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-to-one and onto, Theorem 2.4, page 36, implies CT has an inverse, T, in Mx. Now, CTT = l = TCT means CT is an inverse of T. By Theorem 2.4 the only elements in Mx which have inverses are those which are one-to-one onto mappings. Therefore T E Sx and T is the required inverse of CT. The proof that Sx is a group is complete. We will call an element of Sx a permutation of X, or, for short, a permutation. In the particular case where X = {1,2,...,n}, we write Sx = Sn. Sn is called the symmetric group of degree n. ISnl is calculated as follows. If CT E Sn, then 1CT can be one of n elements. 2CT can be 2CT ¥= 1CT. one of n - 1 elements, as 1CT has been chosen and CT must be one-to-one; so 3CT can be one of n - 2 elements, as 1CT and 2CT have been chosen and CT must be one-to-one; so 3CT is not equal to 2CT or 1CT. Continuing in this way, we conclude that there are n·(n-1)(n-2) ••. ·2·1 = n! elements of Sn, i.e. ISnl = n! The elements of Sa, for example, are Sec. 3.3] THE SYMMETRIC AND ALTERNATING GROUPS 57 2 0 2 ;) -e 2 i) 2 3 0'1 - 0'2 1'1 1 (! 2 ;) =0 2 ;) 3 1'2 1'3 2 (! 2 ~) = (; 2 ;) 1 Here we are using the notation of Section 2.4c, page 37, To :find the multiplication table, we must compute the products. As an example, we calculate 0'11'1' 0'11'1 (lO'~TI 2 20' I l' I 3~TJ ( 2~1 2 31'1 l:J U 2 ~) 2 1'2 To do this calculation mentally, we think as follows: 1-+ 2 (in 0'1)' 2 -+ 3 (in 1'1) Write down 2 -+ 3 (in 0'1)' 3 -+ 2 (in TJ Put 2 beneath |
2 to get (! 2 2 3) Only one other number can appear, namely 1. Thus 0'11'1 (! 2 i) 2 1'2 The multiplication table for S3 is 0'1 0'1 0'2! 7'3 7'1 7'2 0'2 0'2! 0'1 7'2 7'3 7'1 1'1 1'2 1'3 1'1 1'2 1'3 1'2 1'3 1'1 1'3! 0'1 0'2 1'1 0'2! 0'1 1'2 0'1'0'2!! 0'1 0'2 1'1 7'2 7'3 The reader should check some of the entries, Note that 0'11'1 = 1'2 and 1'10'1 = 7'3' so that 0'11'1 # 7 10'1' Hence S3 is not commutative. Problems 3.17. Calculate a{1, {1a, a-I, {1-1, (am-I, and ({1a)-1 if Solution{1 = 5 4 ~) and 1 2 3 4 5 6) {1= 1 3 5 6 24 ( 4 5 2 6 ~), (1a = ( 1 2 3 4 5 6) 2 6 4 1 3 5 To find a-I, we note that x(aa- 1) = x and hence a-I must carry Xa to x. Now we determine is taken onto 1. 6a = 1, so we must have 1a- 1 = 6. Next, since 1a = 2, 2a- 1 = 1. which x Proceeding in this way, we obtain 1 2 3 4 5 6) 6 1 2 54 3 ( 58 GROUPS AND SUBGROUPS [CHAP. 3 An easy method of calculating the answer mechanically follows. Take Interchange the rows, 1 2 3 4 5 6) 23 65 41 ( 2 3 6 5 4 1) 123456 ( Rearrange the columns so that the top row reads 1 2 3 4 5 6, 1 2 3 4 5 6) 6 1 2 543 ( This method is conceptually the same as the first. To find f3 -1, interchange the rows to obtain and rearrange to get 1 3 5 6 2 4) 1 234 5 6 ( 1 2 3 4 5 6) 4 |
1 3 2 5 (6'(f3a)-1 3.18. a 3 1 2 4 = C 2 Verify that a(f3y) = (af3)y where 3 4!) G 2 3 4 (~ a(f3y) = (af3)y = 2 3 4 1 2 4 Solution: 3 2 5 f3 y 2 1 5 = G 2 3 4 :) (~ 2 3 4 :) C 2 :)(~!)G 2 3 4!)!) 3 4!) 3 4!) 3 2 3 2 3.19. Is the subset R = {I, (71' (72} a subgroup of 8 3? Solution: (For this notation see above.) R =I' 0. From the multiplication table of 8 3, page 57, the product of any two elements in R is it follows that xy-1 E R for any x, y E R. Hence (71 1 = (72 and 021 = (71' again in R. Since R is a subgroup of 8 3• 3.20. Find all elements of 8 1 and 8 2 and exhibit multiplication tables for these groups. Solution: 8 1 contains one element I = (~). I CJ is a multiplication table for 8 1• There are two ele- I ments in 8 2: I = (~!) and f3 = (~ ~). The multiplication table for 8 2 is f3 ;tEtB Sec. 3.3] THE SYMMETRIC AND ALTERNATING GROUPS 59 3.21. Find the elements of 8 4, Solution!) G 2 3 ~) G 2 3 :) 2 3 :) (~ G 2 3 :) =c 2 3 ~) 4 1 3 2 1 2 (11 (12 (13 (14 (15 (16 (17 (18 (19 71 72 4 2 1 4 1 4 G 2 3 :) G 2 3 ~) G 2 3 :) G 2 3 D G 2 3 :) = G 2 3 :) 3 1 3 4 4 2 73 74 75 76 77 78 2 4 2 1 4 3 G 2 3 ~) 2 3 :) (~ G 2 3 ~) 2 3 :) (~ G 2 3!) G 2 3!) 3 1 1 3 1 2 "'I "'2 "'3 "'4 "'5 "'6 1 3 2 1 2 3 ~) ~) ~) ~) G 2 3 ;) G 2 3 :) 4 3 3 2 2 4 We give the |
multiplication table for future reference. THE SYMMETRIC GROUP OF DEGREE 4 (11 (12 (13 (14 (15 (16 (17 (18 un 71 72 73 74 75 76 1'7 78 "'I "'2 "'3 "'4 "'5 "'6, (11 (12 (13 (14 (15 (16 (17 (18 (19 71 72 73 74 7,5 76 77 78 "'I "'2 "'3 "'4 "'5 "'6 (11 (12 (13, 76 "'2 72 74 "'5 78 (17 "'3 (14 "'I (19 "'4 (16 "'6 71 (18 77 73 (15 75, (11 "'4 (18 "'3 "'I (15 "'6 74 77 76 71 78 73 72 75 (17 "'5 (16 (14 "'2 (19 (12 (13 (13, (11 (12 73 "'5 (14 78 "'3 71 (15 "'5 (18 "'2 (1s 74 "'4 75 (15 (16 (16,, (14 77 71, 76 "'2 75 "'I (16 "'4 (17 "'6 (14 "'3 (19 74 (15 72 76 (18 78 "'4 73 "'2 (19 "'I (11 (13 "'6 (17 "'5 72 75 (18 (12 74 77 (14 "'6 (12 "'I 75 73 72 78 71 77 76 74 (19 (13 "'4 "'3 (11 (17 (17 76 "'6 77 71 "'I 74 (18 (19 (15 77 "'3 72 (13, "'3 "'I (19 "'5 "'2 (17 "'6 (11 73 71 (12 (18 78 7S (11 (13 "4 (16 "2 "5 (14 (12 72 78 75 73 (15 (18 "2 (15 "5 "3 (12 "4 (19, (17 78 76 77 75 74 72 73 71 "6 (11 (14 (16 (13 "1 (19 (19 72 "1 73 78 "S 75, (17 (18 (14 71 (14 78 "3 "1 74 (17 "2 75 (13 72 "1 73 0"9 (11 76 "2 (16 77 "3 72, "2, "5 (16 "4 (11 (13 "3 (15 76 71 74 77 (12 (12 76 77 (15 (18 73 (11 O"s (19 "6 "4 "5 71 |
78 (15 (18 74 75 (12 (14 (17 (13 "5 "6 "4 73 (19 72 "1 "5 77 (13 (14 76 "4 (15 75 74, (12 78 71 (18 (16 "6 "2 (11 (17 "3 74 "4 75 (16 (17 71 "1 "'5 78 (11 77 (12, 73 72 (18 (15 76 (13 "3 "S (19 0"4 "'2 75 (16 74 "4 (19 78 "S (13 71 "2 73 (15 (18 77 7S, (12 72 "5 (14 "1 (17 "3 (11 7S "S 77 (17 "2 72 (11 "4 73 (1~ (18 78 71 (12, 75 74 (15 "3 (19 "5 (13 "1 (16 77 (17 76 "6 (13 73 "'5 "3 72 (16 (12 74 75 (18 (15 71 78, "4 78 78 "3 71 (14 "6 75 (19 (11 74 "5 76 (18 (15 72 73 (12, 77 "1 73 (19 72 74 (17 71 (15 "6 (12 (16 (13 (11 (14 "3 "5 "2 "4 "2, "'2 (15 "5 (18 72 (11 76 75 (13 71 (19 (14 "3 "6 (17 (16 "4 "1 78 "1 (11 "2 (19 (14 "4 (17 "1 (16 (13 77 75 78 76 (18, 73 77 (12 74 "3 71 (14 78 (12 "4 75 (16 74 (18 "4 "3 (18 72 (16 77 (11 (17 "6 "2 "5 "1 (19 (13 76 74, (15 75 7.... (12 73 (14 76 "5 "6 (17 (13 (11 (19 "1 "2 77 78 (15, 71 72 "5 (18 "2 (15 77 (13 73 78 (11 74 "6 "4 (16 (19 "'I "3 (14 (17 75 (12 76 72 I 71 "6 77 (17 76 75 0"9 78 (12 "1 (15 "4 "s "2 "3 (14 (13 (11 (16 (18 73 74 71 72 I 60 GROUPS AND SUBGROUPS [CHAP. 3 h |
. Even and odd permutations We are interested in a special subgroup of Sn, the alternating group of degree n, usually denoted by An. This subgroup An is obtained from Sn by singling out certain elements. To begin with, consider S3. Let u = (~ ~ ~ ). Then 2u - 1u 3u - 1u 3u - 2u 2-1. 3-1. 3-2 3-2 1-2 1-3 2-1'3-1'3-2 1 If T (! 2 2 ~), then IT 3T - 2T IT 3T - 2T 2-1. 3-1 --3=--------=c2- -_. __._2-3 1-3 1-2 2 - 13 - 13 -2 -1 We say u is even and T is odd. More generally, let us call u E Sn even (or an even permutation) if 2u - 1u 3u - 1u 3u - 2u..... n - 1 nu - 1u nu - 2u nu. n - 2..... n - (n - 1)u (n - 1) 1 On the other hand, we call u E Sn odd (or an odd permutation) if 2u - 1u 3u - 1u 3u - 2u --,--2----,-1-· 3 - 1. 3 - 2 nu -1u nu - 2u n-1 n-2 nu - (n-1)u (n -1)...• n - -1 The definition of even or odd is written more briefly as u is even if II ku - ~u i<k k - '/, 1 u is odd if II ku - i<k k - i iu = -1 We shall show that an element in Sn is either even or odd, i.e. II ku - ~u = ±1. i<k k - '/, Corresponding to each factor k - i in the denominator, we will find a factor in the numerator which is either k - i or -(k - i). As u is a permutation, there exist unique lu - mu = k - i appears in integers l, m such that lu = k, mu = i. the numerator. If l < m, the factor mu -lu = -(k - i) appears in the numerator. The ~u is thus ± |
