S such that Sa oF sa'. 0'0 f3 = a' ° f3 means that s(a ° (3) = s(a' ° (3). Hence (sa)f3 = (sa')f3. As f3 is one-to-one, Sa = Sa'. Therefore we have a contradiction and a must be equal to a'. Prove that a: S -> T (T oF 0) is onto iff for every set U and every pair of mappings and f3': T -> U such that a ° f3 = a 0 f3', it follows that f3 = f3'. (Hard.) f3: T -> U |
Solution: Let us assume that for every set U and every pair of mappings f3 and f3' of T into U such that it follows that f3 = f3'. Say a is not onto, and tl is an element of T which has no a ° f3 = a 0 f3', Sec. 1.5] BINARY OPERATIONS 19 preimage under a. Let U = {1,2} and define the mappings (3 and (3' of T into U as follows: for all in T and Sa 0 (3' = (Sa)(3' = 1, since Sa ¥= t l. Hence a 0 (3 = a 0 (3' and sumption that a 0 (3 = a 0 (3' implies t(3 = 1 t l(3 = 2. Now if S E S, Sa 0 (3 = (Sa)(3 = 1 and (3 ¥= (3'. This contradicts the as (3 = (3'. Thus a must be onto. t(3' = 1 for all t E T, t ¥= tl Conversely, assume a is onto and we can find a set U and two mappings, (3 and (3', of T into U such that a 0 (3 = a 0 (3' and (3 ¥= (3'. (3 ¥= (3' means there is a tl E T such that t l(3 ¥= t l(3'. Further implies (sa)(3 = (sa)(3' more, since a is onto, we can find S E S such that Sa = t l. But a 0 (3 = a 0 (3' or t l(3 = t l(3'. Here we have a contradiction because we chose t l(3 ¥= t l(3'. We therefore conclude (3 = (3'. 1.5 BINARY OPERATIONS a. Definition The idea of a binary operation is illustrated by the usual operation of addition in Z, which may be analyzed in the following way. For every pair of integers (m, n) there is associated a unique integer m + n. We may therefore think of addition as being a brief description of a mapping of Z x Z into Z where the image of (m, n) E Z x Z is denoted by m + n. Any mapping (3 of S x S into S, where S is any non-empty set, is called a binary operation in |
S. We shall sometimes write instead of (s, t)(3 (the image of (s, t) under (3) one of sot, s· t, st, s + t or s x t. We stress that in all these cases the meaning of the various expressions sot, s· t, st, s + t and s x t is simply the image of (s, t) under the given mapping (3 of S x S into S. These notations suppress the binary operation (3, so there is danger of confusion. However, we will work with binary operations and the various notations so Incidentally, we read frequently that the reader will become familiar with the pitfalls. sot as "ess circle tee" s·t as "ess dot tee" or "ess times tee" st as "ess tee" or "ess times tee" s+t as "ess plus tee" s x t as "ess times tee". The notation sot is called the circle notation, the notations s· t and st are termed multi plicative, and the notation s + t is termed additive. We sometimes refer to s· t or st as the product of sand t, and s + t as the sum of sand t. The following problems will help to make the various notations clear. Problems 1.46. Convince yourself that the following mappings are binary operations in P. (i) a: PxP"*P defined by a: (i, j) "* i2, where (i,j) E P. "': PxP"*P defined by a: (i,j)"* i+ j, where (i,j) E P. (ii) (iii) a: PxP""p defined by a: (i, j).... i X j (regular multiplication of integers), where (i,j) E P. (iv) a: PxP""p defined by a: (i, j).... 2i + 3j, (i, j) E P X P. (v) a: PxP""p defined by a: (i, j).... i + j + 1, (i, j) E P X P. 1.47. Which of the following are binary operations in P (throughout (i, j) E P2)? (i) a: (i,j)"" i+ j (ii |
) a: (i, j).... i - j (iii) a: (i,j)"* i+ j (iv) a: (i, j).... i + j + i2 (v) a: (i,j)"" j Solution: In (i), a is clearly a mapping from P X Pinto P. So a is a binary operation in P. (ii) and (iii) do not define binary operations in P because not every element in P X P has an image in P: e.g. in (ii), a: (1,2)"" 1 - 2 = -1 ~ P; and in (iii), a: (1,2)"" 1 + 2 =! ~ P. (iv) and (v) define mappings from P X Pinto P. Hence they define binary operations in P. 20 SETS, MAPPINGS AND BINARY OPERATIONS [CHAP. 1 1.48. Interpret the following (abbreviated) definitions of arbitrary elements of Z. binary operations in Z, where and j denote (i) (iii + j)2 i j - (i X j) (iii) (iv) i 0 j = ij = v) (vi) i· j = ii i + 27j Solution: Throughout, (i, j) E Z2. (i) a: (i"i) --> (i+ j)2 (ii) a: (i, j) --> i j - (i X j) (iii) a: (i, j) --> i + j (iv) a: (i,j)-->i-ixj (v) a: (i,j) --> ii (vi) a: (i,j) --> i + 27j 1.49. Check that the following are binary operations in the plane R2. (i) (ii) (x, y) 0 (x', y') = midpoint of the line joining the point (x, y) E R2 if (x, y) # (x', y'). If (x, y) = (x', y'), define (x, y) 0 (x', y') = (x, y). (x, y) + (x', y') = (x + x', y + y'), where (x, y), (x', y') E R2. (iii) (x |
, y) • (x', y') = (xx', yy'), where (x, y), (x', y') E R2. (iv) (v) (x', y') = (x - x', y - y'), where (x, y), (x', y') E R2. (x, y) (x, y) 0 (x', y') = (x + x', x'y + y'), where (x, y), (x', y') E R2. to the point (x', y') E R2 (Notice here how we have abbreviated the definitions of the binary operations considered.) Solution: (i) 0 is a binary operation since two points determine a unique line and each line has a unique midpoint. (ii)-(v) are binary operations because each has a unique image by virtue of the fact that addition, multiplication and subtraction are binary operations in R. 1.50. Let S = {I, 2, 3} and let a and (3 be the following binary operations in S: a: (1,1) --> 1, (1,2) --> 1, (1,3) --> 2, (2,1) --> 2, (2,2) --> 3, (2,3) --> 3, (3,1) --> 3, (3,2) --> 2, (3,3) --> 1; (3: (1,1) --> 1, (1,2) --> 1, (1,3) --> 2, (2,1) --> 3, (2,2) --> 3, (2,3) --> 3, (3,1) --> 3, (3,2) --> 1, (3,3) --> 2. (i) Is a = (3? (ii) Compute ((1, l)a, 1)(3, ((1,1)(3, l)a. (iii) Compute ((1, 2)a, 3)a, (1, (2, 3)a)a, ((1,2)(3,3)(3, (1, (2, 3)(3)(3. Solution: (i) a # (3 for (2, l)a = 2 and (2,1)(3 = 3. (ii) ((1, l)a, 1)(3 = (1,1)(3 = 1; |
((1,1)(3, l)a = (1, l)a = 1. (iii) ((1, 2)a, 3)a = (1, 3)a = 2; (1, (2, 3)a)a = (1, 3)a = 2; ((1,2)(3,3)(3 = (1,3)(3 = 2; (1, (2, 3)(3)(3 = (1,3)(3 = 2. 1.51. Let S = {I, 2, 3} and let X be the set of all mappings of S into S. (i) Compute IXI. (ii) Verify that the composition 0 of mappings is a binary operation in X. (iii) Is a 0 (3 = (3 0 a for all elements a, (3 EX? Solution: (i) IXI = 27 (see Problem 1.30, page 15). (ii) The composition of mappings a: S --> Sand (3: S --> S is again a mapping of S into S. Therefore 71': X2 --> X, defined by (a, (3)71' = a 0 (3, (a, (3) E X2, is a binary operation. (iii) No. For example, if a: S --> S defined by Sa = 1 for all 8 E S, and (3: S --> S defined by for all s E S, are two mappings of S into S, then s(a 0 (3) = (sa)(3 = 1(3 = 2 and s(3 = 2 s(3 0 a = (s(3)a = 2a = 1. Hence a 0 (3 # (30 C/. 1.52. Let Q* be the set of nonzero rational numbers. Make sense of the remark that division (denoted as usual by -;.-) is a binary operation in Q*. Check whether the following statements hold for all u, b, c E Q*. (i) (ii) u-;.-b = b -;.- u. (u-;.- b) -;.- c = u -;.- (b -;.- c). (iv) If u-;.-b a -:- c, then b = c. (v) If b-;.-u c -;.- u, then b = c. (iii) ((u |
-;.- b) -;.- c) -;.- d = u -;.- (b -;.- (c -;.- d)). Sec. 1.5) BINARY OPERATIONS 21 Solution: Division is a binary operation in Q*; for if w/x E Q* and y/z E Q* where w, x, y, z are False; e.g. 2 -7- 3 # 3 -7- 2. integers, then w/x -7- y/z = wz/xy E Q*. Hence -7- is a mapping of Q* X Q'" -> Q*. (i) (ii) False; e.g. (2 -7- 1) -7- 2 = 1 and 2 -7- (1 -7- 2) = 2 -7- -! = 4. (iii) False; e.g. «1 -7- 2) -7- 2) -7- 2 = k and 1 -7- (2 -7- (2 -7- 2)) = l (iv) True. a -7- b = a -7- c implies ac = ab and, as a # 0, c = b. (v) True. b -7- a = c -7- a implies ab = ca. Hence b = c, since a # O. 1.53. For all a,b,cER, (aob)oc=ao(boc). Let 0 be the binary operation in R defined by a 0 b = a + b + abo Verify that: (i) (ii) For all a, bE R, a 0 b = boa. (iii) Prove that if a#-l, then aob=aoc iff b=c. Solution: (i) (aob)oc (a+b+ab)oc a 0 (b 0 c) = a 0 (b + c + bc) a + b + ab + c + (a + b + ab)c a + b + c + bc + ab + ac + abc a + b + c + bc + a(b + c + bc) a + b + c + bc + ab + ac + abc (ii) a 0 b = a + b + ab = b + a + ba = boa then a 0 b = a 0 c (iii) If b = c, |
for any a. a + c + ac. Therefore b + ab = c + ac, b(l + a) = c(l + a) and, since a # -1, b = C. If a 0 b = a 0 c and a # -1, then a + b + ab = 1.54. Let 0 be the binary operation in R2 defined by (x, y) 0 (x', y') = (xx' - yy', yx' + xy'). Verify that for all (x,y), (x',y'), (x",y") ER2: (i) (ii) «x, y) 0 (x', y')) 0 (x", y") = (x, y) 0 (x', y') = (x', y') 0 (x, y) (x, y) 0 «x', y') 0 (x", y")) Solution: (i) (ii) «x, y) 0 (x', y')) 0 (x", y") (x, y) 0 (x', y') = (x, y) 0 «x', y') 0 (x", y")) (xx' - yy', yx' + xy') = (x'x - y'y, y'x + x'y) = (x', y') 0 (x, y) (yx' + xy')y", (yx' + xy')x" + (xx' - yY')y") (xx' - yy', yx' + xy') 0 (x", y") «xx' - yy')x" (xx' x" - yy'x" - yx'y" - xy'y", yx'x" + xy'x" + xx'y" - yy'y") (x, y) 0 (x'x" - y'y", y'x" + y"x') (x(x'x" - y'y") - y(y'x" + y"x'), y(x'x" - y'y") + x(y'x" + y"x')) (xx'x" - xy'y" - yy'x" - yy"x', yx'x" - yy'y" + xy'x" + xy" |
x') «x, y) 0 (x', y')) 0 (x", y") 1.55. Let 0 be the binary operation in Q defined by (a) a 0 b = a - b + ab, (b) a 0 b a + b + ab 2 (c) a o b a + b 3 (a 0 b) 0 c = a 0 (b 0 c) for all a, b, cEQ Determine which of the above binary operations satisfy (i) (ii) a 0 b = boa for all a, b E Q Solution: (i) (a) (a 0 b) 0 c # a 0 (b 0 c) for some a, b, cEQ; e.g. 2 - 2 + 4 = 4 and 2 0 (0 0 2) = 2 0 (0 - 2 + 0) = 2 0 -2 = 2 - (-2) - 4 = O. (2 0 0) 02 = (2 - 0 + 0) 02 = 2 0 2 = (b) (aob)oc # ao(boc) for some a, b, cEQ; e.g. (1 0 0) 00 (c) 1 0 0 = -!. (a 0 b) 0 c # a 0 (b 0 c) 100 =!. for some a, b, cEQ; e.g. (1 0 0) 00 -!00 =1. and 1 0 (0 0 0) 4 too 1 and 10 (0 0 0) "9 (ii) (a) aob#boa since 100 = 1 and 001=-l. (b) aob (c) aob a + b + ab 2 a+b 3 b+a 3 boa b + a + ba 2 = boa 22 SETS, MAPPINGS AND BINARY OPERATIONS [CHAP. 1 b. The multiplication table So far we have introduced a number of definitions and notations and familiarized our selves with them. The object of this section is to introduce a "table" as a convenient way of either defining a binary operation in a finite set S or tabulating the effect of a binary operation in a set S. To explain this procedure, suppose S = {1, 2, 3} and let p. be the binary operation in S defined by p.: (1,1) -> 1, (1,2) -> 1 |
, (1,3) -> 2, (2,1) -> 2, (2,2) -> 3, (2,3) -> 3, (3, 1) -> 1, (3,2) -> 3, (3,3) -> 2 Then a table which sums up this description of p. is 123 We put the number (2, 3)p. = 3 in the square that is the intersection of the row facing 2 (on the left) and the column below 3 (on the top). More generally, in the (i, j)th square, i.e. the intersection of the ith row (the row labelled or faced by i) and the jth column (the column labelled by j), we put (i, j)p.. A table of this kind is termed a multiplication table because it looks like the usual multi plication tables. One often calls p. a multiplication in S. Thus when we talk about a multiplication /L in a set S, we mean that p. is a binary operation in S. There is a reverse procedure to the one described above. For example, suppose we start out with a table Then there is a natural way of associating with this table a binary operation p. in {1, 2, 3}. We simply define (i, j)p. to be the entry in the (i, j)th place in the table. For example, (1,1)p. = 1, (2,3)p. = 3, (3,2)p.= 3 We shall usually define multiplications in a finite set by means of such tables. Problems 1.56.' Write down the multiplication tables for the following binary operations in 8 == {1, 2, 3}. (i) "': (1,1) --> 2, (1,2) --> 3, (1,3) --> 1, (2,1) -> 3, (2,2) -> 1, (2,3) --> 2, (3,1) --> 1, (3,2) --> 2, (3,3) --> 3. (ii) f3: 8 2 --> 8 defined by (i, j)f3 == 1 for all (i, j) E 8 2 • (iii) y: (1,1) --> 1, (2,2) --> 1, (3,3) --> 1, (1,3) --> 2 |