1. Note that distinct factors k - i, k' - i' in the quotient lU If l > m, the factor - ~u or m; k -'/, -'/, denominator give rise to distinct associated factors ±(lu - mu), ±(l' u - m' u) tor. For if l = l', m = m', we have k = k' and i = i'. in the numera Thus regrouping the factors in the numerator, the product becomes the product of factors ±1 and hence is itself ±1. Therefore every permutation of Sn is either even or odd. There is an easy way of determining whether a permutation u is even or odd. If we are given a row of integers, we call the number of integers in the row smaller than the first integer, the number of inversions. Thus for example the number of inversions in the row 7,4,3,2,1,6,8 is 5. We will use this concept of inversion to find out if a given permutation is even or odd. u 2 1 1u 2u ( Sec. 3.3] THE SYMMETRIC AND ALTERNATING GROUPS 61 To do this we must calculate IT (k~ - ~fF). Much of the calculation is redundant - t<k '/, as we have proved that the result is always 1 or -1. We must only determine the sign. In the denominator we always have positive numbers. In the numerator a negative number kfF - ifF arises if ifF > kfF. For fixed i and varying k, the number of negative factors that arise is the number of k > i for which iu> ku. But this is the number of inversions in the row ifF, (i + l)fF,..., nu. The total number of negative factors is the number of inversions in 1u,2fF,..., nfF + the number of inversions in 2fF,3u,..., nfF + the number of inversions in (n - l)u, nu. Let this total be I. The product of I negative numbers is positive if I is even, and negative if I is odd. So u is even or odd according as I is even or odd. Example 6: Is u = G ~ ~) even or odd? The number of inversions in 3, |
1,2 is 2. The number of inversions in 1,2 is O. Thus the total number of inversions is 2, and hence u is even. Problem 3.22. Is u even or odd? (i) u = (ii) u - (iii) u = 1 4 G 2 3 :) G 2 3 4 ~) G 2 3 4 5!) 5 2 1 6 3 2 4 Solution: (i) Number of inversions in (iv) u (v) u 2 143 143 4 3 Total number of inversions Hence u is even. (ii) Number of inversions in 5 3 2 4 1 324 1 241 4 1 Total number of inversions Hence u is even. (iii) Odd. (iv) Odd. (v) Even. C. The alternating groups 3 1 2 6 7 (! 2 3 4 5 6 ;) G 2 3 4 5 :)(~ 1 4 6 2 2 3 4 5 :) We shall show that the set of even permutations forms a subgroup of Sn. 1. Let An be the set of even permutations in Sn. Then An #~, since the identity permuta tion t E An: IT kt ~t i<k k- '/, I1k-~ = i<k k- '/, 1 2. If u and T E Sn, then IT k(UT) k - i<k i(UT) i IT k(UT) - ~(UT) • ku ku '/, i<k k IT (kU)T ku - i<k ~ifF)T '/,fF ifF ifF (3.1) 62 GROUPS AND SUBGROUPS [CHAP. 3 We will prove that IT (kfI)T kfI - i<k ~ifI)T ~fI IT f!: - mT m<[ l-tn (3.2) To do this we will show that each of the factors ~ =: :T corresponds to one and only one of the factors (krI)T kfI - (ifI)T. ~fI. Correspondmg to each 0 f h f t e actors IT - mT. l _ m m th e right-hand side of (3.2), there exists, since fI is a permutation, unique integers p, q such that PfI = land qfI = m. If p > q, then |
a factor (pfI)T - (qfI)T PfI - qfI IT - nh l-m appears on the left-hand side of (3.2). If p < q, then a factor (qfI)T - (PfI)T (pa)T - (qa)T qfI - PfI pa - qa IT - mT l - m appears on the left-hand side of (3.2). We associate with the factor l~ =: :T the (equal) factor (PfI)T - (qfI)T pa - qa. If P > q, and the (equal) factor (qa)T - (pa)T qfI - P(J'. If p < q. It is easy to see that each of the factors of the right-hand side of (3.2) corresponds in this way to one and only one of the (equal) factors of the left-hand side of (3.2). Hence (3.2) holds. From (3.1) we therefore obtain IT k(fIT) k - i<k i(aT) i IT IT - mT • IT kfI i<k k m<[ l- m ~fI '/, (3.3) It follows from (3.3) and the rules for multiplying +1 and -1, that we have Lemma 3.2: The product of (i) two even permutations is even (ii) two odd permutations is even (iii) an odd permutation and an even permutation is odd (iv) an even permutation and an odd permutation is odd. Accordingly if (J' is even, then a - l is even too, since (J'(J'-l = ~ is even. Thus if a, T E An, l E An. An is therefore a subgroup of 8 n. It is called the alternating group of degree n. fIT- As an illustration let us find the multiplication table for A 4 • From the list of elements of 84 given in Problem 3.21 we determine that the even permutations are = (~ ~ :!) (T2 = (~! ~ :) (T = (1 5 4 (~ T = (1 1 2 T = (1 2 5 The multiplication table is easily written down from the multiplication table of Problem 3.21. Sec. |
3.3] THE SYMMETRIC AND ALTERNATING GROUPS 63 t U2 Us t Us Us Us Us t U2 U2 t Us 7S 73 72 7S 74 77 76 71 7S 77 76 7S 71 73 72 74 76 77 71 7S 72 73 7S U2 Us Us Us 71 72 73 74 7S 76 77 7S 7S 71 74 7S 7S 72 t Us 77 73 Us U2 72 73 77 76 73 72 74 71 7S 7S 7S 71 76 73 77 76 77 7S 74 72 t 71 U2 7S 76 Us 77 Us 7S 74 t U2 U2 t 73 72 Us 74 7S Us 77 72 76 73 Us 7S 71 Us Us Us 77 76 t U2 7S 71 U2 t 7S 74 7S 75 74 71 73 U2 Us 76 72 Us 74 7S Us Us 72 Us 73 71 7S t U2 t 77 74 76 U8 Problems 3.23. Write out a multiplication table for (i) A 1, (ii) A 2, (iii) A 3• Solution: (i) There is only one element in S 1, namely t = (~), and it is an even permutation. Hence a multiplication table for A1 is t [J. Notice A1 is the same group as Sl' (ii) S2 = {G!), G ~)}. G ~) is an even permutation and G ~) is an odd permutaand to where t = (~ ~) is a multiplication table tion. Therefore A2 = {( 1 2)} for A 2• 1 2 t (iii) S3 contains six elements (see example in Section 3.3a). The elements t, U1 and U2 are the even permutations, and a multiplication table for A3 is t U1 U2 U1 U2 t U2 t U1 3.24. Prove An = Sn implies n = 1. Solution: If n > 1, Sn must contain a permutation which interchanges 1 and 2 and leaves everything i7 = i (i = 3,..., n). 7 Et: An' since 7 is an odd permutation, else fixed, i.e. 1r = 2, 27 = 1, and and therefore An'" Sn. By Problem 3.23(i), A1 = Sl' Hence An = Sn implies n = 1. 3.25. Show that the set S |
= {t, U2' Us' us} is a subgroup of A 4• (For notation see page 62.) Solution: If we look at the multiplication table for A 4, we see that the product of any two elements in implies u;uk1 E S because u; = 11;-1 (j = 2,5,8). S is again in S. It is clear that u;, Uk E S Hence S is a subgroup of A 4• 64 GROUPS AND SUBGROUPS [CHAP. 3 d. The order of An We now count the elements of An. Suppose that n ~ 2. Let T be the following element of Sn: We claim T is odd because T: 1 ~ 2, 2 ~ 1, 3 ~ 3,..., n ~ n 2T - lr 3T - lr 3T - 2T. 3 - 1 2 - 1 nT nT. 3 - 2..... n - 1..... n - lr (n - 1 )T (n - 1) -1 Now let (1' (2' •••, (k be the even permutations (i.e. elements of An). Then are all odd; moreover, if (iT = fjT, then Notice that this means there are at least as many odd permutations as even ones. there are exactly the same number of each; for if w is odd, then Indeed W = (WT)T since T2 = t. Since WT is even by Lemma 3.2, every odd permutation is listed in (3.4). Thus there are precisely the same number of even permutations as odd ones. Consequently, as k is the number of even permutations, the number of odd permutations is also k, and the number of odd and even permutations is 2k. But every permutation of Sn is either even or odd, and ISnl = n! Therefore k = n!/2. Putting our results together, we have proved Theorem 3.3: If n is any positive integer> 1, then Sn is of order n!, and An is of order n!/2. Problem 3.26. Check to see whether IAjl (j = 2,3,4) agrees with Theorem 3.3. Solution: IA21 = 1 by Problem 3.23, and 2!/2 = 1. IA31 = 3 by Problem 3.23, and 3! |
/2 = 3. IA41 = 12 by Section 3.3c, and 4!/2 = 12. 3.4 GROUPS OF ISOMETRIES a. Isometries of the line We begin with certain subgroups of SR, the symmetric group on R, the set of real num bers. We think of the elements of R arranged as points on the real line. Then if a, b E R it is clear what we mean by the distance between a and b, namely the absolute value, la - bl, of a-b. We denote the distance between a and b by d(a, b). We define a group I(R), the group of isometries of R, as a subgroup of SR in the follow ing way. 1. Let I(R) be the set of all elements of SR which preserve distance. The elements of this set will be called isometries of R. To put this definition more explicitly, an element a E SR is termed an isometry if and only if d(a, b) = d(aa, ba) for every pair of elements a, b E R. Since the identity mapping L E I(R), I(R) oF 0. Sec. 3.4] GROUPS OF ISOMETRIES 65 2. Suppose (J E I(R). Then of course (J-l E SR. We claim (J-l E I(R). To see this, sup pose a, b E R. Then d(a(J-l, b(J-l) = d((a(J-l)(J, (b(J-l)(J) = d(a, b) as (J is an isometry. Hence d(a, b) = d(a(J-l, b(J-l) which is precisely what we need. Thus (J-l E I(R). Suppose (J, T E I(R); then T- 1 E I(R) and d(a((JT- l), b((JT- l)) = d((a(J)T-l, (b(J)T- l) = d(a(J, b(J) = d(a, b) Thus (JT- l E I(R) and I(R) is a subgroup of SR. Considered as a group in its own |
right, I(R) is called the group of isometries of R. Problems 3.27. Is u an isometry? (In the following, x E R.) (i) u: x --> x + 2. (ii) u: x --> nx, n an integer ¥= 1 or -1. (iii) u: x --> x 2. (iv) u: x --> -x. Solution: (i) First note that u E SR' u is an isometry, since d(x, x') = Ix - x'i and d(xu, x'u) = d(x + 2, x' + 2) = I(x + 2) - (x' + 2)1 Ix-x'i (ii) u is not an isometry, since d(x, x') = Ix - x'i and d(xu, x'u) = d(nx, nx') = In(x - x')1 so that, for x ¥= x', d(x, x') ¥= d(xu, x'u), e.g. x = 2, x' = 1 d(xu, x'u) = Inl. (iii) u is not an isometry, since d(x, x') = Ix - x'i and d(xu, x'u) = d(X2, X,2) = Ix2 so that, for x ¥= x', d(x, x') ¥= d(xu, x'u), e.g. x = 2, x' = O. x,21 implies d(x, x') 1 and (iv) u is an isometry, since u E SR and d(xu, x'u) = dr-x, -x') = I-x - (-x')1 = I-x + x'i = Ix - x'i = d(x, x') 3.28. Set lu = 2, 2u = 1 and xu = x for all x E R excepting 1 and 2. Is u an isometry? Solution: No, since d(I,3) = 13 -11 = 2 and d(lu,3u) = d(2,3) = 1. h. Two points determine an isometry The following lemma gives us |
a method of determining whether the two isometries are the same. Lemma 3.4: If (J, T E I(R) have the same effect on two distinct real numbers a and b, i.e. a(J = aT and b(J = bT, then (J = T. Proof: Let C be any element of R. Then d(c,a) = d(c(J,a(J) = IC(J-a(J1 = d(cT,aT) and hence Assume C(J =/= CT. Since a(J = aT, C(J - au = +(CT - aT) implies Cu - CT = au - aT = 0, Cu = CT, contrary to our assumption. Therefore i.e. Cu - au = -(CT - aT) = -CT + aT, i.e. Cu + CT = 2(au) Similarly, Cu + CT = 2(bu). Hence 2a(J = 2bu and au = bu. But u is a permutation. Con sequently a = b, contrary to the hypothesis that a and b are distinct real numbers. We conclude that Cu = CT and, since C is any element in R, u = T. 66 GROUPS AND SUBGROUPS [CHAP. 3 Using this lemma it is possible to describe the elements of I(R). Let eT E I(R) and let OeT = a. Now d(OeT, leT) = d(a, leT) = la -lITI = d(O, 1) = 1. Hence (i) (ii) lIT = a ± 1 a - leT = ±1 or lIT = a + 1 and OeT = a. Let eT* be the member of I(R) defined by mapping r E R to IT* is clearly an isometry. Then eT* and eT agree on 0 and 1. Hence IT = eT*. r + a. leT = a -1 and OIT = a. Let eT* be the member of I(R) defined by mapping r E R to - r + a. Then IT and IT* agree on 0 and 1. Hence IT = IT*. Thus if eT E I(R), rIT = £r + OIT where £ is either 1 or |
-1. Geometrically, if reT = r + a, it "moves" the real line a units to the right. If reT = - r + a, the real line is inverted about the origin and then moved a units to the right. We come now to an interesting subset of 1= I(R) which we will prove is a subgroup. IT E I, neT E Z whenever n E Z}. Let eT, l' E (I: Z). Let OIT = a, 01' = b. Let (I: Z) = {eTl a and b must be integers. The effects of IT and l' are rIT 1'1' £r + a where 'Y]r + b where £ = 1 or -1 'Y] = 1 or -1 r(Tp,) = ('Y]r+b)p,='Y]('Y]r+b)-'Y]b=r and sim Let p,:r~'Y]r-'Y]b. Then p,=T-1, for ilarly 1'(p,T) = r. Hence 'Y](a - b) E Z. Thus if n is an integer, n(eTT- I ) E Z. Therefore ITT- 1 E (1: Z) and (I: Z) is a subgroup of I. r(eTT- I ) = (£'Y])r + 'Y](a - b). Clearly £'Y] is ±1, and Problem 3.29. Determine whether the following sets of mappings, with composition as the binary operation, are groups. (i) The set of all mappings of the plane of the form Ta: (x, y) -+ (x + a, y + a) with (x, y) E R2, a E R Notice that Ta is defined for each real number a. (ii) The set of all Ta with a E Z. (iii) The set of all Ta with a E Q. (iv) The set of all mappings of real numbers of the form Il-a: x -+ ax with a ¥o 0, a E Q (v) The set of all mappings of the plane R2 of the form Il-a: (x, y) -+ (ax, ay) with a E Q (vi) The set of mappings |
of (v) but with a E Q':', i.e. a¥< 0. Solution: (i) We show first that Ta is a permutation of R2. Ta is a matching of R2 -+ R2, for if (x, Y)Ta = (Xl> yI)T a' (x - a, Y - alTa = (x, y). Hence Ta is a one-to-one onto mapping and so Ta is a permutation of R2. T-a = 1'-1. The set of all Ta is not empty, and TaT;1 = TaT-b = Ta-b since (X,Y)TaT;1 = (X'Y)TaT~b=(x+a,y+a)T_b=(x+a-b,y+a-b)=(x,Y)Ta_b' Thus the set of all Ta is a subgroup of SR2 and so it constitutes a group. (x, y) = (Xl> YI)' If (x, y) E R2, there exists (x - a, Y - a) E R2 and then (ii) Using (i), we need only consider whether TaT;1 = Ta-b belongs to the given set, which it clearly does. (iii) This follows by arguing as in (ii). (iv) It is easy to verify that each Il-a is a permutation of R. Also, fJ.afJ.;1 = ll-af1.1/b = f1.a/b' Hence the set of all fJ.a. a¥< 0, a E Q, constitutes a subgroup of the group of all permutations of R and itself forms a group. fJ.(1/al = (ll-a)-I. Now Sec. 3.4] GROUPS OF ISOMETRIES 67 (v) The set of fla is not a group, as flo has no inverse. For flo maps all points onto the point (0,0), and by Theorem 2.4, page 36, the only elements of M R2 which have an inverse are the one-to-one and onto mappings. (vi) The set of all fla in this case forms a group. The set is non-empty. If fla is an element, then fll/a = |