, (3,1) --> 2, (2,3) --> 1, (3,2) --> 1, (2,1) --> 3, (1,2) --> 3. Solution: (iiiiii) 1 2 3 Sec. 1.5] BINARY OPERATIONS 23 1.57. Does the following table define a binary operation in {I, 2, 3}? In {1, 2, 3, 4}? Solution: The table does not give a binary operation in {I, 2, 3} since (1,3) --. 4 fl: {I, 2, 3}. The table also does not define a binary operation in {I, 2, 3, 4}, because (1,4), (2,4), (3,4), etc., have no images. 1.58. Write down explicitly the binary operations in {I, 2, 3, 4} defined by the following tablesi) (ii) (iii) Solution: (i) (1, j) --. j (j = 1,2,3,4); (j, 1) -> j (j = 2,3,4); (2,2) -> 3; (2,3) -> 4; (3,2) -> 4; (3,3) -> 1; (3,4) --> 2; (4,3) -> 2; (4,4) --> 3; (4,2) -> 1; (2,4) -> 1. (ii) (1, j) -> j (j = 1,2,3,4); (j,l) -> j (j = 2,3,4); (2,2) --. 4; (2,3) --> 1; (3,2) -> 1; (3,3) --> 4; (3,4) --> 2; (4,3) -> 2; (2,4) -> 3; (4,2) -> 3; (4,4) -> 1. (iii) (i, j) -> i (i = 1,2,3,4 and j = 1,2,3,4). 1.59. Rewrite the three binary operations in Problem 1.58, using (a) circle notation, i.e. write (i, j)f3 as i 0 j, (b) additive notation, i.e. write (i |
, j)f3 as i + j, (e) multiplicative notation, i.e. write (i, j)f3 as i· j. Solution: (a) Problem 1.58(i): 10 j = j (j = 1,2,3,4); j 0 1 = j (j = 1,2,3,4); 202 = 3; 303 = 1; 404 = 3. Problem 1.58(ii): 1 0 j = j (j = 1,2,3,4); j 01 = j (j = 2,3,4); 202 = 4; 3 0 3 = 4; 304 = 4 0 3 = 2; 204 = 4 0 2 = 3; 404 = 1. Problem 1.58(iii): i 0 j = i (i = 1,2,3,4 and j = 1,2,3,4). (b) Problem 1.58(i): 1 + j = j (j = 1,2,3,4); j + 1 = j (j = 1,2,3,4); 2 + 2 = 3; 3 + 3 = 1; 4 + 4 = 3. Problem 1.58(ii): 1 + j = j and j + 1 = j (j = 1,2,3,4); 2 + 2 = 4; 3 + 3 = 4; 4 + 4 = 1. Problem 1.58(iii): i + j = i (i = 1,2,3,4 and j = 1,2,3,4). (e) Problem 1.58(i): 1· j = j and j'1 = j (j = 1,2,3,4); 2·2 ::: 3; 2' 3 = 3' 2 = 4; 3· 3 = 1; 3' 4 = 4·3 = 2; 4' 4 = 3; 4·2 = 2' 4 = 1. Problem 1.58(ii): l'j = j and j. 1 = j (j = 1,2,3,4); 2· 2 = 4; 2· 3 = 3·2 = 1; 3' 3 = 4; 3·4 = 4' 3 = 2; 2·4 = 4' 2 = 3; 4' 4 = 1. Problem 1.58(iii): i' j = i (i = 1,2,3,4 and j = |
1,2,3,4). 24 SETS, MAPPINGS AND BINARY OPERATIONS [CHAP. 1 A look back at Chapter 1 We began with a few remarks about sets. We then introduced the idea of cartesian products. This led to the idea of an equivalence relation on a set. Then the notion of a mapping was defined, followed by the definition of a binary operation. In this book we are mainly concerned with binary operations in sets. At this stage the reader may wonder what one could possibly say about binary operations in a set. Without some specialization we can say very little. In Chapter 2 we begin to place re strictions on binary operations. Supplementary Problems SETS 1.60. 1.61. The sets S;(i:=1,2,...,n) are such that SiCSi+1 (i:=1,2,...,n-1). Find SlnS2n···nSn and Sl uS2u··· USn- Let E:= {n I n E Z and n even}, 0:= {n I n E Z and n odd}, T:= {n I n E Z and n divisible by 3}, and F:= {n I n E Z and n divisible by 4}. Find (i) En T, (ii) Eu T, (iii) TnFnEnO, (iv) TUF, (v) OnF, (vi) On T. 1.62. Given A = {-3, -2, -1, 0, {1, 2, 3,}} and B {-3, -1, {1,3}}. Find AuB and AnB. 1.63. Prove Sn(TuU):= (SnT)u(SnU). 1.64. Let A:= {-5,-4,-3,...,3,4,5}, B:= {-4,-2,0,2,4}, C:= {-5,-3,-1,1,3,5}, D:= {-4,4}, E := {-3, -2, -1, O}, F := O. Which. if any, of these sets take the place of X if (i) X - C := 0, (iv) XCB and X is not a subset of E, (ii) XnB:=C, |
(iii) XCC but X is not a subset of A, (v) XnCcA, (vi) Xu(BnD):= A? 1.65. Prove Sc T if and only if (TnC)uS:= Tn(CuS) for every set C. '::ARTESIAN PRODUCTS 1.66. Prove S X (TuW) := (S X T)u(S X W) for any sets S, T and W. 1.67. 1.68. Let P be the set of positive integers and S:= P2. Show that E = {p I p := ((r, s), (t, w)) E S2 and r + w := S + t} is an equivalence relation on S. Find the E-class determined by (4,7). Find the equivalence relation, E, on Z: q E Z}, {n I n:= 1 + 4q for q E Z}, {n I n := 2 + 4q for q E Z}, and {n I n:= 3 + 4q (ii) if every equivalence class consists of a single integer; {q, -q} for each q E Z. for for q E Z}; (iii) if the equivalence classes are (i) if the equivalence classes of E are {n I n:= 4q 1.69. If E and F are equivalence relations on S, is (i) EnF, (ii) EuF an equivalence relation on S? 1.70. What is wrong with the following argument: E is a non-empty subset of S2 which has the symmetric and transitive properties. If (a, b) E E, then by the symmetric property, (b, a) E E. But by the implies (a, a) E E. Therefore E is also reflexive. transitive property, (a, b) E E and (b, a) E E 1.71. (a) If P is the set of positive integers, show that E:= {(a, b) I (a, b) E p2 and a divides b} is reflexive and transitive but not symmetric. (b) Find an example of a subset E of p2 which is both symmetric and transitive but not reflexive. (c) Find an example of a |
subset E of p2 which is reflexive and symmetric but not transitive. CHAP. 1] SUPPLEMENTARY PROBLEMS 25 MAPPINGS 1.72. Show that the following define mappings of Z into Z. (i) a: X -> {X/~XI if x=o if x~o Ixl is the absolute value of x, i.e. Ixl if x""o if x<O {-= and r is the number of distinct primes dividing x if Ixl = ° or 1 if Ixl ~ ° or 1, (ii) {3: x -> (iii) y: x --> {(_OI)r n if x<o if x=o if x>O (iv) 8: x -> sin2 x + cos2 X 1.73. Which of the mappings a, (3, y, I) of the preceding problem are equal? 1.74. 1.75. 1.76. 1.77. Find subsets S(~ 0) and T(~ 0) of the real numbers R such that the mappings (i) a: x -> cos x, (iii) y: x -> tan x are one-to-one mappings of S onto T. Do a, {3, y define (ii) {3: x -> sin x, and mappings of R into R? Let E = {n I n E P and n even}. Define a: E -> E by na = n for all nEE, and {3: E -> E by n{3 = 2n for all nEE. Find an infinite number of mappings a', {3' of Pinto P such that a~E = a and {3(E = {3. Suppose a is a mapping of a set S into a set T and, for any subset W of S, Wa = {t I t E T and t = 8a (i) (A uB)a = AauBa; (ii) (A n B)a C; Aa n Ba; and (iii) A C; B If A and B are any subsets of S, show: implies Aa C; Ba. for some 8 E W}. Let S be a subset of a set T, and a: T -> W. Prove: (ii) a onto implies a is onto. lS ( |
i) a one-to-one implies a is one-to-one; lS COMPOSITION OF MAPPINGS 1.78. Suppose a: S -> T is onto and {3: T -> U is onto. Show a 0 {3 is an onto mapping. 1.79. Given a: S -> T is one-to-one and {3: T --> U is one-to-one. Prove a 0 {3 is one-to-one. 1.80. (ii) {3: x -> sin (sin (x)), and (iii) y: x -> yl - x 2 define mappings of non-empty (i) a: x -> sin (X2), subsets of the real numbers R into R. First, find an appropriate subset in each case. Secondly, write a, {3 and y as the composition of two mappings, giving in each case the domain and codomain of each mapping defined. BINARY OPERATIONS 1.81. Let S be a set and d the set of all subsets of S, i.e. d union define binary operations on d. {A I A C; S}. Show that intersection and 1.82. How many different binary operations can be defined on a set of 3 elements? 1.83. Consider the set, F, of mappings Ii (i = 1,2,...,6) of R - {O, I} into R defined for each x E R - {O, I} by: 11 :x->x; 12:x->-I--; la:x->--; 14 :x->-; I s :x->--I; 16 :x->l-x. Show that composition of mappings is a binary operation on F and write a multiplication table for the operation. x-I x 1 -x x x- 1 x Chapter 2 Groupoids Preview of Chapter 2 In this chapter we define a set G together with a fixed binary operation 0 to be a groupoid. As we remarked at the end of Chapter 1, there is little one can say about binary operations without making restrictions. The first restriction is tllat of associativity. A groupoid with set G and operation 0 is said to be associative if glo (gz 0 ga) = (gi 0 g2) 0 ga for all gl, g2, ga in G. Such a groupoid is called a semigroup |
. In order of increasing specialization we have the concepts of groupoid, semigroup, and group. To define a group we need the concepts of identity and inverse. Hence we discuss these ideas. We introduce the semigroup Mx of mappings of X into X. The importance of Mx is that, but for the names of the elements, each semi group is contained in some Mx. Two other important concepts we deal with are homomorphism and isomorphism. Homomorphism is a more general concept than isomorphism. There is an isomorphism between two groupoids if they are essentially the same but for the names of their elements. 2.1 GROUPOIDS a. Definition of a groupoid Consider the set Z of integers. Z has two binary operations, addition (+) and multi plication (x). The set Z is one thing, a binary operation in Z is another; the two together constitute a groupoid. Repeating this definition in general terms. Definition: A groupoid is a pair (G, p,) consisting of a non-empty set G, called the carrier, and a binary operation p, in G. We shall mostly use a multiplicative notation when dealing with groupoids. Thus we write g' h or simply gh for (g, h)p" g, h E G. This notation has been used in Chapter 1 in our consideration of binary operation. As an example, let G = {I, 2, 3} and let p, be the binary operation in G defined by the following table Then the pair (G, p,) is a groupoid. Suppose we use multiplicative notation' ; then 1. 1 = 1, 1· 2 = 3, 2· 2 = 1, 3· 2 = 2, etc. 26 Sec. 2.1) GROUPOIDS 27 These products look bizarre unless we recall that the notation employed is a shorthand version of (1,1)/.L = 1, (1,2),u = 3, (2,2)" = 1, (3,2),u = 2, etc. If we use the expression "the groupoid G," where G is a set, it is understood that we have already been given a binary operation I' in G, and that we have been talking about the groupoid (G,,u). Suppose now that (G,,u) is a groupoid. If we use the circle notation for,u, i.e. we write |
goh instead of (g,h),u, we shall sometimes write (G,o) to refer to the groupoid (G,,u). Sim ilarly we write (G, e), (G, +), (G, x) if we employ g" h, g + h, g x h respectively for (g, h)p.o Example I: Let G == {1,2} and let /1 be the binary operation in G defined as follows: (1,1)/1 == 1, (1,2)/1 == 2, (2,1)/1 == 1, (2,2)/1 = 2 If we use the circle notation, we have 101 = 1, 1 0 2 == 2, 201 = 1, 2 0 2 == 2 The pair (G, /1) or (G,o) is then a groupoid. Example 2: Let 8 be the set of all mappings of {I, 2, 3} into {I, 2, 3}. Then (8,0) is a groupoid, where 0 is interpreted as the usual composition of mappings. The composition makes sense, for if ", {3 E 8, then,,: {I, 2, 3} ~ {I, 2, 3} and f!: {I, 2, 3} ~ {I, 2, 3} Therefore" 0 {3, the composition of " and {3, is defined by a", 0 {3 == (a",){3, a E {I, 2, 3}, and is once again a mapping of {I, 2, 3} into itself. (8,0) is indeed a groupoid. Example 3: Let 0 be the binary operation in Q, the rational numbers, defined by a 0 b == a + b + abo Then (Q,o) is a groupoid, because for every pair of rational numbers a and b, a 0 b defines a unique rational number a + b + abo Example 4: Let R2 be the plane. Further, let there be a cartesian coordinate system in R2 and let C be the disc of radius 1 with center at the point (2,0) of the coordinate system. Consider the region R2 - C, the unshaded area in the diagram. We term any path beginning and ending at 0, which does not meet any point of C (i.e. it is entirely in R2 - C), |
a loop in R2 - C. By a path we mean any line which can be traced out by a pencil without raising the point from the paper. For example, 1 and mare loops in R2 - C. Let L be the set of all such loops in R2 - C. Then there is a natural binary operation in L which we denote by "; thus if 11,12 E L, then 11 0 12 is the loop obtained by first tracing out 11, followed by tracing 12 • This type of groupoid (L,o) is of considerable importance in modern topology. 28 GROUPOIDS [CHAP. 2 Example 5: Let F be the set of all mappings of R, the real numbers, into R. Consider two elements a, {3 E F. We define the mapping a + (3: R --. R by a(a + (3) = aa + a{3, a E R. a + {3 is clearly a unique mapping of R into R and hence is an element in F. Therefore + is a binary operation in F and (F, +) is a groupoid. Notice that (F, +) is not the groupoid with F as the carrier and the composition of mappings as the binary operation. Problems 2.1. Are the following groupoids? (i) (ii) (8,0) where 8 = {1, 2, 3, 4} and i 0 j = 1 for i and j elements of 8. (Z, -), the set of integers with the usual subtraction of integers as binary operation. (iii) (P, -), the set of positive integers with the usual subtraction as binary operation. (iv) (Q, +), the set of rational numbers with the usual binary operation of division. (v) (Z, +), the set of integers with the usual binary operation of division. Solution: (i) (8,0) is a groupoid since 0 is clearly a binary operation in 8. (ii) (Z, -) is a groupoid; for if a, b E Z, then a - b is a unique element of Z. (iii) (P, -) is not a groupoid since a - b fl P for all a, bE P. Therefore is not a binary operation in P. (iv) (v) (Q, +) is not a groupoid because a + 0 is not defined for |
any a E Q and hence + is not a binary operation in Q. (Z, +) is not a groupoid since a + b fl Z for all a, bE Z, e.g. 2 + 3 fl Z. Therefore division is not a binary operation in Z. 2.2. Is (Z,o) a groupoid if 0 is defined as (i) a o b = v'a+ b, (ii) a o b = (a + b)2, (iv) a 0 b = 0, (v) a 0 b = a? Solution: (iii) a o b = a- b - ab, All but (i) define a binary operation in Z. Therefore (Z,o) is a groupoid in (ii) through (v). The multiplication 0 in (i) does not define a binary operation in Z since v' a + b is not always an integer. 2.3. Let 8 be any non-empty set and T the set of all subsets of 8. Are (T, n) and (T, u) groupoids? Solution: Both intersection n and union U are binary operations on T, for the intersection or union of two subsets of 8 is again a unique subset of 8. Thus (T, n) and (T, u) are groupo ids. h. Equality of groupoids Two groupoids are equal if and only if they have the same carriers and the same binary operation. Remember, a binary operation was defined as a mapping and two mappings are equal if and only if they have the same domain and codomain, and the image of each element is the same under both mappings. Thus the groupoids described in Examples 1-5 are all different. Problems 2.4. Are any two of the groupoids in Problems 2.1-2.3 equal? Solution: No. 2.5. Which of the following pairs define equal groupoids? (i) (Z, +) and (Z, 1'), where (a, b)/L = a + b. (Z, -) and (Z, 0), where a 0 b = a-b. (ii) (iii) (Z,o) where a 0 b = a for all a and b in Z, and (Z, x) where a X b = b for all a and b in Z. Solution: The groupoids in (i) are clearly the |
same. So too are the groupoids in (ii). In (iii) (Z,o) is not the same as (Z, X); for if a#- b, a 0 b #- a X b. Sec.2.2J COMMUTATIVE AND ASSOCIATIVE GROUPOIDS 29 2.2 COMMUTATIVE AND ASSOCIATIVE GROUPOIDS Definition of commutative and associative groupoids Let (Z, +) be the groupoid of integers under the usual operation of addition. Then and for all a, b, c E Z. a+b b+a (a+b)+c a + (b + c) Similarly if (Z,·) is the groupoid of integers under multiplication, and for all a, b, c E Z. a' b b· a (a'b)'c a • (b' c) The analog of (2.1) and (2.3) in an arbitrary groupoid (G,o) is aob=boa for all a, bEG. Similarly the analog of (2.2) and (2.4) III (G,o) is (aob)oc = ao(boc) (2.1) (2.2) (2.3) (2.4) (2.5) (2.6) for all a, b, c E G. We term a groupoid satisfying (2.5) commutative or abelian, and a groupoid satisfying (2.6) associative or a semigroup. Thus a semigroup is an associative groupoid. Of course it is not clear that there are non-commutative groupoids, i.e. groupoids which are definitely not commutative, and similarly it is not clear that there are non associative groupoids. We settle the issue now. Let G = {1,2} and let 0 be the following binary operation in G: 1 2 1!tTl 2~ Then (G,o) is a groupoid. Observe that 102 = 1 but 201 = 2, so G is not commutative. Furthermore, (2 0 1) 02 = 202 = 1 but 20 (1 0 2) = (2 0 1) = 2, so G is also non-associative, i.e. G is not a semigroup. For the most part we shall use the |
multiplicative notation for a groupoid (G, fL) and simply talk about the groupoid G. If the groupoid is commutative we will use the additive notation instead of the multiplicative notation, since we are accustomed to addition as a commutative binary operation, e.g. in the integers. The o'rder of a groupoid (G, fL) is the number of elements in G and is denoted by IGI; (G, fL) is infinite if IGI is infinite, and finite if IGI is finite. Problems 2.6. Which of the groupoids in Examples 1, 2, 3 and 5 are commutative and which are associative? Solution: The groupoid of Example 1 is not commutative, since 102 = 2 and 201 = 1, but is associative. To show (G, 0) is associative we must examine the following 8 cases: (a) 1 0 (1 0 1) = 1 0 1 = 1, (1 01b) 2 0 (1 01) = 201 = 1, (2 0 1) 0 1 = 1 01 = 1 (c) 1 0 (2 0 1) = 101 = 1, (1 02) 01 = 2 0 1 = 1 (d) 1 0 (1 02) = 1 02 = 2, (1 01) 02 = 1 02 = 2 (e) 2 0 (2 0 1) = 201 = 1, (2 0 2) 01 = 2 0 1 = 1 (I) 2 0 (1 02) = 2 02 = 2, (2 0 1) 02 = 1 0 2 = 2 (g) 1 0 (2 0 2) = 1 02 = 2, (1 02) 02 = 2 0 2 = 2 (h) 20 (2 0 2) = 2 02 = 2, (2 0 2) 02 = 2 02 = 2 30 GROUPOIDS [CHAP. 2 The groupoid of Example 2 is not commutative: for if a is defined by 1a = 1, 2a = 3 and 3a = 2, then 1a 0 f3 = (la)f3 = 1f3 = 2, 1f3 0 a = (lf3)a = f3 0 a. Since the binary composition of mappings is an associative binary and f3 2a = 3. Hence a 0 f3 ¥ operation (Problem 1.41, page 18), ( |
8,0) is an associative groupoid. is defined by 1f3 = 2, 2f3 = 1 and 3f3 = 1, The groupoid in Example 3 is both commutative and associative. a 0 b = a + b + ab = b + a + ba = boa, since addition and multiplication are commutative binary operations in Q. Also, a 0 (b 0 c) = a 0 (b + e + be) = (a -!- (b + c + be)) + alb + c + be) = a + b + e + be + ab + ae + abc and (a 0 b) 0 e = (a + b + ab) 0 e = a + b + ab + c -I- (a + b + able = a + b + ab + e + ae + be + abc. U sing the associative and commutative properties of addition and multiplication in the rationals, we see ao(boe) = (aob)oe. In Example 5 the groupoid (F, +) is commutative because a(a + (3) = aa + af3 = af3 + aa = a(f3 + a), a, f3 E F and a E R (here we use the fact that a", af3 E R and addition is a commutative binary operation in R). (F, +) is also a semigroup, for a«" + (3) + y) = ala + (3) + ay = aa + (af3 + ay) = a" + a(f3 + y) = a(cr + (f3 + y)) (here we use the associativity of addition in R). 2.7. Construct an example of a commutative groupoid of order 3. Solution: Let 8 = {a, b, e} and the binary operation 0 be defined by the multiplication table (8,0) is clearly a commutative groupoid. 2.8. Show that the set Q* of nonzero rational numbers with binary operation the usual division of rational numbers, is a groupoid. Is it commutative? Is it associative? Solution: If a b an d are any wo e emen SO " ', t I t f Q'" d e th e en b + d = be ad. a IS a umque e |
emen m. I t'Q* (~ ¥ 0 since a, b, e, d ¥ 0). Therefore division is a binary operation in Q*. However, division t + (1 + 1) = t· = i + t) nor associative (e.g. t + i = 2 ¥ is neither commutative (e.gt + i) + i)· 2.9. Which of the groupoids in Examples 1-3, 5 and in Problems 2.7 and 2.8 are finite? Solution: The groupoid of Example 1 is clearly finite of order 3. In Example 2 the set 8 of all mappings of {1, 2, 3} into itself contains 27 elements, and so is finite. In Example 3, (Q,o) is not finite as there are an infinite number of rational numbers. In Example 5 the set F is infinite. To show that F is not finite we construct an infinite number for all i = j. Therefore we have found an infinite number of of mappings of R into R as follows. Let Pi: R -> R (i = 1,2,3,... ) be defined by rpi = r r ¥ i E Rand iPi = O. Clearly Pi = Pj different elements in F. Notice the Pi are not all the elements of F. iff In Problem 2.7 the groupoid has only three elements and is therefore finite. In Problem 2.8, since there is an infinite number of nonzero rational numbers, Q* is not finite. 2.3 IDENTITIES AND INVERSES IN GROUPOIDS a. The identity of a groupoid Let G be a groupoid written mUltiplicatively. An element e in G is called an identity element of G if eg ge g Sec. 2.3] IDENTITIES AND INVERSES IN GROUPOIDS 31 for every g E G. For example in the multiplicative groupoid Z of integers, 1 is an identity element. A less natural example is the groupoid {I, 2, 3} with multiplication table given by The element 3 is an identity element of G since 3. 1 = 1 = 1. 3, 3· 2 = 2 = 2. 3, 3' 3 = 3 = 3· 3 One might ask whether or not a groupoid can have more than one identity element. The following theorem settles this question. Theorem 2.1: If a groupoid G has an identity |
element, it has precisely one identity element. In other words if e and e' are identity elements of G, then e = e'. Proof: Since e is an identity element of G, ee' = e'. But e' is also an identity element of G. Hence ee' = e and so e = e'. It is instructive to reformulate the proof of Theorem 2.1 in a different notation. Thus we revert to the notation (G, p,) and instead of the multiplicative notation gh we write (g, h)p,. Suppose e and e' are identity elements. Then as e is an identity element, (e, e')p, = e'. But e' is also an identity element. Hence (e, e')p, = e; and because the image of any element under the mapping is unique, e = e'. Problem 2.10. Which of the following groupoids have identity elements? (i) The groupoid (Z, +) under the usual operation of addition. (ii) The groupoid of nonzero rational numbers under division. (iii) The groupoid of complex numbers under multiplication. (iv) The groupoid of all mappings of {I, 2, 3, 4} into itself under the composition of mappings. (v) The groupoid with carrier {1,2} and multiplication table 1 2 1[2J!J 2~ (b) 1 2 1~ 2~ (e) 1 2 :8:Hj (a) 1 2 (d) :8!E Solution: (i) (ii) (Z, +) has the identity element 0, since 0 + z = z + 0 = z for all z E Z. If this groupoid had an identity element e, then e +- q = q for all q in the groupoid. In particular, e +- e = e and so e = 1. But 1 +- 2 = 1/2 oF 2. Hence there is no identity element. (iii) Recall that a complex number is any number of the form a + bi where a, b E Rand (a + bi)(1 + Ot) = a + Oai + i = R. 1 + Oi is the identity element of this groupoid, since Obi2 + bi = a + bi = (1 + Oi)(a + bi). (iv) The identity mapping " defined by j, = j, j E |
{1, 2, 3, 4}, is the identity element of the group oid since j(u 0,) = (ju), = ju = (j,)u = j(, 0 u). (v) Only (d) has an identity element, namely 1. b. Inverses in a groupoid If we use multiplicative notation for groupoids, we shall for the most part reserve the symbol 1 for the identity element. 32 GROUPOIDS [CHAP. 2 In the multiplicative groupoid of nonzero rational numbers, one talks about inverses; In gen for example, t is the inverse of 2, the determining factor being 2 X t = 1 = t x 2. eral if G is any groupoid with an identity 1, we term h an inverse of g (h, g E G) if gh = 1 = hg Clearly if h is an inverse of g, then g is an inverse of h. Examples follow. Example 6: Let G = {a, b, e} be the groupoid with multiplication table Then a is the identity element of G, as inspection of the table shows. Furthermore, be = a = eb and aa = a = aa, so that e is an inverse of b, b is an inverse of e, and a is its own inverse. a, band e have no other inverses. Example 7: Let G be the groupoid of mappings of {I, 2, 3} into itself. Then the identity mapping jL = j (j = 1,2,3) is the identity element of the groupoid (see Prob L defined by lem 2.10(iv». Now let U E G be defined by 7" E G, defined by IT = 3, 27" = 1, 37" = 2, lu = 2, 2u = 3, 3u = 1. Then is an inverse of u because lu 0 7" = (lu)7" = 27" = 1, 2rr 0 T = (2U)T = 37" = 2, 3u 0 7" = (3u)7" = IT = 3 Similarly, 7"U = L. Unlike identities, inverses in groupoids are not always unique. For example, let which implies U7" = L. G = {I, 2, 3, 4} be the groupoid with multiplication table Here 1 is the identity element in G. |