fl;;1 because (x, Y)(flafl1!a) = (ax, aY)fll/a = (x, y) and (x, Y)(fll/afla) = (x, y). By Theorem flafl/; I = flafll/b = 2.4 fla is a one-to-one onto mapping and thus fla is a permutation of R2. Now (x, Y)flafl/; I = (ax, aY)fl/; I = (~x, ~ Y) = (x, Y)fla/b' Hence the set of all fla is a sub fla/b since group of SR2. c. Isometries of the plane Let E be the set R2 = R x R. If (XA, YA) = A, (XB, YB) = B are two elements of E, we define the distance between A and B as V(XA - XB)2 + (YA - YB)2 and denote it by d(A, B). Recall that if we interpret A and B as points of the Euclidean plane with coordinates (XA, YA) and (XB, YB) relative to a given cartesian coordinate system, then the formula for d(A, B) is the ordinary distance between A and B. The interpretation of E as the Euclidean plane is not an essential part of our argument. Logically we do without it and work abstractly with the ordered pairs (x, y). a ESE, the symmetric group on E, is called an isometry if for any A, B points of E, d(A, B) = d(Aa, Ba). Theorem 3.5: The set I of all isometries of E forms a subgroup of SE. Proof: I is not empty, since the identity mapping is an isometry. We need to show only that aT-I E I whenever a, TEl. Consider the effect of T- I. As T ESE, there exist, for each pair of points A, BEE, A', B' E E such that A'T = A, B'T = B. Then d(A', B') = d(A'T, B'T) = d(A, B) since T is an is |
ometry. But AT-I = A', BT- I = B'. Hence d(AT- I, BT- I) = d(A, B) and T- 1 is an isometry. Consequently d(AaT-t, BaT-I) = d(Aa, Ba) d(A, B) and so aT- 1 E I. Hence I is a subgroup of SE. In the next four paragraphs we shall argue informally. Let us imagine that the Euclidean plane E is covered by an infinite rigid metal lamina S. If we move S so that it still covers E, we can define an isometry induced by that movement. Let the points of E be denoted P, Q, R,... and the points of S be denoted A, B, C,.... In the initial position suppose that A lies on top of P, B on top of Q, C on top of R,.... After the sheet S is moved, the point A is on top of another point of E, say PI; the point B is on top of, say, the point Ql; C is on top of, say, R1;......-s- Metal lamina ~ • P '"-->-Euclidean plane--->"" • Q • R A B C • • • • • • PI Q1 Rl (a) Initial position. (b) After a movement. Fig. I. The isometry induced by a movement. 68 GROUPS AND SUBGROUPS [CHAP. 3 Let e: E -7 E be defined by Pe = P l, Qe = Ql, Re = R l, • • • • Then we assert e is an isometry. For if P, Q are any two points of E, with, in the initial position, A of S on top of P, and B of S on top of Q, then we have d(A, B) = d(P, Q). After S is moved, A lies on top of, say, P l, and B on top of Ql. Hence d(A, B) = d(P l, Ql) and so d(P l, Ql) = d(P, Q). Using this informal approach, we now describe three particular isometries: 1. Rotation about a point. Let 0 be any point of S. Rotate S about 0 |
through an angle \jr. Then the isometry induced by this movement of S is called the rotation about 0 through an angle \jr. _____'-5 - Metal lamina -_~ (a) Initial position. (b) A rotation through an angle.y. Fig. 2 2. Reflection in a line. Let us choose a line in E and turn S over this line and back to E. The isometry obtained in this way is called the reflection in XY. (a) Initial position. (b) Rotating about XY. (c) Final position. Fig. 3. A reflection in line XY. 3. Let us choose a line XY. Let e be an isometry corresponding to a movement of S for which XeYe, the line joining Xe and Ye, is parallel to XY. Then e is called a translation. We now describe formally these three types of isometries. As a simplification we describe only rotations about the origin and reflections about the axis OX. 1. A translation Ta,b is the mapping defined by (x, Y)Ta,b = (x + a, Y + b) It can be shown that for each a, b, Ta,b is an isometry and that (Ta,b)-l = T -a,-b' Problem 3.30 for details.) (See 2. A counterclockwise rotation about the origin through an angle e is the mapping p. defined by = (x cos e - Y sin e, x sin e + Y cos e) For each e, p. is an isometry and (p.)-l = p_.' 3.32 shows why p. is called a rotation through e.) (See Problem 3.31 for proof. Problem Sec. 3.4] GROUPS OF ISO ME TRIES 69 3. Define a reflection in OX to be the map a y where (x, y)ay = (x, -y) This is readily seen to be an isometry and it is easy to show that (a)-l = ay (Problem 3.33). Since t is the product of two reflections, we call it a reflection. This is a convenience which will simplify the statement of Theorem 3.8. Problems 3.30. Show that 'Ta,b is an isometry and that ('Ta,b)-l = 'T-a,-b' Solution: First we must show that 'Ta,b |
ESE' so we must show that it is a one-to-one onto mapping. (x, Y)'Ta,b = (x', Y')'Ta,b clearly implies (x - a, y - b)'Ta,b = (x,y) and so 'Ta,b is onto. Hence 'Ta,b ESE' Is 'Ta,b an isometry? If A = (XA,YA) and B = (XB,YB), then (x, y) = (x', y'). If (x, y) E E, then d(A,B) and 'Ta,b is an isometry. Similarly, 'T-a,-b'Ta,b =, and so ('Ta,b)-l = 'T-a,-b' (x,Y)'Ta,b'T-a,-b = (x+a,y+b)'T-a,-b = (x,y). Hence 'Ta,b'T-a,-b=" 3.31. Show that p. is an isometry and that (p.) -1 = P _. • (Hard.) Solution: We must show first that p • x cos (J Y sin (J x sin (J + y cos (J - is an element of SE' (x, y)p = (x', y')p • • implies - x' cos (J y' sin (J x' sin (J + y' cos (J (3.5) Multiply the first equation by cos (J and the second by sin (J and add to obtain x(cos2 (J + sin2 (J) + (-y cos (J sin (J + y cos (J sin (J) = x'(cos2 (J + sin2 (J) + (-y' cos (J sin (J + y' cos (J sin (J) Since cos2 (J + sin2 (J = 1, x = x'. Similarly by multiplying the first equation of (3.5) by sin (J and the second by cos (J and subtracting the first from the second, we find y = y'. Hence p. is one-one. Is it onto? (a, b) E R2? That is, does there exist a solution to In other words, can we find (x, y) such that (x, y)p = |
(a, b) • for any x cos (J y sin (J x sin (J + y cos (J - a b (3.6) for x, y E R? We solve these equations for x and y using the same stratagem as above, i.e. multiply the top equation by cos (J and the bottom by sin (J and add to obtain x(cos2 (J + sin2 (J) = a cos (J + b sin (J x = a cos (J + b sin (J or Multiply the first equation of (3.6) by sin (J and the second by cos (J and subtract the first from the second to obtain y = y(sin2 (J + cos2 (J) b cos (J - a sin (J On substituting these values of x and y in (3.6), it is easily seen that they satisfy the equation. (The reader who knows that the condition for existence of a solution to (3.6) is can verify the existence of a solution to (3.6) immediately.) c~s (J I sm(J Finally, is p. an isometry? If A = (XA,YA) and B = (XB'YB)' then -sin(J I # 0, cos (J 70 GROUPS AND SUBGROUPS [CHAP. 3 d(Ap,Bp ) e e Y[(XA cos 0 - YA sin 0) - Y[(XA - Xn) cos 0 - (XB cos 0 - YB sin 0)]2 + [(xA sin 0 + YA cos 0) (YA - YB) sin of + [(XA - x B) sin 0 + (YA - YB) cos of (XB sin 0 + YB cos 0)]2 Y(XA - XB)2 [cos2 0 + sin2 0] + (YA - YB)2 [cos2 0 + sin2 0] Y(XA - XB)2 + (YA - YB)2 = d(A, B) and thus P e is an isometry. (x cos 0 - Y sin 0, x sin 0 + Y cos o)p -e [(x cos 0 - Y sin 0) cos (-0) - (x sin 0 + Y cos 0) sin (-0), (x cos 0 - Y |