Moreover, 2·2 = 1 = 2·2 and 2·3 = 1 = 3·2; thus 2 has two inverses, 2 and 3. Notice that in this groupoid, 4 is not an inverse of 2 even though 2·4 = 1. The definition of an inverse requires both 2·4 and 4·2 to equal 1. Problems 2.11. Consider the groupoids in Problem 2.10 which have identity elements. What elements in each of the groupoids have inverses? Solution: (i) Any integer z has an additive inverse, namely -z; for z + (-z) = 0 = (-z) + z. (iii) All nonzero elements in the groupoid of complex numbers under multiplication have inverses; i.e. if a + bi (a and b not both zero) is any element in the groupoid, then (1 )(a-bi) a + bi a - bi a2 + b2 = a2 + b2 - 1 a + bi = a - bi a b. a2 + b2 ~ is an element in the groupoid and a ~ bi (a + bi) = 1 = (a + bi) a ~ bi. (0 + Oil has no in verse, since (0 + Oi)(a + bi) = (0 + Oil. Sec. 2.4] SEMIGROUPS WITH AN IDENTITY ELEMENT 33 (iv) Only mappings T which are one-to-one and onto have inverses. Say T is a one-to-one mapping of {I, 2, 3, 4} onto itself, defined by h = a, 2T = b, 3T = C and 4T = d, where a, b, c, dare the elements of {I, 2, 3, 4}. We define the mapping u which will be an inverse of T, by au = 1, bu = 2, CU = 3 and du = 4. u is a mapping of {I, 2, 3, 4} onto {I, 2, 3, 4}, [:ince {a, b, c, d} = IT ° U = (IT)u = au = 1, 2T ° U = bu = 2, 3T ° U = CU = 3, and 4T ° u = du = 4. Hence {I, 2, 3, 4}. TOU = t. We must also |
show UOT = t. Now aUOT = IT = a, bUoT = 2T = b, CUOT = 3T = C, du ° T = 4T = d. As {a, b, c, d} is {I, 2, 3, 4}, u ° T = t. If T is not one-to-one then there are at least two elements, a and b in {I, 2, 3, 4} which are mapped onto the same element, c, by T, i.e. aT = C and bT = C. Now if u is an inverse of T, then arT ° u) = a and b(T ° u) = b thus CU = a and cu = b. But under a mapping or each element has a unique image. Hence we have a contradiction. So T has no inverse. (aT)u = CU = a and (bT)U = CU = b; (a # b) (v) Both 1 and 2 have inverses since l'1 = 1 = 1 • 1 and 2·2 = 1 = 2' 2. 2.12. Find the identity of the groupoid (Q,o) where a ° b What elements have inverses? a + b + ab (see Example 3, page 27). Solution: o is the identity of (Q,o), since 0 ° a = 0 + a + Oa = a and a ° 0 = a + 0 + Oa = a. To find such that a ° x = 0 = x ° a. Now an inverse for an element a E Q, we must find an + ax, so that x must satisfy the equation a + x + ax = O. If a = -1, we obtain -1 + x x = -1 = 0; thus if a = -1, there is no x such that aox = O. If a # -1, the equation a + x + ax = 0 can be solved, giving x = - 1 ~ a; thus a ° 1 ~aa = O. Since (Q,o) is commutative, 1 ~aa ° a = 0 so that 1 ~aa is an inverse of a. Hence all elements of Q, except -1, have inverses. 2.13. Find the identity of the groupoid (F, +) in Example 5 and show that every element has an inverse. |
Solution: for all r E R, The mapping w: R --> R defined by rw = 0 is the identity of (F, +). For a(a + w) = aa + aw = aa + 0 = aa = 0 + aa = aw + aa = a(w + a) implies a + w = a = w + a. If f3 is an inverse of a, then a + f3 = wand a(a + (3) = aw for all a E R. But a(a + (3) = aa + af3 and aw = O. Thus aa + af3 = 0 or -(aa) = af3. Consequently if f3 is an inverse of a, the image of a E RundeI' f3 must be the negative of the image of a under a. We therefore define f3 is a mapping of R into R, since -(aa) is an inverse f3 for the mapping a by af3 = -(aa), a E R. a unique element of R. Furthermore, a(a + (3) = aa + af3 = aa + (-(aa)) = 0 = aw implies a + f3 = w. (F, +) is a commutative groupoid. Hence a + f3 = w = f3 + a and f3 is an inverse of a. for all a E R 2.4 SEMIGROUPS WITH AN IDENTITY ELEMENT a. Uniqueness of inverses Suppose G is a groupoid with an identity 1 and suppose h is the inverse of g: gh = 1 = hg. It is tempting to employ the notation used when dealing with real numbers and write g-l for an inverse of g. The trouble with this notation is that in a groupoid an element may have more than one inverse, as we have already seen in Section 2.3b. However, the associative law on a groupoid prohibits this as we see from Theorem 2.2: Let G be a semigroup with an identity element 1. If g E G has an inverse, Proof: it has precisely one; i.e. if hand h' are inverses of g, then h = h'. (since h1 = h for all h E G) (gh' = 1, since h' is an inverse of g) = h(gh') h = h1 (by associ |
ativity) (hg = 1, since h is an inverse of g) = (hg)h' = 1h' = h' 34 GROUPOIDS [CHAP. 2 Theorem 2.2 entitles us to denote the inverse of an element g in a semigroup, written multiplicatively, by g-l. Note that if g and h have inverses, then gh has an inverse, namely h-1g- 1. (Notice the reversed order.) For, (gh)(h-lg-l) = ((gh)h-1)g-1 = (g(hh-l))g-l = (gl)g-l = gg-l = 1 Similarly, (h-lg-l)(gh) = 1. Problems 2.14. Let G = {I, 2, 3, 4}. The binary operations on G given by the following tables make G into a groupoidabe) 1 2 3 4 Which of the groupo ids are semigroups? Which have an identity? Which elements have inverses? Solution: (a) In order to see that in this case G is a semigroup, we must check associativity, i.e. we must show a(be) = (ab)e for every a, b, e E G. Notice that when either a, b or e is 1, a(be) is clearly equal to (ab)e; e.g. if e = 1, (ab)1 = ab = a(bl). Since 4g = 4 = g4 for any g E G, then (ab)e = 4 = a(be) if either a, b or e is 4. Therefore we need only check the products when a, band e have values 2 or 3. If two of the three elements a, b, e are equal to 2 and the other is equal to 2 or 3, then (ab)e = 4 = a(be) because 2· 2 = 4 and 2· 3 = 2 = 3' 2. The follow ing calculations take care of the remaining cases: 3(3 • 33' 3)3 3(2 • 33' 2)3 2(3 • 32 • 3)3 3(3 • 2) = 3 • 2 = 2 = (1 • 2) = (3 • 3)2 1 is the identity element of G. The only elements which have in |
verses are 1 and 3; these inverses are unique. (b) The associative law does not hold, since 4(2' 3) = 4· 4 = 1 and (4·2)3 = 2· 3 = 4. 1 is the identity, and 1,2,3 and 4 have inverses. Notice that the inverses are unique even though G is not a semigroup. (c) G is a semigroup. Associativity follows from the fact that ab = a for all a, bEG; hence (ab)e = a = a(be) for any a, b, e in G. G has no identity element; therefore no element has an inverse. 2.15. Let G be the groupoid with carrier Q, the set of rational numbers, and binary operation 0 defined by a 0 b = a + b - abo Is the groupoid (Q, 0) a semigroup? Is there an identity element in (Q,o)? Which elements of the groupoid have inverses? Solution: Notice a + b - ab is a unique rational number, so that (Q,o) is a groupoid. Now if a, b, care any elements in (Q,o), (aob)oe and (a + b - ab) 0 e = a + b - ab + e a + b + e - ab - ae - be + abc ao(b+e-be) = a+b+e-be-a(b+e-be) a + b + e - be - ab - ae + abc (a + b - able Hence (Q,o) is a semigroup. The identity of (Q,o) is 0 since a 0 0 = a + 0 - aO = a and 0 0 a = 0+ a - Oa = a. Using an analysis similar to that in Problem 2.12, we find that for a # 1 the inverse of a is a ~ 1 and that a = 1 has no inverse. To check that a ~ 1 is the inverse of a, Sec. 2.4] SEMIGROUPS WITH AN IDENTITY ELEMENT 35 a a O - a-I a + a a -- 1 ( a) a - \ a-I a + -a(a-l) a-I o Similarly, a_loa = O. a 2.16. Let G |
be the groupoid with carrier Q and binary operation 0 defined by a 0 b = a - b + abo Is the groupoid (Q,o) a semigroup? Is there an identity element in (Q,o)? Which elements of the groupoid have inverses? Solution: (Q,o) is clearly a groupoid. ao(boe) a o (b - e + be) a - (b - c + be) + a(b - e + be) a - b + e - be + ab - ae + abc and (a 0 b) 0 e = (a - b + ab) 0 e = a - b + ab - e + (a - b + able = a - b + ab - e + ae - be + abc (0 0 0) 0 1 = These two expressions are not the same for all values of a, band e. For example, 0 0 1 = -1 while 0 0 (0 0 1) = 0 0 -1 = 1. Hence (Q,o) is not a semigroup. Furthermore, (Q,o) has no identity element; for if e were an identity, then col = 1 and eo 0 = 0, since 1,0 E Q. e = O. But eo 1 = e - 1 + e = 1 This is clearly impossible. Therefore (Q,o) has no identity element. c = 1, and 0 = eo 0 = e - 0 + 0 implies 2c = 2 or implies 2.17. If Q is replaced by Z, the integers, as the carrier for the groupoids in Problem 2.15, are the solutions the same? (Hard.) Solution: (Z,o) with ° defined as in Problem 2.15 is a semi group with an identity, since the argu ment for the associativity depended only upon the associativity and commutativity of addition and multiplication in Q. These laws also hold in Z. The same is true for the proof that 0 is the identity of (Z,o). However, the inverse of an element a(# 1) E Z would be a-I E Z. Our problem a then is: for which integers a # 1 is ~1 an integer? a- a Let a-I = r, an integer. Then a = 1'(a -- 1). Clearly r = 1 is impossible. (a) Assume first that a> |
1. Then r must be positive, and so l' '" 2. Hence a"" 2(a - 1) and thus 0 "" a - 2. Therefore a = 2. If a = 2, then -~-1 = 2 is an integer. (b) Now assume a "" O. If a = 0, then ~1 = 0 is an integer. If a < 0, r must be positive and"" 2. Then a = ria - 1) "" 2(a - 1) and ao "" a - 2, which is impossible. Thus 0 and 2 are the only elements with inverses. 0.- 2.18. Let G be the mappings of P, the positive integers, into P. Determine whether G is a semigroup with an identity element if the binary operation in P is (i) the composition of mappings, (ii) the addition of mappings, where" + (3 is defined by a(" + (3) = a" + a(3, a, (3 E G and a E P. Solution: (i) The composition of mappings is an associative binary operation (see Problem 1.41, Page 18). Then G with the binary operation of composition of mappings is a semigroup. The mapping j(. ° T) = (j.)T = • defined by j o. = j for all jT = (jr). = j(T °.). In Problem 2.6 we showed that the addition of mappings in the set F of all mappings of R into R is an associative binary operation. The argument here is similar. Thus G is a semigroup. then,,+ (3 = (3. Thus if j E P, then this is a,,: P --> P and, since 0 ~ P, If " were an identity element in G and if (3 E G, jet = O. But "E G, j(" + (3) = jet + j(3 = j(3; hence contradiction. Thus (G, +) has no identity. (Compare with Problem 2.13.) is the identity of G; for if T E G, j E P then (ii) 2.19. Let a, b, e be three elements in a semigroup which have inverses. Prove that a(bc) has an inverse and that this inverse is (c- 1 b- 1 )a- 1• |
Solution: We need only verify that {a(bc)}{(c- 1 b- 1)a- 1 } = a{(bc)[(c- 1 b- 1)a- 1]} = a{[(bc)(c- 1 b- 1)Ja- 1 } = a{la- 1} = aa- 1 = 1 and similarly that {(c- 1 b- 1)a- 1}{a(bc)} = 1 to prove the result. Note our use of the associative law. 36 GROUPOIDS [CHAP. 2 b. The semigroup of mappings of a set into itself A particularly important semi group is the set Mx of all mappings of a given non-empty set X into itself, where the binary operation is composition of mappings. We repeat the fl: X ~ X and definition of the composition of mappings in this special case. Suppose y: X ~ X, i.e. fl, y E Mx. We define fl 0 y to be the mapping of X into X given by X(fl 0 y) = (Xfl)y for all x EX It is clear that 0 is a binary operation in Mx. We shall use the multiplicative notation fl' y, or simply fly, instead of fl 0 y. We now show Mx is a semigroup with an identity. Theorem 2.3: If X is any non-empty set, Mx is a semigroup with an identity element. Proof: We begin by proving Mx is a semigroup. Let ft, y, p E Mx and let x EX. Then using the definition of composition of mappings, Since x is any element of X, (fly)P and fl(yP) have the same effect on every element of X. Thus (fly)P = fl(yP)' and so Mx is a semigroup. To show that Mx has an identity element, let t: X ~ X be defined by Xt = x for all x EX. Then if fl E Mx, X(tfl) = (Xt)fl = Xfl = (Xfl)t = X(flt) Thus tfl = fl = flt for all fl E Mx; hence t is an identity element of Mx. The proof of the theorem is complete. Not every element in Mx necessarily has an inverse. For example, if X = {I, |
2, 3} and u E Mx is defined by lu = 1, 2u = 1, 3u = 1, then u has no inverse; for if uy = t, we have 1 = It = l(uy) = (luh = ly and 2 = 2t = 2(uy) = (2uh = ly, so that 1 would have two dis tinct images under y, which contradicts the assumption that y is a mapping. The subset of Mx consisting of all those elements which have inverses is very important. We characterize these elements in the following theorem. Theorem 2.4: An element in Mx has an inverse if and only if it is one-to-one and onto. Proof: Suppose fl has an inverse y; then fl is onto. For if x E X, then x = Xt = x( Yfl) = then (Xy)fl and Xy E X is a pre image of x. Moreover, fl is one-to-one. For if Xfl = Yfl, Therefore, since y is a mapping, Xfl = Yfl implies x = Y; in other words, fl is one-to-one. To prove the converse, we assume fl is one-to-one and onto. Define y, which will be shown to be an inverse of fl, as follows: if x E X, we define Xy = Y where Y is that element of X which is the preimage of x under fl. To check that the definition of y is meaningful,. observe that as fl is onto there certainly is at least one element Y E X such that Yfl = x. But fl is one-to-one, i.e. distinct elements have distinct images. So Y is the unique element such that Yfl = x. To conclude we show y is the inverse of fl. Let x E X and Xfl = y. Then X(fly) = (xflh = Yy = x by the definition of y, and so flY = t. Since fl is onto, each x E X must have a preimage fi E X, i.e. fifl = x. Then X(yfl) = (fifl)(yfl) = ((Yflh)fl = (Y(fly))fl = (Yt)fl = fifl = x and so Yfl = t. Thus y is the inverse of fl and the |
proof of Theorem 2.4 is complete. Sec. 2.4] SEMIGROUPS WITH AN IDENTITY ELEMENT 37 c. Notation for a mapping When X is finite, say X = {aI,...,an }, there is a convenient way of denoting any u E M x, namely That is, we place below each element ai of X its image under u; e.g. if X = {1, 2, 3} and IT E X is defined by lIT = 1, 2IT = 1 and 3IT = 2, then IT is represented by (i = 1,2,..., n) (~.~ :) Notice that every element of X has a unique image under an element u in Mx; therefore the top row (al a2 a3... an) will contain all the elements of X and under each element will appear its unique image under IT. All elements of X need not appear in the bottom row, as we see from the example above. Problems 2.20. (a) Write the element (1 in M x, X = {I, 2, 3, 4, 5, 6}, in the notation introduced above, when (1 is defined by (i) 1(1 = 1, 2(1 = 4, 3(1 = 5, 4(1 = 6, 5(1 = 2, 6(1 = 6 1(1 = 1, 2(1 = 1, 3(1 = 1, 4(1 = 1, 5(1 = 1, 6(1 = 1 (ii) (iii) 1(1 = 6, 2(1 = 5, 3(1 = 4, 4(1 = 3, 5(1 = 2, 6(1 = 1 (b) What elements of M x, X = {a, b, c, d, e}, are represented by the following? (; b c d :) (: b (; b c d :) c d :) (iii) (ii) (i Solution: (a) (i) (~ 2 3 4 5!) G 4 5 6 2 (ii) 2 3 4 5 :) 1 1 1 1 (iii) (~ 2 3 5 4 4 3 5 2 ~) (b) The mappings defined by (i) a --+ b, b --+ a, c --+ c, d --+ d, e --+ e (ii) a --+ a, |
b --+ e, C --+ d, d --+ C, e --+ b (iii) a --+ e, b --+ a, C --+ c, d --+ b, e --+ a 2.21. Exhibit all elements of Mx when (i) X = {I}, (ii) X = {I, 2}, (iii) X = {aI' az}. Solution: (i) There is only one element in M{l},namely L: 1 --+ 1, the identity mapping. (ii) The following are the elements of M{l, 2}: G ~), G ~ ), G ~), G ~). (iii) 2.22. Write out the multiplication tables of the three semigroups in Problem 2.21(i) and (ii). (i) LO (ii) The identity of M{l, 2} l'S L -- (11 22)' 38 [CHAP. 2 Let Ul C 2\ \ 1)' °2 '1 } \., Hr;d (T:, 2\ ;2 2/ (f, U;; Then the mUltiplication table is "1 "1 a u a;~ L _____ -'- <7, (11 0'1 The mUltiplication is calculated as follows. Sinn' 1 is the identity, 1 leaves every element of jO'I = 1, M{1,2} unchanged; hence the ficst row and first column are easily written down. Since j = 1,2, and if u E M, 1,", then k(UUl) = (ku)ul = 1; thus 0'0'1 = Ul' Hence the second column consists of Ul' '-,'Similarly UU3 = U:l, so the last eoiumn consists of U:l' We must still calculate 0'1U2' U20'2 and 0'3U2' Now each element of {1, 21 is taken to 1 by Ul' and 1 is taken to 2 by U2, so UIU2 20'2 = 1 and 20'2U2 = takes every element to 2; herce 0':1' Now jU3u2 = 20'2 = 1. Thus (2u2)U2 = luz = 2; hence <TZU:! 1. U30'Z = 0'1' |
0'4 = G 2 3 2 Jz -- \ 1 3 (1 2!) G 2!) <T5 = G 2!) G 2 :) as 0'3 1 2 1 2.26. What are the inverses of the elements in Problem 2.25? Solution: Theorem 2.4 explains how to find the inverse of a mapping which is one-to-one and onto. For : 2 ---? 1. u3: 3 ---? 3, hence is its own inverse. Similarly a;-l = al, u;l = a2, 0'.;1 = u5, US-I = u4. example, to find the inverse of aa, we note 0':] takes 1 -> 2. Hence u- l 3 0'-1: 3 ---? 3. But then O'a 3 US-I = 0'6' Sec. 2.4] SEMIGROUPS WITH AN IDENTITY ELEMENT 39 2.27. Prove that the subset of elements of Mx which have inverses is a semigroup with identity under the usual compositions of mappings. Solution: Let 8 be the elements of Mx which have inverses. Is 0, the usual composition of mappings, a binary operation in 8? In other words, is (a, (3) --> a 0 f3 a mapping of 8 X 8 --> 8? If a 0 f3 E 8, (a 0 (3) 0 (f3- 1 0 a-I) = the answer is yes. So we ask: if a, f3 have inverses, does a 0 f3? Note that a O (f3of3-l)oa- l = aOloa- l = I. Hence a O f3 has the inverse f3- lo a- l. As 1 is its own inverse, 1 E 8. As Mx satisfies the associative law, so does 8. Hence 8 is a semigroup. d. The order in a product There is one way in which the associative law makes it easier to work in a semigroup than in a non-associative groupoid. Suppose S is a semigroup and let aI, az, a3 E S. There are two ways in which one can multiply aI, a2 and a3 together (in this order): (ala2)a3 and al(a2a3). The point |
of the associative law is that these products coincide. Suppose now aI, az, a3, a4 E S. Then we can multiply aI, az, a3, a4 together (in this order) in the following five ways: It is conceivable that some of these products give rise to different elements of S. However, the associative law, which of course involves products of only three elements, prohibits this. (ala2)(a3a4) = al(a2(a3~)); To see this consider first (alaZ)(a3a4). By the associative law, hence the second product coincides with the first. In fact all of the products equal the first. As a second illustration consider ((ala2)a3)~. Here we have, as desired, In general, we have Theorem 2.5: Let S be a semigroup and let aI, a2,..., an E S. Then any two products of aI, az,..., an coincide, when the al appears first in each product, az second,..., and an last. Proof: Assume the contrary, that not all possible products of aI,..., an in that order are equal. We may assume that n is the first integer for which two different products give rise to different elements. Let x and y be these two different products. N ow x = uv and y = UlVl for some u, v and some Ul, VI. Suppose U is the product of the elements aI,..., ar and v the product of the elements ar+l,..., an, while Ul is the product of aI,..., as and VI is the product of as+l,..., an. Without loss of generality we may suppose that s ~ r. If s = r then U = Ul and v = VI, since n is the first integer for which there exist two unequal then U = (al... as)(as+l... ar) while VI = products of the same elements. (as+l... ar)(ar+l... an). Hence x = uv = {(al... as)(as+l... ar)}(ar+l... an) while y = UlVl = (al... as){(aS+l... ar)(ar+l... an)}. By the associative law for the three elements (al.. |
. as), (as+l... ar) and (ar+l... an), we have x = y, contradicting the assumption that not all possible products are equal. Hence the result follows. If s < r, It follows from Theorem 2.5 that in a semigroup, if we are given the order of a product, the bracketing is immaterial. Thus we write simply alaZ... an, without brackets, for the product of aI, az,..., an in this order. For example, if m is any positive integer, we write a' a'...• a for the product of m a's; a useful abbreviation for such a product is am. If n is a second positive integer, then am. an = am+n since am. an is simply the product of m + n a's. Similarly, (am)n = amn• 40 GROUPOIDS [CHAP. 2 For example, if S = {I, 2, 3} following table, then 13 = 3, 23 = 3, 33 = 3. 1 is the semigroup with binary operation given by the.5 HOMOMORPHISMS OF GROUPOIDS AND CAYLEY'S THEOREM a. Definition of a homomorphism Before we give the formal definition of a homomorphism of one groupoid into another, we will give an example. Let (R,·) be the groupoid of positive real numbers with the binary operation of ordinary multiplication. Let (R, +) be the groupoid of real numbers with binary operation the usual addition inside R. Let us define a mapping 0: R ~ R by xO = loglOx. Recall that loglO(xy) = loglOx + loglOY; hence 0 is an example of a homomorphism. The formal definition is (xY)O = xO + yO. Definition: A homomorphism from a groupoid (G, a) into a groupoid (H, (3) is a mapping o : G ~ H which satisfies the condition ((gl, g2)a)O = (g10, g20)(3 Usually groupoids are written in multiplicative notation. The definition then takes the form: A homomorphism of (G,·) into (H,·) is a mapping 0: G ~ H such that (g1g2)O = g10g20 (2.7 |
) for all g1, g2 in G. (2.7) is often expressed as: "0 preserves multiplication." Problems 2.28. Let G be the semigroup of integers under the usual addition of integers and let H be the semigroup of even integers under the usual addition. Verify that the mapping 8: G --> H defined by 8 : 0 --> 20 for all 0 EGis a homomorphism of G into H. Solution: First we must check to see if 8 is a mapping. Since 20 is a unique integer, 8 is clearly a mapping. Let 01> 02 E G. Then U18 + U28 = 201 + 202 = 2(U1 + 02) = (01 + 02)8. Therefore 8 preserves multiplication and is a homomorphism. 2.29. Let G be the semigroup of integers under the usual multiplication of integers, and let H be the semigroup of even integers under the usual multiplication. Is CT: G --> H defined by CT: U --> 20 for all U EGa homomorphism of G into H? Solution: As in the preceding problem, CT is a mapping. But CT is not a homomorphism; for if U1> U2 E G, (01U2)CT = 20 1U2 and U1CT02CT = 201202 = 40102. Hence for all 01' U2 E G. (0102)CT # (U1CT)(02CT) then 2.30. Let (G, +) and (H, +) be the semigroups of Problem 2.28. Find a homomorphism of (G, +) into (H, +) which is not equal to 8. Solution: Define 'Tn: G --> H by 'Tn: 0 --> no where n is a fixed even integer. 'Tn is a mapping of G into H, since no E H for any even integer n and no is unique. For each n, 'Tn is a homomorphism because Sec. 2.5] HOMOMORPHISMS OF GROUPOIDS AND CAYLEY'S THEOREM 41 implies (Ul + U2)Tn = n(Ul + U2) = nUl + nU2 = UITn + U2Tw Tn ¥= 9, since 19 ¥= h n if Ul' U2 E G n ¥= 2. Therefore there is an infinite number of different hom |
omorphisms between (G, +) and (H, +). Notice Tn is an onto mapping if n = 2, because any element in H is of the form 2q, q an integer, and qT2 = 2q. But when n ¥= 2, Tn is not onto since 2 has no pre image under T. For if qTn = 2, q E G, then nq = 2 or q = 2/n (if n ¥= 0) and 2/n e: G. 2.31. Let G and H be as in Problem 2.28. Verify that the mapping u: G... H defined by u: U... U2, for all U E G, is not a homomorphism. Solution: Let Ul, U2 E G. Then (Ul + 02)U = (Ul + U2)2 = ui + U; + 2UIU2 and Ulu + U2u ur + U~. Hence u is not a homomorphism for (Ul + U2)U ¥= UIU + U2U for all Ul' U2 E G. 2.32. Is the mapping u of G into H, G and H as in Problem 2.28, defined by Uu = 0 for all U E G, a homomorphism? Solution: If Ul' U2 E G, then (Ul + U2)U = 0 and Ulu + U?JI = 0 + 0 = o. Hence u is a homomorphism. 2.33. 2.34. 2.35. Let G be the semigroup of positive integers P under the usual addition, and let H be the semi group of positive integers P under the usual multiplication. Show that the mapping 'T/: G... H defined by U'T/ = 2g for all U EGis a homomorphism. Solution: 'T/ is clearly a mapping. Let U1> U2 E G. Then (Ul + u2h = 2 g, + g2 = 2 9 '2 g2 = Ul'T/U2'T/. Hence 'T/ is a homomorphism. Let G = ({1, 2, 3}, a) and H = ({a, b, e}, f3) be the groupoids with binary operations a and f3 defined by the multiplication tables : 2 3 a b a f3 Which of the following mappings are hom |