sin 0) sin (-0) + (x sin 0 + Y cos 0) cos (-0)] [x(sin 2 0 + cos2 0), y(sin2 0 + cos2 0)] (x, y) and so PeP- e = t. Similarly P-eP e = t. 3.32. Show that (0,0) is equidistant from (x, y) and (x, Y)P Ll joining (x, Y)Pe and (0,0), and the line L2 joining (x, y) and (0,0), is 0 when ° === 0 === 11", or'and that the smallest angle between the line e 211" - 0 when 11" === 0 === 211". Solution: As we have shown in Problem 3.31, d«O, 0), (x,y)) = d«O,O)P, (x,y)p) = d«O, 0), (x,y)p). We remind the reader of the formula for calculating the angle between two lines meeting at the origin. If the one line Ll has end point (al' bl) and the other L2 has end point (a2' b2), then the cosine of the angle between Ll and L2 is given by e y(a2 + b2 )(a2 + b2 ) 2 1 1 2 Put (a l, bl ) = (x, Y), (a2' b2) = (x, y)p o. Then if fl is the smallest angle between the lines Ll and L 2, cos fl = (X2 + y 2) cos 0 Y (X2 + y 2)(X2 + y 2 ) Hence fl = 0 if ° === 0 === 11", and fl = 211" - 0 if 11" === 0 === 211". Show that Uy is an isometry and that u; = t. Solution: If A = (XA' YA) and B = (XB' YB), 3.33. = cos 0 d(Auy, Buy) Y(XA - XB)2 + (-YA - (-YB))2 = d(A, B) and so uy preserves distance. Obviously clearly onto, for u y - (x, y)uy = (x', Y')u y (x, -y)uy = (x, y). Hence uy is an is |
ometry. 2_ t. (x, y) = (x', y'). And uy is implies (x, y)uyuy = (x, -y)uy = (x, y). Thus 3.34. Show that rotations about the origin form a subgroup of I. Solution: Po, the rotation through zero degrees, is the identity; hence the set of rotations is not empty. If Pe,Pq, are two rotations, (pq,)-l = p(_q,)" Put -¢ = '1', as it is annoying to carry the minus sign. Is P (p )-1 = P P.•. a rotation? e ~ e q, (x cos 0 - y sin 0, x sin 0 + y cos o)p,!! «x cos 0 - y sin 0) cos '1' - (x sin 0 + Y cos 0) sin '1', (x cos 0 - (X (cos 0 cos '1' - y sin 0) sin '1' + (x sin 0 + y cos 0) cos '1')) sin 0 sin '1') - y(sin 0 cos '1' + cos 0 sin '1'), x(cos 0 sin '1' + sin 0 cos '1') + y(cos 0 cos '1' - sin 0 sin '1')) (x cos (0 + '1') - Y sin (0 + '1'), x sin (0 + '1') + y cos (0 + '1')) (x, Y)P e +,!! Hence PeP,!! = P(e+'!!l' and Pe(prp)-l is a rotation. Thus the rotations form a subgroup of I. Sec. 3.4] GROUPS OF ISOMETRIES 71 3.35. Show that the reflections about OX form a subgroup of I. Solution: We have only two elements, for there are only the two reflections, and (1y. Since we have shown that (1y = (1-1, we quickly verify that the two possible products of a reflection and the inverse of a reflection is Yagain a reflection. Hence the set of reflections forms a subgroup of I and is of order 2. 3.36. Show that the set of translations forms a subgroup of I. Solution: Ta.b(Tc.d)-l = Ta.bT |
-c.-d = Ta-c.b-d is easily verified as (x,Y)Ta,bT-c,-d = (x+a,y+b)T_c.-d = (x+a-c,y+b-d) (x, y)Ta-c,b-d and thus the set of translations forms a subgroup of I. 3.37. Find an element of SE that is not an isometry. Solution: Let (1 E SE be defined by (x, Y)(1 = (x, y) except that (0,0)(1 = (1,0) and (1,0)(1 = (0,0). It then 1 = d(A, C). is easy to verify that (1 ESE' Now if A = (0,0), B = (1,0) and C = (0,1), d(A(1, C(1) = d(B, C) = V2 ~ d(A, C). Hence (1 is not an isometry. d. Isometries are products of reflections, translations and rotations We will prove in this section that an isometry is determined uniquely by its action on any three points not all on a straight line. This enables us to prove that every isometry is expressible as the product of a reflection, a translation and a rotation. (Note, however, that there are isometries that are neither reflections, rotations, nor translations.) Lemma 3.6: Let u E I. Let A, Band C be three noncollinear points in E. Let Au = A', Bu = B' and Cu = C'. Then A'B'C' is a triangle congruent to ABC. Proof: d(A', B') = d(A, B), d(A', C') = d(A, C) and d(B', C') = d(B, C). Thus we have two triangles with corresponding sides equal, and so 6.ABC is congruent to 6.A'B'C'. Lemma 3.7: If u and T, elements of I, have the same effect on three points A, B, C which do not lie on a straight line, then u = T. Proof: Let D be any point of E. Let d(D, A) = a, d(D, B) = band |
d(D, C) = c. Let A' = Au = AT, B' = Bu = BT and C' = Cu = CT. Then the distance of Du from A' is a, from B' is b, and from C' is c. Similarly DT is a distance a from A', b from B', and c from C'. Geometrically it is possible to see that Du = DT, for both Du and DT lie on the inter section of three circles: 0 1 with center A' and radius a; O2 with center B' and radius b; and 0 3 with center C' and radius c. Two circles intersect in two points at most. Hence 0 1 and O2 determine two points. It is possible for Du and DT to be these two points. But we shall show that as Du and DT must lie on 03, they must be the same point. If this is not so, then 01, 02 and 0 3 have two points in common. We shall prove geometrically then that A, B, C must lie on the same straight line. Let three circles with centers A', B' and C' inter sect in two points P and Q. Without loss of gen erality we may assume that B' lies between A' and C'. A' P = A'Q and B'P = B'Q; hence A' B' lies on the perpendicular bisector of PQ. Similarly B'C' lies on the perpendicular bisector of PQ. Therefore A' B'C' is a straight line. But 6.ABC is congruent to 6.A' B'C' by Lemma 3.6. Hence ABC lies in a straight line. But we assumed this was not so. This contradiction proves Lemma 3.7. C' 72 GROUPS AND SUBGROUPS [CHAP. 3 Theorem 3.8: Every isometry of E is expressible as the product of a reflection, a rotation and a translation. We present an intuitive proof first. The formal proof below follows exactly the same steps. Intuitive proof: Let a E I. Let A = (0,0), B = (1,0) and G = (0,1) as shown in Fig. 1. Our idea is first to find a-I as a product, "0/, of a translation, a rotation and a reflection. It is easy then to prove a is the product of a reflection |
, rotation and translation. So we must find a "0/ which is the product of a translation, rotation and reflection such that u"o/ = t. To check that a"o/ = t, we need only prove that for the three noncollinear points A, B, G the effect of t and a"o/ is the same (Lemma 3.7). We build "0/ in three stages so that we bring (a) Aa to A; (b) Ba to B; (c) Ga to G. Let Au = (a, b). C A o Bu- -_ Au Cu C ~, BUT_ a -b = B' / /8 B B X o A = AUT_a,-b X -CUT_a,-b C B= BUT_a,-bP-O A = AUT_a,-bP-O ~------~----~---X o CUT -a, -bP-O = C' Fig. 1 Fig, 2 Fig. 3 Apply the translation T -a, -b which moves each point a distance d(A, Ao-) parallel to the line joining Aa and A, so that AaT -a, -b = A. Let B' = BaT -a, -b' Then B' A is at an angle 8, say, to OX and, since T -a, -b and a are isometries, is of length 1. See Fig. 2. Apply the rotation through an angle of -e that takes B' onto B. Let G' = GaLa, -b p_.' Then since a, T -a, -b and p_. are isometries, G' is at a distance 1 from A and a distance V2 from B. Hence G' is either as in Fig. 3, in which case let p. be the reflection in OX, or in which case let p. be the identity reflection. Let "0/ = P.-a,-bP_.P.' Since else G' = G, A, Band G are mapped to A, Band G by a"o/, a"o/ = t by Lemma 3.7. Hence, using our remark on the inverse of a product in Section 2.4, page 34, and the theorem is true. a = "0/-1 = p.-I(p_.)-I(La,-b)-1 = |
p.p. Ta,b Formal proof: We follow the same three steps and use the same notation. Thus A = (0,0), B = (1,0), G = (0,1) and a E I. (a) Suppose Aa = (a, b). Then AaT -a,-b = A. (b) B' = BaT -a, -b must be a distance 1 from A, since 1 = d(A, B) = d(AuLa, -b' BaL a, -b ) = d(A, B'). Hence B' is of the form (cos e, sin e) for some angle e. [This is a well known fact of coordinate geometry. All we must check is that if B' = (d, e) with d2 + e2 = 1, the equations cos e = d and sin e = e can be solved for e.] Then B' P-o (cos e, sin e)p_. (cos e (cos -e) - sin e (sin -e), cos e sin (-e) + sin e cos (-e)) (cos2 e + sin2 e, 0) = (1,0) = B Also, A p _. = A. Sec. 3.4] GROUPS OF ISOMETRIES 73 (c) C' = CaT_a._bP_. is a distance 1 from A and v'2 from B, since AaT_a,_bP_. = A, BaL a, -bP-. = B, and aLa, -b p_. is an isometry. Thus C' = (0,1) or (0, -1). Let fJ- be the identity if C' = (0, 1), and let fJ- be the reflection ay if C' = (0, -1). Put 'IfF = T_a,-bP-ofJ-' Then Aa'IfF=A, Ba'IfF=B and Ca'IfF=C. Hence a'IfF=L by Lemma 3.7, and a = 'IfF- 1 = fJ-Ta.bP •. e. Symmetry groups Given a figure in the Euclidean plane, we shall define a group which we will call the symmetry group of the figure. The word symmetry is not sufficiently precise |
for the needs of Chemistry and Physics; so instead of talking about symmetry, we talk about symmetry groups. In comparing two figures, it usually turns out that what one would normally think of as the more symmetrical figure has a symmetry group of greater order than the less symmetrical figure. Also, the symmetry group of what we would normally call a non-symmetrical figure usually turns out to be of order 1. More generally, we will be concerned with subsets of the Euclidean plane. (Clearly, a figure is a subset of the Euclidean plane.) Theorem 3.9: Let S be any subset of the Euclidean plane. The set, denoted by Is, of all a E I such that (i) S E S t E S, forms a subgroup of I, called the symmetry group of S. (An element of Is, therefore, is characterized by its mapping elements of S, and only elements of S, into S.) implies Sa E Sand (ii) ta E S implies Proof: Is ¥= 0, as the identity mapping of the Euclidean plane into itself is in Is. If IJ, T E Is, is aT- 1 E Is? We first prove T- 1 E Is. If S E S, (ST- 1)T = S E S. Because T E Is, (ii) implies ST- 1 E S. Thus T- 1 satisfies (i), i.e. S E S implies ST- 1 E S. To show T- 1 also satisfies (ii), let tT- 1 E S. Then (tT- 1)T = t E S, since T satisfies (i). Hence T- 1 satisfies (ii), and T- 1 E Is. Now we show aT- 1 E Is. Let S E S. Then Sa E Sand SaT- 1 E S, since a and T- 1 are If taT- 1 E S, since T- 1 satisfies (ii), we have ta E S. t E S and consequently aT- 1 also satisfies (ii). in Is. Therefore aT- 1 satisfies (i). Furthermore, since a E Is, Hence aT- 1 E Is and Is forms a subgroup of I. (ii) implies Problems 3.38. Find a plane figure S such that U = {O' I 0' E I and for all 8 E S, 80' E S} does not form a subgroup |
of I. (In other words, in Theorem 3.9 we cannot drop condition (ii).) Solution: Let S be the infinite half-line starting at (1,0), i.e. S = {(x, 0) I x "'" I}. Then Tl,O E U. Now Tt:~ = T_l,O' But T_l,O moves (1,0) E S to (0,0) (/:. S. Hence U is not a subgroup. 3.39. Find the orders of the symmetry group of (i) (ii) 1 ~rl~~~ ____ 1~0 ______ -;D L B 11 C 11 I G Y 1 i 1 -----12- ---iii) IK 1+1 M~L Y?IN 74 GROUPS AND SUBGROUPS [CHAP. 3 (Hint: Use Theorem 3.8 and argue intuitively. A useful intuitive approach for this problem is to cut a cardboard figure corresponding to each figure, label its vertices on both sides, and draw its perimeter onto a sheet of paper. The isometries are obtained on moving each cardboard figure so that it lies on the drawn perimeter.) Solution: (i) Let S = ABCDEF. Let the images of A, B, C, D, E and F under (J E Is be denoted by A', B', C', D', E' and F'. By Theorem 3.8 it is easy to see intuitively that the image of the plane figure S will be the congruent figure A' B'C'D'E'F', since Theorem 3.8 states that every element of I is a product of a reflection, rotation and translation. But if a rigid body is rotated, translated or reflected it retains the same shape. Now A' B' must lie along AB, as all other sides of S are either smaller or larger than d(A', B') = d(A, B), which means F' must lie on F and hence E' on E, etc. Thus (J must be the identity. Accordingly Is = {,} and is of order 1. (ii) Let S = IJGH. Now IIsl is at least 8, for we have as members of Is: Sl: A reflection in the diagonal GI, GS1 = G, HS 1 = J, IS 1 = I. S2: A reflection in |
the diagonal HJ, GS2 = I, HS2 = H, IS2 = G. sa: A reflection in OX, GSa = J, HSa = I, ISa = H. S4: A reflection in OY, GS4 = H, HS4 = G, IS4 = J. S5: A clockwise rotation about 0 of 0 0 GS5 = G, HS 5 = H, Iss = I. S6: A clockwise rotation about 0 of 90 0 GS 6 = H, HS 6 = I, IS 6 = J. S7: A clockwise rotation about 0 of 180 0,, GS7 = I, HS 7 = J, IS 7 = G. S8: A clockwise rotation about 0 of 270 0 GS8 = J, HS 8 = G, IS 8 = H.,, That all of these are distinct is clear. Could IIsl be greater than 8? Let s E Is. d(G, I) = d(Gs, Is) = V2. Only two pairs of points of S are a distance V2 apart, namely G, I and H, J. Therefore the line GsIs is one of the diagonals of S. Similarly the line HsJs is a diagonal of S. As s is a permutation of S, distinct points of S are mapped by s to distinct points. This means that at most the following possibilities arise: (1) GsIs is GI, HsJs is HJ or JH (2) GsIs is IG, HsJs is HJ or JH (3) GsIs is JH, HsJs is GI or IG (4) GsIs is HJ, HsJs is GI or IG Since these represent eight distinct cases and each involves the movement of three points not in a straight line, by Lemma 3.7 at most one isometry could correspond to anyone case. Then IIsl ~ 8. But we have already exhibited eight elements of Is. Hence IIsl = 8. (iii) Let S be the triangle KLM. Then Is contains the following elements: (a) (b) (J1, the identity mapping of I, K(J1 = K, L(J1 = L, 1I1(J1 = M. (J2' the reflection in KN, K(J2 = K, L(J2 = M, M |
(J2 = L. Hence IIsl:=O 2. Let (J E Is. Since d(M, L) = V2 and the only two points of KLM which are a distance V2 apart are Land M, then either Sec. 3.4J GROUPS OF ISOMETRIES 75 (a) Mu = M, Lu = L. Then as K must be a distance 1 from both M and L, Ku = K. or (b) Mu = L, Lu = M. Then as K must be a distance 1 from both Mu and Lu, i.e. from M and L, and K is the only point of S which is a distance 1 from both Land M, Ku = K. Hence!/sl ~ 2, by Lemma 3.7. Therefore Ilsl = 2. f. The dihedral groups Let S be a regular n-gon, n> 2, e.g. one of the figures below. We will show that in any isometry of S, vertices are taken to vertices. This will make it easy to determine the order of the symmetry group of a regular n-gon, n > 2. We will take the following geometrical lemma for granted. 0 1 1 1 0 1 1 1 Lemma 3.10: Every regular n-gon can be circumscribed by one and only one circle. We call the center of the circumscribing circle of an n-gon its center. Lemma 3.11: The center of a regular n-gon S is taken onto itself by any element of Is. Proof: Since every point of S is within a distance r, say, from the center 0, and II is an isometry, then every element of SII is within a distance r from 01I. Also, there are points of SII which are exactly a distance r from 01I, as there are points of S which are exactly a distance r from O. But SII = S. Hence the circle with radius r and center OII is a circum scribing circle of S. But by the previous lemma there is only one circumscribing circle of S. Thus OII = O. Lemma 3.12: If S is a regular n-gon and II E Is, then vertices of S are taken onto ver tices of S by II. Proof: If A is a vertex of Sand 0 is the center of S, OII = 0 by Lem |