omorphisms? (a) 1... a, 2... b, 3'" e (b) 1... a, 2'" a, 3'" a (e) 1..... a, 2... b, 3..... b (d) 1... b, 2... e, 3'" e (e) 1... b, 2... b, 3'" b (I) 1'" e, 2'" a, 3... b Solution: We use u to indicate the mapping in each case. (a) 1u2u = ab = a and (1' 2)u = 2u = b. u is not a homomorphism. (b) 1u2u = aa = e and (1, 2)u = 2u = a. u is not a homomorphism. (e) 1u2u = ab = a and (1' 2)u = b. u is not a homo morphism. (d) 1u3u = be = a and (1' 3)u = 3u = e. u is not a homomorphism. (e) u is a homomor phism since the image of 1, 2 and 3 is b, so that (ij)u = b for any i, j E {1, 2, 3}, and iuju = bb = b. (I) 1u2u = ea = e and (1' 2)u = 2u = a. u is not a homomorphism. Let G be the semigroup of positive integers under the usual addition. Determine which mappings of G into G are homomorphisms: (i) u: n..... 2n + 1, (ii) u: n..... 2n2, (iii) u: n'" 1. Solution: (i) (1·1·) (nl + n 2)u = 2(nl + n2) + 1 and nlC1 + n2u = 2nl + 1 + 2n2 + 1 = 2(nl + n2) + 2. Hence (nl + n 2)u ¥= nlu + n2u, and so u is not a homomorphism. ( + ) nl (n 1 + n2)u, and so u is not a homomorphism. n2 = 2nl + 2n2 + 4nln2. nlu + n2u = 2 |
nl + n2. 2 22Th n2 u = 2( +)2 en nlu n2u ¥= n 1 + 2 2 (iii) (nl + n2)u = 1. nlu + n?JI = 1 + 1 = 2. Thus nlu + n2u ¥= (nl + n2)u and hence u is not a homomorphism. 42 GROUPOIDS [CHAP. 2 b. Epimorphism, monomorphism, and isomorphism Three special types of homomorphism arise naturally. 1. A homomorphism of groupoid G into groupoid H may be an onto mapping. 2. A homomorphism of a groupoid G into a groupoid H may be a one-to-one mapping. 3. A homomorphism of a groupoid G into a groupoid H may be both onto and one-to-one. We give these three types of homomorphisms special names. Definition: Let () be a homomorphism of a groupoid G into a groupoid H. Then 1. 2. 3. (See Section 1.3a, page 12, for the definition of G().) () is called an epimorphism if () maps the carrier of G onto the carrier of H, i.e. G() = H. () is called a monomorphism if () is a one-to-one mapping of the carrier of G into the carrier of H. () is called a isomorphism if () is both an epimorphism and a monomorphism, i.e. () is one-to-one and onto. If there is an isomorphism from groupoid G onto the groupoid H, then we say G and H are isomorphic, or G is isomorphic to H, and write G "'" H. Problems 2.36. Let G be the groupoid of integers with addition as binary operation, and H the even integers with addition as the binary operation. Let Tn for n an even integer be the homomorphism (Problem 2.30) defined by UTn = nu, for U E G. When is Tn an isomorphism, monomorphism or epimorphism? Solution: If n =F 0, Tn is one-to-one since UTn = U'T n implies nu = nu' and so U = U'. TO is not one-to one, so it is not |
a monomorphism. If Tn is onto, there exists U E G such that UTn = nu = 2. Then U = 2/n and n = ±2. Hence T±2 are the only epimorphisms. Thus T±2 are isomorphisms and Tn is a monomorphism when n =F O. 2.37. G and H are finite groupoids and IGI =F!HI. Show that G cannot be isomorphic to H. Solution: Let I): G -> H be an isomorphism and let Ul'..., Un be the (distinct) elements of G. Then Ull), U21),..., Unl) are distinct (since I) is one-to-one) and are all the elements of H (since I) is onto). Hence IHI = n, which contradicts IGI =F IHI. Thus there exists no isomorphism I): G -> H. 2.38. Prove that if G, H, K are groupoids, then: and H "" K, then G "" K. In other words" """ is an equivalence relation. (i) G "" G; (ii) if G "" H, then H "" G; (iii) if G "" H (Hard.) Solution: (i) Let L: G -> G be the mapping defined by UL = U for all U E G. L is a one-to-one epimorphism. Hence it is an isomorphism, and so G "" G. (ii) Let '" = G -> H be an isomorphism. Then we define a mapping /3: H -> G as follows: Let h E H. As '" is one-to-one and onto, there exists a unique U E G such that U'" = h. Put h/3 = U· Note that h/3'" = h. Now /3 is onto G, for if U E G, u'" = hE Hand h/3 = U by definition. Also /3 is one-to one; for if h l/3 = h2/3, then h l/3", = h2/3'" and so hl = h 2• Finally /3 is a homomorphism. Let hl' h2 E H. Suppose Ul'" = hl' U2'" = h 2• Then h l/3 = Ul, h2/3 = U2 |
0 = 1/2n, then 1"q = 1!2n and so 1" = 1/2m. But 1/2m is not an int2ger. Thus 0 is not an epimorphism and there is no isomorphism. 2.40. Let (Z,o) be the groupoid with bimlry operation 0 defined by and let (Z, *) be the groupoid with binary operation * defined by (1 0 Ii,= (J '- Ii -i- ali (1.' b = (1 + b --- ali Is (Z, 0) ~ (Z, *)? Solution: Let u: (Z, 0)-> (Z, *) be defined by (1u ccc --a for a E Z. u is clearly a mapping. (a + b + ab)u = -(a + b + ab) Hence u is a homomorphism. u is onto, for if a E (Z, *), au = bu implies -a = -b and and (1(J * bu = ---a..,. --b,cc (--a) + (-b) - (1 = li. Therefore a is one-to-one and (a 0 b)u = (-a)(-b) = -(a + b + ab). then -a E (Z,o) and (-a)u = a. Now (Z, 0) ~ (Z, *). 2.41. Is M{1} ~ M{I,2}? Is M{l,~} :::::: M:l,~.:l)? Solution: IM{!} I = 1 and!M:l.~} i = 4 (see Problem 2.21, page 87). Therefore, by Problem 2.37, M{!} cannot be isomorphic to.M; 1. 2:' From Problem 2.25, the subset of elements in M{l, 2, 3) which have inverses has 6 elements. Thus 121'1: 1. 2: I is less then the order of M{l. 2. 3)' and so M{l,2} morphic to M{l.2,:lJ' is not iso 2.42. Give an example of two groupoids of order two whieh are not isomorphic. Solution: a b (1 a a taB b a |
(J. c d c icTdl d~ These two groupoids are not isomorphic, since there are only two one-to-one mappings, namely (ab)o = ao = c while is not an isomorphism, for 0: a --' c, b --' d and "': a --. d, b --> c. Now {/ aobo = cd = d. '" is not an isomorphism, for (aa)", = af = d while a",a", = dd = c. 2.43. Prove that the mapping 0: a + ib --' a numbers under addition with itself. ib is an isomorphism from the groupoid C of complex 44 GROUPOIDS [CHAP. 2 Solution: o is onto; for if x E C, x = a + ib. Now (a + i(-b))o = a + ib = x. Also 0 is one-to-one, since (a j + ibj)o = (az + ib 2 )0 implies aj - ib j = a2 - ib 2 and hence aj = a2 and bj = b2. Finally 0 is a homomorphism, for [raj + ib j) + (a2 + ib 2)]0 = (a j + a2) - i(b j + b2) = (aj + ibj)o + (a2 + ib2)o 2.44. Is the homomorphism 0 of the groupoid C of complex numbers under the usual multiplication of complex numbers to the groupoid of real numbers under the usual multiplication defined by 0: a+ib --> la+ibl = +ya2 +b2 an epimorphism or monomorphism? Solution: It is well known that if Xj, X2 are two complex numbers, then IXjX21 = IXjllx21. The calcula tions are j j 2 2 y(a 2 + b2 )(a2 + b2 ) v' (aja2 - bj bz)2 + (b ja2 + bZaj)2 'a\a2 - b jb2 + i(b ja2 + b2aj)1 i(aj + ibj)(az + ibz)1 (XjX2)0 = xjOX20 and so 0 is a homomorphism. On the other hand Then ib)lI, |
so II is not one-to-one. Finally 0 is not onto, for it is always the case that Ixl ~ 0, and thus there exists no x such that xo = -1. (a + ib)1I = (-a - 2.45. Let (P,') be the groupoid P under the usual multiplication of positive integers and (R, +) the group oid R under the usual addition of real numbers. Is the mapping II of (P,') into (R, +) defined by II: a --> logjO a an epimorphism, monomorphism or isomorphism? Solution: As in Section 2.5a, II is a homomorphism. If all = bll, then logjO a = IOgio b and hence a = b. Thus II is a monomorphism. Since 0 = loglo 1 < loglo 2 < 10giO 3 < "', there is no integer such that loglO x = -1. Hence II is not onto. Therefore II is a monomorphism but not an epimorphism nor an isomorphism. c. Properties of epimorphisms We will show in this section that if 0 is an epimorphism from the groupoid G to the groupoid H, then H shares some of the properties of G. Theorem 2.6: Let 0 be an epimorphism from the groupoid G to the groupoid H. Then (a) if G is a groupoid with an identity 1, so is Hand 1() is the identity of H. Furthermore if f is an inverse of g in G, then (() is an inverse of gO in H. (b) if G is commutative, so is H. (c) if G is a semigroup, so is H. Proof: (a) Let hE H. We shall prove 1() is the identity of H, i.e. h·1() = h = 1(). h. As 0 is an epimorphism, 0 is onto and we can find an element g in G such that gO = h. Then and gO·1() (g·l)() 1(). gO (1' g)O gO gO h h Thus 10 is the identity of Hand H is a groupoid with an identity. Now suppose g E G has an inverse f. Then gf = 1 = fg |
. Therefore gO' {() = (gf)O = 10 = (fg)O = {(). gO which means {() is the inverse of gO in H. Sec. 2.5] HOMOMORPHISMS OF GROUPOIDS AND GA YLEY'S THEOREM 45 (b) Suppose G is commutative. We show H is commutative. To this end let h, h' E H. Because () is onto, we can find g,g' E G for which g() = hand g'() = h'. Hence hh' = g(). g'() = (g. g')() = (g'. g)() = g'(). g() = h'h and so H is commutative as claimed. (c) To show H is a semigroup, we must prove that multiplication is associative in H. Let h, h', h" E H. Then we can find g, g', g" E G such that g() = h, g'() = h' and g"() = h". Since G is associative we have, as required, (hh')h" (g(). g'())g"() = [(gg')()]g"() = ((gg')g")() = (g(g'g"))() = g()[(g'g")()] = g()(g'(). g"()) h(h' h") d. Naming and isomorphisms In our study of groupoids we will take isomorphic groupoids to be essentially the same. To explain why, we will describe a "naming process," beginning with an example. Let G be the groupoid with binary operation·, elements 1 and a, and multiplication table 1 a 1~ a~ We define a new groupoid G by relabeling the elements of G. Let G consist of the elements a, (3 and have multiplication table f3 a~ f3~ What we have done is to call the elements of G by different names. In general if G is any groupoid, we can form a new groupoid G by renaming the elements of G. Thus for each g E G we take a new element g, ensuring only that 17"= g if f 7"= g, i.e. don't use the same name twice. If fg = h, then we define multiplication of elements of G by log = h. |
It is easy to prove that G is a groupoid with this multiplication. We are not interested in distinguishing between groupoids which differ only because their elements have different names. Considering groupoids to be the same if they are isomorphic overcomes this snag. To see this we will show that the G constructed from G above by renaming is isomorphic with G. We must find a one-to-one onto homomorphism (). Define g() = g, i.e. the image of g under () is the new name of g. () is one-to-one onto, as one and only one g corresponds to each g. Also (fg)() = h where fg = h. But te 0 g() = l o g = h, by definition of the multiplication of G. Hence (fg)e = fB 0 gB, and G is iso morphic with G. Thus isomorphism gets rid of the difficulty of obtaining a new groupoid on simply renaming. We look at the problem from another point of view. Suppose F and G are two iso morphic groupoids, and that () is an isomorphism between F and G. Then we will apply our renaming process to show that G and F, F suitably renamed, cannot be distinguished either as regards their elements or the way they multiply. Let us as before rename each element f E F, 1. But we shall choose 1 to be te. This is a proper renaming since 1= g means that fB = gB, and, as B is one-to-one, f = g. So we have not used the same name twice. As before, if fg = h we define l o g to be h. Thus F becomes a groupoid with respect to the binary operation o. 46 GIWUPOIDS [CHAP. 2 How does F compare with G? F has the same elements as G. But do these elements multiply the same way? Suppose 11 and 12 are hvo elements of F. Then II was previously called ft, and 12 was called h If!d~ = /:;. then \ye defined 11 0 12:= 11. Now 11.12 are also elements of G; in fact, 11 = /111, 12:= he. The product Id2 in G is therefore /le!20 = (fd2)(} = /:;0 = ll, since 0 is a homomorphism. Hence the product 11 0 |