ma 3.11. d(OII, AlI) = d(O, AlI) = d(O, A). Hence AlI is a distance r from 0, where r is the radius of the circumscribing circle C. The only points of S on the circumference of C are vertices. But AlI is an element of S on the circumference of C. Thus AlI is a vertex. The symmetry group of the regular n-gon is called the dihedral group of degree n. We can now calculate the orders of the dihedral groups. Let the vertices of a regular n-gon S with center 0 be AI,..., An (in a clockwise direction). Let II., 1 ~ j ~ n, rotate S about 0 in a clockwise direction through an angle 27rU -1) J n radians ( = 3!O U -1) degrees) so that Al<1j = A j regular pentagon is shown below. • As an example, the effect of <13 on the 76 GROUPS AND SUBGROUPS Let, be the reflection about the line through Al and 0, so that AI' = AI' A The effect of, on the regular pentagon is shown. [CHAP. 3, = An' 2 o T - ~-~---4Aa o The following diagram illustrates the effect on a regular pentagon of the reflection, fol lowed by the rotation O"a' T - The elements 0"1' •••, O"n',0"1' •••, TUn are all distinct. For certainly O"j"l'= O"k' j"l'= k, as j = k. AlO"j "1'= AlO"k' j"l'= k. But TO"j = O"j implies,= 0"1' the identity, contrary to assumption. Finally, TO"j = 'O"k implies O"j = O"k' then AlTO"j = AlO"j = AlO"k' Thus If TO"j = O"k' 'O"j = O"k implies So there are at least 2n possible elements of the dihedral group of degree n. But we can easily show that there are no more than 2n. For if 0" E Is, S the regular n-gon, then there are n possibilities for AlO". As vertices are |
taken to vertices, AlO" is one of AI' •••, An' A 20" has only two possibilities once AlO" is determined as d(AlO", A ), and A 20" 0" are determined, AiO", i = 3,4,..., n are also deter must also be a vertex. Once AlO" and A mined. Hence there are at most two elements 0" E Is which map AlO" to A j • Thus there are at most 2n elements of Is, and so II sl = 2n. 2 0") = d(Al' A 2 2 Let D n denote the dihedral group of degree n. Problems 3.40. Find Da and its multiplication table. Solution: The elements of Da are the O'j. TO'j above. Note that O'j0'2 - O'j+ 1 if 1 "'" j "'" 2. and O'a0'2 = 0'1' Also note that T- l = T and O'iT = T20'iT = TTO'iT. Now TO'IT = 0'10 since 0'1 is the identity; T0'2T = O'a. as AlT0'2T = Al0'2T = A2T = Aa; and A2T0'2T = Aa0'2T = AlT = AI' So O'aT = T0'2 and TO'a = 0'2T. Ac cordingly the mUltiplication table is as follows: 0'1 0'2 O'a T 0'2 T 0'2 0'2 O'a 0'1 O'a 0'1 0'2 T0'2 TO'a T0'2 T0'2 TO'a T T T T0'2 TO'a TO'a T T0'2 T0'2 TO'a 0'1 O'a 0'2 0'1 T O'a 0'2 TO'a T T0'2 0'2 0'3 0'1 Sec. 3.5] THE GROUP OF MC)BIUS TRANSFORMATIONS 77 3.41. Show that the following are subgroups of D3: the same as in the preceding problem.) (0 {0"1}, (ii) {0"1' 0"2' 0"3}, (iii) |
{r0"2'0"1}' (Notation is Solution: It is only necessary to check in each case that the set is not empty and if g, h belong to the set, gh- 1 belongs to the set. It is easy to calculate gh- 1 from the multiplication table of Problem 3.40. 3.42. Find D 4, the symmetry group of the square, and its multiplication table. Solution: Notice that we have already found the elements of D4 in Problem 3.39(ii). We will, however, j = 1,2,3,4, for the rotations and r for the reflection. for 1 '"'" j '"'" 3, use the notation of Section 3.4f, i.e. O"j' j = 1,2,3,4. Now O"j0"2 = O"j+ 1 Accordingly the elements of D4 are O"j and rO"j' 0"40"2 = 0"1, and O"iO"j = O"jO"i for all i and j. Also r- 1 = r, O"ir = rrO"ir, and rO"lr = 0"1' We show r0"2r = 0"4' A1r0"2r = A10"2r = A 2r = A 4, A2r0"2r = A40"2r = A 1r = Alo and Aar0"2r = A30"2r = A4r = A 2• Furthermore A 10"4 = A 4, A 20"4 = Alo and A30"4 = A 2• Since r0"2r and 0"4 have the same effect on the three points A 1, A2 and A 3, r0"2r = 0"4 by Lemma 3.7. The following calculations facilitate the construction of a multiplication table for D 4• r0"3r = rO"~r = (r0"2r)(r0"2r) = O"! = 0"3, and 0"4 = 0"30"2 implies 0"3r = rr0"3r = r0"3, and 0"4r = rr0"4r = r0"2' O"~ = 0"3 r0"4r = r0"30 |
"2r = (r0"3r)(r0"2r) = It is now easy to implies 0"30"4 = 0"2' Hence 0"2r = rr0"2r = r0"4, construct the table: r0"2 0"1 0"2 0"3 0"4 r r0"2 rO"a r0"4 0"2 0"2 0"3 0"4 0"1 r0"4 r r0"2 r0"3 0"3 0"4 0"4 0"1 0"1 0"2 0"2 0"3 r0"3 r0"4 r r0"2 r0"2 rO"a r0"4 r r r0"2 r0"3 r0"4 0"1 0"2 0"3 0"4 r0"2 r0"3 r0"4 r r0"3 r0"4 r r0"2 r0"4 r r0"2 r0"3 0"4 0"3 0"2 0"1 0"2 0"3 0"4 0"1 0"2 0"3 0"4 0"1 3.5 THE GROUP OF MOBIUS TRANSFORMATIONS a. Defining the group The complex numbers can be represented as points of the Euclidean plane E, the com plex number z = x + iy corresponds to the point with coordinates (x, y). Instead of inquiring (as we did in Section 3.4c) what are the permutations of E that preserve distance, we inquire what are the permutations of E that preserve both angles and their orientation. These are called conformal mappings. It can be shown (see Ford, L. R., Automorphic Functions, Chelsea, 1951) that the mappings az + b u = u(a, b, c, d): z ~ cz + d where a, b, c, d are fixed complex numbers such that ad - be # 0, preserve angles and their orientation. But u(a, b, c, d) is not always a mapping of E to E. Two things can go wrong. If c # 0, then: (i) Zu is not defined if z = -d/c, as then the denominator becomes 0. (ii) There is no complex number that maps to a |
/c. For suppose Zu = a/c, then az + b = (cz + d)a/c and b - ad/c = ° and hence be - ad = 0, the very condition we assumed did not hold. 78 GROUPS AND SUBGROUPS [CHAP. 3 It seems as if we have been cheated in our efforts to argue analogously to Section 3.4c in order to prove the a for various a, b, c, d form a group, because not all the a are permutations of E. However, by adding an extra element 00 to E and forming E u {oo} = E we can over 00 is any object outside E. It is customary to write 00 for historical come these difficulties. reasons. The reader is cautioned that just as the symbol x can have different meanings (e.g. x is sometimes a number and sometimes an element of a group or a groupoid, etc.), so 00 has different meanings. The 00 we introduce should not be confused with the 00 in such expressions as lim 1 = 0; it is logically distinct. x-+cc x Our idea is to extend a to E in order to patch up the difficulties (i) and (ii) above so that a E SE5, the symmetric group on E. az + b (a) If c = 0, define Za = CZ + d for any complex number z, and put OOa = 00. (b) If c # 0, put az + b Za = CZ + d for Z ¥ -d/c, z E C, the complex numbers. Put (3.7) (-d/c)a = 00 and OOa = a/c In (a) we have no real problem. Having had to add an extra element, we just let it map In (b) we neatly get rid of both difficulties (i) and (ii) above, for we have both to itself. defined (-d/c)a and found an element to map to a/c. M will denote the set of all mappings of E to E defined by (3.7). We will show that M is a subgroup of S if' leaving most of the checking of details to the problems. First, each of the a(a, b, c, d) is a member of S E (Problem 3.45). Next, the inverse of a(a, b, c, d) is |