12 of two elements in F is the same element as the product fJ-2 inside G. Thus a groupoid isomorphic with a groupoid P is indistinguishable from a suitable renaming of P as far as the elements and the way they multiply are concerned. For this reason we do not distinguish betvveen groupoids that are isomorphic. e. Mx and semigroups The importance of Mx is explained by the following theorem, which says that an iso morphic copy of any semigroup S is contained in some l11x. Theorem 2.7: (Cayley's Theorem): Let S be a semigroup with identity. Then there is a monomorphism of S into Ms' (The semigroup S is an abbreviation for the semigroup (S,,1) where It is a binary operation. Ms is the semigroup of all mappings of the set S into itself, with binary operation the composition of mappings.) Proof: Let s E S and let Ps: S ---> S be defined by xPs:= xs, xES. Here xs is the It is clear that Ps is a mapping of S into S, i.e. Ps EMs' product of x and s in S, i.e. (;1',.'1)/1. Let e: S ---> Ms be defined by sO = ('s' We shall show that e is a monomorphism. First we have to check that 0 is a homomorphism. Let s, s' E S. (.'1.'1')0 =----= Pss' and slis' 0 - Ps Ps' Now if xES, xPsPs'= (xpJps' = (:r8)ps' = (;1'8).'1' = xPss" As this is true for all xES, Pss' = PSPS" Hence (ss')e = ses'e. Secondly we must show that 0 is one-to-one. Suppose 8e = s'(); In particular, if 1 is the identity of S, Ips = Ips.. But then Ips = 1· s = s = 1· PS' = 1· s' = s' and so s = s' and () is a monomorphism. This completes the proof of Theorem 2.7. then P.• - Ps" As an illustration of the proof, let S = {aI, |
a2, a:l] be the semigroup given by the follow ing multiplication table: °a a3 al a2 a 1 0.'2 a':l °2 °:1 ° 1! I I I Notice that al is the identity element of S and that S is a semigroup. Now The mapping e is defined by ale = Pal' a2e = P"2' a:le = Paa'8 is a monomorphism. It can be checked directly that CHAP. 2] SUPPLEMENTARY PROBLEMS 47 Let Se = {se I s E S} where e is the monomorphism of Theorem 2.7; then S e= se and hence Ms contains an isomorphic copy of S. In Section 2.5d we pointed out that, but for naming, S and an isomorphic copy were the same. Thus we see that in a rough way every semigroup with identity appears in some Mx. Hence the importance of Mx. A look back at Chapter 2 We defined a groupoid, associative groupoid (called a semigroup), and commutative groupoid. We showed that in a groupoid the identity is unique, while inverses are unique in a semigroup. We defined a mapping e: (G, a) ~ (G, (3) [(gl, g2)a]e = [(glO,g2e)1(3. If e is one-to-one and onto, we called it an isomorphism of (G,a) onto (G,(3). We proved Cayley's theorem, that each semigroup has an isomorphic copy in Mx for some suitable X. to be a homomorphism if Supplementary Problems GROUPOIDS 2.46. Let G = {I, -I}. Is (G,') a groupoid if • is the usual multiplication of integers? 2.47. Show that (G,') is a groupoid when G = {I, -1, i, -i}, i = A, and' is the usual multiplication of complex numbers. 2.48. Suppose Gn is the set of all integers divisible by the integer n. For which n is (Gn,.), with the usual multiplication of integers, a groupoid? 2.49. Let!;, i = 1,2,3,4,5,6, be the set G of mappings of R - { |
O, I} into R defined for each x E R - {O, I} by Il:X~X; 12:x~-I-; 13:X~--; 14:X~-; 15:x~--I; 16:x~1-x. Suppose Gij= 1 -x x-I x x x- 1 x {f;, I j } and that 0 is the composition of mappings. Determine which of the following are groupoids: (i) (Gl,2, 0), (ii) (Gl,3, 0), (iii) (Gl,4' 0), (iv) (Gl,5' 0), (v) (G l,6, 0), (vi) (G 5, 6,·0). 2.50. Let F = {h/2,!3} and H = {hlz,hfs} where Ii are the mappings defined in Problem 2.49. Prove (F,o) is a groupoid while (H,o) is not a groupoid (0 is the composition of mappings). COMMUTATIVE AND ASSOCIATIVE GROUPOIDS 2.51. Let R* be the set of nonzero real numbers. Define the binary operation 0 on R* by a 0 b = lal b for a, bE R*. Prove (R*,o) is an associative groupoid but not a commutative groupoid. Hint: lallbl = labl· 2.52. Define the binary operation' on G = R X R as (a, b) • (e, d) = (ae, be + d). Is (G,') a commuta tive or an associative groupoid? 48 2.53. 2.54. 2.55. GROUPOIDS [CHAP. 2 The binary operation a on R is defined by a: (a, b) ---. la - bl commutative but not associative. for a, b E R. Show that (R, a) is Suppose we define a binary operation + on R by a + b = the minimum of a and b (a, b E R). Show that (R, +) is both associative and commutative. Let G = {a I a: R ---. R}. For a, (3 E G define the mapping a * (3 = a |
0 (3 (30 a where 0 is the for all x E R. Prove: usual composition of mappings and (i) (G, *) is a groupoid; (ii) (G, *) is neither associative nor commutative; (iii) (a * (3) * a = a * ((3 * a) for all a, (3 E G; (iv) a * (3 = (-(3) * a where -(3 is the element of G defined by -(3: x ~ (x(3) for all x E R. (Hard.) (30 a) = x(a 0 (3) - x(a 0 (3 - x((3 0 a) 2.56. Let G be the set in Problem 2.55. For a, (3 E G define a' (3 = a 0 (3 ; (3 0 a where 0 is the usual composition of mappings and x 2 - 2 a 0 (3 + (3 0 a _ X(a 0 (3) + x((3 0 a) is a groupoid; a, (3 E G. (ii) (G,') is commutative but not associative; for all x E R. Prove: (iii) (a' (3) • a = a' ((3 • a) (i) (G,') for all INVERSES IN GROUPOIDS 2.57. 2.58. Let R + be the set of all non-negative real numbers. Define a * b = ya2 + b2 (Ya 2 + b2 for all a, b E R + is the positive square root). Find an identity in (R+, *). What elements have inverses? For all a, (3 E G = {a I a: Z ~ Z}, let a X (3 be the mapping defined by x(a X (3) = Xa' x(3 where x E Z and, is the usual multiplication of integers. Is (G, X) a groupoid? Does it have an identity? What elements have inverses? 2.59. Let G = {a I a: Z ---. Q, Q the rational numbers}. Define a X (3 as in the preceding problem. Does (G, X) have an identity? What elements have inverses? 2.60. Which of the groupoids in Problems 2 |
.57,2.58 and 2.59 are commutative and which are associative? 2.61. Define the following binary operation + in R+, the non-negative real numbers: a + b = the maximum of a and b, a, bE R+. Does (R+, +) have an identity? What elements have inverses? 2.62. Let (G, *) be the groupoid of Problem 2.56. What is the identity of (G, *)? Find an infinite number of elements which have inverses. 2.63. Show that (G, *), the groupoid of Problem 2.55 has no identity. SEMI GROUPS WITH AN IDENTITY 2.64. Which elements of (G,.), where G = {1, -1, i, -i}, i = R, and' the usual multiplication of com plex numbers, have inverses? 2.65. Let G = {/1,!2,!3,!4,!5,/s} of Problem 2.49. Prove that G with the binary operation 0 of com position of mappings, is a semigroup with an identity. Find the inverse of each element in (G,o). 2.66. Let G = {(a, b, c, d) I a, b, c, dE Z}. Define (a, b, c, d)· (a', b', c', d') = (aa' + cb', ba' + db', ac' + cd', be' + dd') Show that (G,') is a semigroup with an identity. Show that the subsets H = {(1, 0, 0, 1), (-1,0,0, -1)} and F = {(1, 0, c, 1) IcE Z}, with the binary operation of (G,') restricted to Hand F respectively, are semigroups with an identity. Find the inverses of the elements of Hand F. 2.67. Let G = {a I a a mapping of {1, 2, 3, 4, 5} into {1, 2}}. For a, (3 E G, position of mappings. Is (G,o) a semigroup with an identity? let a 0 (3 be the usual com CHAP. 2] SUPPLEMENTARY PROBLEMS 49 2. |
68. For a,[3EG = {al a: {1,2,3,4,5}--+{1,-1}}, define the mapping aX[3 by n(aXf3)=na·n[3, where n E {1, 2, 3, 4, 5} and' the usual multiplication of integers. Is (G, X) a semigroup with an identity? If so, what elements have inverses? HOMOMORPHISMS OF GROUPOIDS AND CAYLEY'S THEOREM 2.69. Let H = {(1,0,0,1), (-1,0,0,-1)} be the groupoid defined in Problem 2.66. Show that H is isomorphic to the groupoid (G,') where G = {1,-1} and· is the usual multiplication of integers. 2.70. If F = {(1, 0, c, 1) IcE Z} is the groupoid defined in Problem 2.66 and (Z, +) the groupoid of integers under addition, prove the two groupoids are isomorphic. 2.71. Which of the groupoids of Problem 2.49 are isomorphic? 2.72. Let G = {f1,!2,!3,hf5,!6} be the groupoid of Problem 2.65, (H,') the groupoid with H = {1,-1} and· the usual multiplication of integers. Find all possible homomorphisms of G into H. Show that there is no homomorphism of (G,o) into (F,o) where F = {fl' i 2, i3} and 0 the usual composi tion of mappings. (Hard.) 2.73. Can the groupoid (G, *) of Problem 2.55 be a homomorphic image of the groupoid (MR' 0) where 0 is the usual composition of mappings? 2.74. Suppose G = {a I a: Z --+ Z}, X the binary operation defined in Problem 2.58 and 0 the usual com position of mappings. Show that 'l': (G, x) --+ (G, 0) defined by 'l': a --+ a is not a homomorphism. (See Theorem 2.6, page 44.) Chapter 3 Groups and Subgroups Preview of Chapter 3 We define a group as a semigroup |
with an identity in which every element has an inverse. The object of this chapter is to show that the concept of a group is natural. This is done by providing illustrations of groups which arise in various branches of mathematics. The most important concepts of this chapter are group and subgroup. 3.1 GROUPS Definition As we remarked in the preview, a semigroup with an identity in which every element has an inverse is termed a group. We repeat the definition in more detail: Definition: A non-empty set S together with a binary operation in S is called a group if (i) there exists an identity element (usually denoted by) 1 E S; words, a' 1 = a = 1· a Recall that the identity is unique by Theorem 2.1, page 31. for all a E S in other (ii) for every choice of the elements a, b, c E S, (a'b)'c = a'(b'c) Thus (i) and (ii) are the conditions for (S,') to be a semigroup with identity. (iii) every element a E S has an inverse in S, i.e. there is an element b E S such that a·b = 1 = b'a This element b is often denoted by a- 1 2.2, page 33. • The inverse is unique by Theorem Whenever we define a group we shall follow the pattern: I. Define a set S (# ~). II. Define a binary operation in S. III. Verify that the groupoid (S,·) contains an identity element. IV. Verify that the groupoid (S,·) is a semigroup, i.e. is associative. V. Verify that every element of S has an inverse. The number of elements of S, lSI, is called the order of the group. page 29.) We will exhibit groups of infinite and finite order. problems.) (Compare Section 2.2, (See following examples and 50 Sec. 3.1] GROUPS 51 Examples of groups of numbers Example I: The additive group of integers. Let Z be the set of integers. I. II. Let + be the binary operation of addition in Z. III for every n E Z. Thus (Z, +) has an identity element. IV. If l, m, n are integers, i.e. (Z, +) is a semigroup. (l+m |
) + n = l + (m+n) V. If nEZ, then -n in Z has the property n + (-n) = 0 = (-n) + n i.e. -n is an inverse of n in (Z, +). Thus we have shown that the groupoid (Z, +) is a group. This group is usually referred to as the additive group of integers. Example 2: The additive group of rationals. Let Q be the set of rational numbers. I. II. Let + be the binary operation of addition in Q. III for every a E Q, so 0 is an identity element for (Q, +). IV. If a, b, cEQ, then (a + b) + c = a + (b + c). V. If a E Q, then -a in Q has the property a + (-a) o (-a) + a. Example 3: The additive group of complex numbers. The description of this group is left to the reader. Example 4: The multiplicative group of nonzero rationals. 1. Let Q* be the set of nonzero rational numbers. II. Let 0 be the binary operation of multiplication, i.e. the usual multiplication of rational numbers. III. The rational number 1 is clearly an identity in the groupoid (Q*,o). IV. If a,b,c E Q*, then V. If zEQ*, soisl/aand (a 0 b) 0 c a 0 (b 0 c) a o! = 1 a 1 -oa a Thus every element of Q* has an inverse. Example 5: The multiplicative group of nonzero complex numbers. This group is very similar to that in Example 4. We shall go through the usual five stages in setting up and describing the group. I. Let C* be the set of all nonzero complex numbers. Thus C* = {x I x = a + ib where x =1= 0 + iO and a, bE R} Recall that i2 = -1. II. We define multiplication of complex numbers as follows: (a + ib)(c + id) = (ac - bd) + i(ad + be) This is a binary operation in C* since (ac - bd) + i(ad + be) ment in C* (not both ac - bd and ad + be can be zero). III* and it is |
clearly an identity in (C*, 0). is a unique ele 52 GROUPS AND SUBGROUPS [CHAP. 3 IV. Suppose a + ib, e + id, e + if E C':'. Then [(a + ib)(e + id)](e + if) On the other hand, (a + ib)[(e + id)(e + if)] [(ae - bd) + i(be + ad)](e + if) [(ae - bd)e - (be + ad)f] + i[(be + ad)e + (ae - bd)f] (a + ib)[(ee - df) + i(de + ef)] [a(ee - df) - b(de + ef)] + i[b(ee - df) + a(de + ef)] It follows from these two computations that (a + ib)[(e + id)(e + if)] = [(a + ib)(e + id)](e + if) V. We have to check the existence of inverses. Thus suppose a+ ib E C*; not both a and b are zero. Hence a2 + b2 # 0 and so then Moreover, a a2 + b2 - b. t a2 + b2 E c:' 1 (a + ib) (a 2 : b2 - i a 2! b2 ) Thus we have proved (C*,o) is a group and we term this group the multiplicative group of nonzero complex numbers. Problems 3.1. Is (S,o) a group if (i) S = Z and 0 is the usual multiplication of integers? (ii) S = Q and 0 is the usual multiplication in Q? (iii) S = {q I q E Q and q > O} and 0 is the usual multiplication of rational numbers? (iv) S = {z I z E Z and z = v'2} and 0 is the usual multiplication in Z? (v) S = Rand 0 is the usual addition of real numbers? (vi) S = Z and 0 is defined by a 0 b = 0 for all a, b in Z? Solutions: (i) The identity element is the integer 1. integer z in Z such that z 0 5 = 50 z = 1. (S,o) is not a group because 5 E Z but there is no (ii) Again the |
identity is the number 1. There is no q E Q such that q 0 0 = 1. Hence (S,.o) is not a group. (iii) (S,o) is a group. Clearly S # 0 and 0 is a binary operation on S. q 01 = 10 q = q for all q E S; hence 1 is an identity. Multiplication of rational numbers is associative and every element in S has an inverse; for if q E S, 1 1 then - E Sand _0 iv) S = 0 since V2 fl Z. Therefore (S,o) is not a group. (v) (S,o) is a group. S # 0 and addition is an associative binary operation on S and r + (-r) = 0 = (-r) + r for all rES. (S,o) is not a group because there is no identity element in S. (vi) 3.2. Let S be the set of even integers. Show that S is a group under addition of integers. ~~: Let a = 2al and b = 2b 1 be any two elements in S. a + b = 2(al + b1) is a unique element in S; thus addition is a binary operation on S. Associativity of addition in S follows from the as sociativity of addition in Z. 0 = 2 0 0 is an identity element in S. If a E S, then -a E S since a = 2al implies -a = 2(-al). Hence a has an inverse in S, as a + (-a) = (-a) + a = o. \ 3.3. Let S be the set of real numbers of the form a + bV2 where a, b E Q and are not simultaneously zero. Show that S becomes a group under the usual mUltiplication of real numbers. Sec. 3.1] GROUPS 53 Solution: (a + bV2)(e + dV2) = (ae + 2bd) + (eb + ad)V2. If neither a + bV2 nor e + dV2 is zero, i.e. (0 + 0V2), then their product cannot be zero. Hence the product of elements in S belongs to S. 1 = 1 + 012 IS an IdentIty for S. MultIplymg... " 1 a + bV2 a - bV2 |
a2 - 2b 2 1 a + bV2 a -:0----::-::-::a2 - 2b2 - by - - - -, we obtain a - bV2 a - bV2 b (a2 - 2b 2) V2 Hence 1 a + bV2 E S. The associativity holds because it is true for multiplication of real numbers. 3.4. Let S be the set of complex numbers of the form a + bH where a, b E Q and are not both simultaneously zero. Show that S becomes a group under the usual multiplication of complex numbers. Solution: (a + bR )(e + dH ) = and cannot be zero if its factors are not zero; hence the product of two elements of S belongs to S. 1 = 1 + OM is an identity for S. is certainly an inverse for a + bH. MUltiplying top and bottom by a - b 0, we get (ae - 5bd) + (be + ad) H 1 M a + b -5 1 a -bM a 2 + 5b 2 a -b a2 + 5b2 + a2 + 5b2 R and so 1 _,----;; E S. The associativity of multiplication in S follows from associativity of multi a + bv-5 plication of complex numbers. 3.5. Let m be any fixed positive integer and let S = {O, 1, 2,..., m -I}. Define a binary operation in S by aob a 0 b = r a+b if a+b<m if a + b = m + r, 0 ~ r < m Prove that (S,o) is a group of order m. (Hard.) Solution: If a, b E S, then a 0 b is uniquely defined and belongs to S. a 0 0 = 00 a = a, so 0 is an identity. Note that a 0 b = a + b - 8m where 8 is 0 or 1, for any a, bE S. So b 0 e = b + e - 0lm where 01 is 0 or 1. a 0 (b 0 e) = a + (b 0 e) - 02m where 02 is 0 or 1. Then a 0 (b 0 e) = a + b + e (01 + 02)m where both 01 and 02 could be 0 or 1. Hence Similarly a 0 |
(b 0 e) (aob)oe 11m where 'II is 0 or 1 or 2 '12m where '12 is 0 or 1 or 2 Now 0 ~ a 0 (b 0 e) < m and 0 ~ (a 0 b) 0 e < m. Suppose 'II > '12; then a 0 (b 0 e) ~ a + b + e - ('12 + l)m = a + b + e - '12m - m because 'II is at least '12 + 1. But b 0 e) < 0; this contradicts 0 ~ a 0 (b 0 e). Hence contradiction. Thus a 0 (m - a) = (m - a) 0 a = 0; hence m - a is an inverse to a. Thus S is a group. '12m < m and the above equation implies that leads in a similar way to a then m - a E Sand 'II ~ '12' and a 0 (b 0 e) = (a 0 b) 0 e. '12 > 'II If a E S, 'II 7J2 3.6. Let SeC (C the set of complex numbers) be the set of all mth roots of unity, where m is a fixed positive integer. Prove that under the usual multiplication of complex numbers, S becomes a group of order m. Solution: Recall that a complex number x is an mth root of unity if xm = 1 and that there are exactly m distinct roots of unity, viz. ei2rrTlm, T = 1,2,..., m; also, If a, b E S, then ab is uniquely defined. Since (ab)m = amb m = 1, ab is an mth root of unity and hence ab E S. l' a = a' 1 = a, so 1 is an identity. Associativity is true, since it is true for\!omplex numbers in general. If a E S, then (l/a)m = l/am = 1; thus l/a E Sand l/a is the inver~e of a. e irrx = cos x + i sin x. 54 3.7. GROUPS AND SUBGROUPS [CHAP. 3 Let S c; C (C the set of complex numbers) be the set of all roots of unity. Describe one way of making S into a group. Solution: Use as binary operation the |
usual multiplication of complex numbers. If a, bE S and a is an mth root of unity, b an nth root of unity, then ab is an mnth root of unity because (ab)mn = amnbmn = (am)n(bn)m = Inlm = 1. Hence ab E S and is, of course, uniquely defined. 1 E S and acts as an identity. l/a E S and is the inverse of a. Associativity holds for multiplication of complex numbers. Thus S is a group with respect to the usual mUltiplication of complex numbers. 3.8. The following table defines a binary operation. Is the resultant groupoid a group? 1 2 ffij 1 2 Solution: We need to check only (a) associativity, (b) existence of identity and inverse. (a) To check associativity, we have the following possible questions: (a) Does 1 • (1·1) equal (1' 1) • 1? (b) Does 1 • (1, 2) equal (1' 1) • 27 (e) Does 2' (1 '1) equal (2' 1) • I? (f) Does 2' (1' 2) equal (2 '1) • 2? (c) Does l'(2 '1) equal (1.;) • 1? (d) Does l'(2, 2) equal (1' 2) • 2? (g) Does 2' (2' 2) equal (2' 2) • 2? (h) Does 2' (2 '1) equal (2,2) • I? Checking all these products, we see that associativity holds. (b) 1 acts as an identity. The inverse of 1 is 1, the inverse of 2 is 2. Hence the table defines a group. 3.9. Write the multiplication table for the group of Problem 3.5 with m = 3 and m = 4. Solution: If m = 3, the multiplication table is o 1 2 If m = 4, the multiplication table is.2 SUBGROUPS Definition Let (G,') be a group with binary operation' and let H be a non-empty subset of G. Then we say H is a subgroup of G if the operation' restricted to H is a binary operation in H which makes H into a group. Sec. 3.2] SUBGROUPS 55 For |
example, if G is the group with m = 4 of Problem 3.9, then the subset H = {0,2} is a subgroup of G. For when the operation 0 in G, as defined in the multiplication table for G, is restricted to H, it is a binary operation in H, i.e. 000 = 0 E H, 002 = 2 E H, 200 = 2 E H, and 202 = 0 E H. H is a group because: H oF 0; 0, the identity, is in H; the operation 0 restricted to H is an associative binary operation (since the operation in G is associative); and every element in H has an inverse in H. The following lemma facilitates proving a subset of a group is a subgroup. Lemma 3.1: Let (G,') be a group. Then a subset H of G is a subgroup of G iff (i) H 0/= 0 and (ii) if a, b E H, then ab- 1 E H Proof: If H satisfies these conditions, then H is a group with respect to the binary op eration. For if H 0/= 0,. then there exists a E H. Hence aa- 1 = 1 E H. Also, if b E H then 1b- 1 = b- 1 E H. Hence a, b E H )-1 = ab E H. Associativity is true in H, as it is true in G. Thus' is an associative binary operation on H, 1 E H, and the inverse of every element of H is an element of H. Therefore (H,·) is a subgroup. implies a(b- 1 Conversely if H is a group with respect to., then clearly H satisfies conditions (i) and (ii) above. Problems 3.10. Let (Q, +) be the additive group of rationals. Is Z a subgroup of Q? Is P a subgroup of Q? Solution: Clearly Z # 0, and Z C Q. If a, b E (Q, +) then the binary operation is +, and the inverse of b is -b. So, if a, bE Z, we ask in accordance with Lemma 3.1, whether a + (-b) = a - b E Z. It is, and so Z is a subgroup of (Q, +). Clearly P # 0 and Pc Q |
. If a, bE P, is a + (-b) = a - bE P? No, for P does not contain negative numbers; and if a = 1 and b = 2, then a - b is negative. 3.11. Is Q a subgroup of (C, +), the additive group of complex numbers? Solution: Q is a subgroup of (C, +). Q # 0. Q c C; for if a E Q, a = a + Oi E C. If a, bE Q, then a + (-b) = a - bE Q. Hence 3.12. Is Z - {O} a subgroup of (Q'~,.), the multiplicative group of nonzero rational numbers? Solution: 1 is the identity. 3 E Z - {O}, but 3 has no inverse in Z - {O}. Therefore Z - {O} is not a subgroup of (Q*,.). 3.13. Is Q*, as above, a subgroup of (C''',.), the multiplicative group of nonzero complex numbers? Solution: Q* # 0. a, b E Q* implies ab- 1 is a nonzero rational. Thus Q* is a subgroup of (C*, '). 3.14. Is Q* a subgroup of the group of real numbers of the form a + bV2, a, b E Q and a, b not simul taneously zero, under multiplication? Solution: Q* # 0. If a E Q*, then a = a + OV2 E {a + bV2 I a, bE Q and not both zero}, since a # O. implies ab- 1 is a nonzero rational. Thus Q* C {a + bV2 I a, bE Q and not both O}. a, bE Q':' Hence Q* is a subgroup. 3.15. Prove that the intersection of two subgroups Hand K of a group G is a subgroup. Solution: 1 E H, for as H is not empty, there is an element h E H. But then H contains hh -1 = 1, the then identity of G. Similarly, 1EK. Hence 1EHnK and HnK#0. g, hE Hand gh- 1 E H. Also, gh- 1 E K. Thus gh- 1 E HnK and HnK is a |
subgroup of G. If g,hEHnK, ) 56 3.16. GROUPS AND SUBGROUPS [CHAP. 3 By considering the group of Problem 3.5 with m = 6, show that the union of two subgroups is not necessarily a subgroup. Solution: That {O,3} and {O, 2, 4} are subgroups is easily verified. But U = {O, 2, 3, 4} = {O, 3} U {O, 2, 4} is not a subgroup as 304 = 1 il U. Therefore 0 is not a binary operation in U. 3.3 THE SYMMETRIC AND ALTERNATING GROUPS a. The symmetric group on X Let X be any non-empty set. A very important group arises from the set Sx of all one-to-one mappings of X onto X, called the symmetric group on X. We describe this group in the usual five steps. I. Sx is the set of all matchings of the non-empty set X with itself. Clearly, Sx r;;;; Mx (see Section 2.4b, page 36). II. If CT, T E Sx, then we define CTT to be the composition of the mappings CT and T. Here we must verify that CTT is a matching of X with itself. Suppose x E X; then as T is onto, we can find x" E X such that X"T = x. But CT is also onto, so we can find x' E X such that x' CT = x". Consequently and hence CTT is onto. If X(CTT) = Y(CTT), then (XCT)T = (YCT)T by the definition of the composition of mappings; this means, since T is one-to-one, that XCT = YCT. This in turn implies, since CT is one-to-one, that x = y. Therefore CTT is also one-to-one. Thus composition of mappings is a binary operation in Sx. III. Clearly the identity mapping l: x -'; x is in Sx and is an identity element of Sx. IV. The groupoid (Sx,') is a semigroup, since composition of mappings is associative. (Theorem 2.3, page 36.) V. Let CT E Sx. Since CT is one |
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