given by a(-d, b, c, -a) (Problem 3.44). Finally, for some choice of a3, b3, C3, d 3 (Problem 3.46). Note that a(l, 0, 0, 1) is the identity mapping (denoted by L). Hence the product of an element of M and the inverse of an element in M belongs to M. Thus M is a subgroup of S E' It is called the group of Mobius transformations. Problems 3.43. Determine the image of (i) i, (ii) 1 + 2i, (iii) "', and (iv) -1/3 under 0'(2,1,3,1). Solutions: (i )'2i + 1 to' = 3i + 1 (ii) (1 + 2') 1. t 0' = 3(1 + 2i) + 1 = 13 - 26 t 2(1 + 2i) + 1 9 (iii) 000' = 2/3 (iv) -1/30' 3.44. Show that O'(-d, b, c, -a) is the inverse of O'(a, b, e, d), given ad - be # 0. Solution: Case (a): e = 0. zO'(a, b,O,d) O'(-d, b,O,-a) = [(az:b)(_d) + b]/(-a) = z 00 O'(a, b, 0, d) O'(-d, b, 0, -a) = 00 O'(-d, b, 0, -a) = 00 Case (b): e # 0. z = -die. z O'(a, b, e, d) O'(-d, b, c, -a) (i) 00 O'(-d, b, e, -a) -die z Sec. 3.5] THE GROUP OF MOBIUS TRANSFORMATIONS 79 (ii) Z ¥ -dlc or "". Then Z u(a, b, c, d) u(-d, b, c, -a) [ ( ~; ~!) (-d) + b J/[ (~; ~!) c - a J bcz + bd - adz - bd azc + bc - acz - ad (bc- ad)z |
bc - ad z "" u(a, b, c, d) = alc and (alc) u(-d, b, c, -a) = "" Hence u(a,b,c,d)u(-d,b,c,-a) =,. Similarly we can show that u(-d, b, c, -a) u(a, b, c, d) =,. Hence u(-d,b,c,-a) = u(a,b,c,d)-l. 3045. Prove u(a, b, c, d) E S"E for any choice of a, b, c, d such that ad - bc ¥ O. Solution: In Problem 3.44 we have seen that each u(a, b, c, d) has an inverse. By Theorem 2.4, page 36, any mapping of a set into itself which has an inverse is a one-to-one onto mapping. Therefore u(a, b, c, d) is one-to-one and onto, and so u(a, b, c, d) is an element of S E' U6. Show that where a3 = ala2 + b2c l' b3 = a2 bl + b2d v ca = alc2 + d2Cv and da = blC2 + d ld 2. Prove also that a3d3 - b3C3 ¥ O. (Hint: This is very much an endurance test.) Solution: Let u(al, bl, Cl, d l ) = 0'1, u(a2, b2, C2, d 2) = 0'2, and u(a3' b3, C3, d 3) = 0'3' Note that (alaZ + b2Cl)(b l C2 + d l d2) - (b l a2 + b2d l )(alc2 + d2Cl) = (aldl - b lc l)a2d 2 - (aldl - b lCl)b2C2 = (aldl - blC1)(U2d2 - b2C2) ¥ 0 Hence 0'3 is a Mobius transformation. Now if z E E satisfies (A) if C1 ¥ 0, z ¥ -d1/c1 (B) if |
Cz ¥ 0, zU1 ¥ -~/c2 (C) (D) if C3 ¥ 0, z ¥ -d3le3 z ¥ 00 then (u1a2 + b2Cl)Z + (a2 b1 + b2d 1) (alc2 + d 2C1)Z + (b1C2 + d 1d 2) Thus except for restrictions (A), (B), (C) and (D), there is nothing more to prove. Since 0'10'2 and 0'3 are permutations, we may ignore one of these cases, say (D). This obtains because if for all com plex numbers z, ZU1U2 = zUa, then"" can only be mapped to one element of E by 0'10'2 and ua. Ac cordingly we shall not consider Z = 00 in the following case by case analysis. Case (a), C1 = O. We have two possibilities: (i) c2 = 0, (ii) C2 ¥ O. c2 = O. Then ca = u1C2 + c1dz = O. Thus (A), (B) and (C) do not restrict z, and we can conclude 0'10'2 = 0'3' c2 ¥ O. We first show Z = -d3/ca if and only if ZO'l = -d2Ic2' Z = -dalca implies (i) (ii) A simple computation shows that ZU1 = (a1z + b1)/d1 = -d21c2 zUl = -d2Ic2' zU1 = -d2/c2' Thus ZUlU2 = 00 = zU3 in this case, and so 0'10'2 = O'a' then ZU1U2 = (-d2Ic2)u2 = 00 and, as Z = -da/c3' Zua = 00. Z = -da/ea -a1(b 1C2 + d 1d 2) + u1C2b1 a1c2d l implies Z = -da/ca. Now if implies = 80 (i) GROUPS AND SUBGROUPS [CHAP. 3 Case (b), CI 7'= o. Again there are two possibilities: C2 = o. Then c |
3 = cldz and, as a2d2 - C2b2 7'= 0 it follows that C37'= O. Note also that -d31c3 = -dld2/cld2 = -dl/cl. Hence we need consider only the possibility Z = -dlic l • If Z = -dj/cb then, as c2 = 0, z"I"2 = 00"2 = 00 while z"3 = 00. Hence "1<72 = <73. (i) C2 = 0, (ii) C2 7'= O. implies d 2 7'= 0, (-d jic j)(ala2 + c l b2) + (bla2 + d jb2) bjC2 + d jd 2 al(-ajd l + blcl) cI(bjC2 + d jd 2) From aj/cI = -d2/c2 we have d 2 = (-a lic j)c2 and a2( -at d l + b j Cj) a2 Now Z<71 = -d2/c2 = al/cl only if Z = 00, and we need not consider this case. Hence <71<72 = <73. (f3) C3 7'= o. Then al/cl 7'= -d2Ic2. If Z = -dl/c2' Z"I<72 = 00<72 = a21c2. (aja2 + b2cI)(-d jic j) + (a2bl + b2d j) (ajc2 + d2cI)(-d l ic j ) + (b1C2 + d l d 2) a2(-a l d j + blcl) c2(-a l d l + bjcj) then Z = -d3/c3; for (alz + bj)/(clz + d l ) = -d21c2 a2 C2 and, as implies alc2z + c3 = a l c2 + d2Cl 7'= 0, Finally if Z<71 = -d2Ic2' b1C2 = -c jd 2z - d l d 2. Hence z = -(b1C2 + d 1d 2)/(alc2 + d 2cI) = |
-d3lc3. Therefore (alc2 + c jd 2)z = -(bIC2 + d jd 2) = 00 3.47. Let <7(a, b, c, d) = <71 be a Mobius transformation and k a nonzero complex number. Show that,,(a, b, c, d) =,,(ka, kb, kc, kd). Solution: Denote <7(ka, kb, kc, kd) by u. If z 7'= 00 or -dlc (if c 7'= 0), Z<71 = az + db = cz + kaz + kb kcz + kd = zc;. To treat the special cases z = 00 or -dlc (when C 7'= 0), we first assume c = O. Then kc = 0, and then kc 7'= O. Therefore 00<71 = ~ = ka = 00 Ci and - 0..<71 = 00 = (-kdlkc) u. Thus for all possible choices of z, we have kc Z<71 = zCi and hence <71 = iT. 00<71 = 00 = 00 u by definition. Secondly, if c 7'= 0, then, as k -7'= 0, c c 3.48. Show that the set of all Mobius transformations {<7(a, b, c, d) I ad - bc = 1}. Solution: Let M = {<7(a, b, c, d) I ad - bc = 1}. If <7 = <7(a, b, c, d) is any Mobius transformation, then, by definition, D = ad - bc 7'= O. From Problem 3.47 above we know that <7 = <7 ( ~, ~,_c_, _ ~). a But _ In _ In yD yD d ad - bc b c _ In _ In = d b = 1. Hence c a yD yD yD VD VD yD <7 E M. Furthermore, any element of M IS ob-. - - viously a Mobius transformation. Thus M is the set of all Mobius transformations. 3.49. Suppose ad - bc 7'= o. Prove <7(a, b, c, d) = L iff a = d and b = |
c = O. Solution: If <7(a, b, c, d) = L, then z = z<7(a, b, c, d). Hence o = O<7(a, b, c, d) = bid implies b = O. 1 = l<7(a, b, c, d) = 1 • ~ implies a = d. If a = d and c = b = 0, then, using the results of Problem 3.47, <7(a, b, c, d) = <7(a, 0, 0, a) = 00 = 00 <7(a, b, c, d) implies c = o. <7(1,0,0,1) = L. Sec.3.5I THE GROUP OF MOBIUS TRANSFORMATIONS 81 b. 2 X 2 matrices In this section we will define the group of two by two matrices and indicate its relation ship to the group of Mobius transformations. An array (3.8) of complex numbers a, b, c, d is called a two by two (2 X 2) matrix. (Since we will only deal with 2 X 2 matrices, we usually omit the adjective 2 x 2.) a, b, c and d are called the entries of matrix (3.8). Two matrices are equal if and only if their entries are the same, i.e. (~ ~) = (~: ~:) if and only if a = a', b = b', c = c' and d = d'. We define the product of two matrices as follows: (a c )(a' c') b' d' b d aa' + cb' ac' + Cd') ba' + db' bc' + dd' ( The product of two matrices is clearly a matrix. A calculation shows (Problem 3.51) matrix multiplication is an associative binary operation. The matrix I = ( 10 01) is the identity matrix, since (~ ~)(~ ~) = (a + 0 0 + c) b+O O+d In order to determine which matrices have inverses, we define the determinant of the matrix A = (~ ~) to be the complex number D(A) = ad - bc. It is easy to show D(A) D(B) = D(AB) for any two matrices A and B (see Problem |
3.52). If A = (~ ~) and D(A) # 0, then D(A) D(A) d -C) ( -b a D(A) D(A) is the inverse of A (see Problem 3.53(i)). If D(A) = 0, then A has no inverse (see Problem 3.53(ii) ). We claim that the set (M { ( ~ ~) I a, b, c, d complex numbers, ad - bc # O} with the operation of matrix multiplication is a group. For if A, B E (M, then, as D(A) D(B) = D(AB), D(AB) # 0 and AB E (M. The determinant of J = (~ ~) is 1. Hence the identity I is in (M. Furthermore if A E (M, then A-I E (M since D(A) D(A -1) = D(J) = 1 implies D(A -1) # O. Therefore (M is a group. We call (M the group of 2 X 2 mat rices over the complex numbers. The relationship between the group of Mobius transformations and the group of 2 x 2 matrices is now evident. For in Problem 3.46 we found 82 GROUPS AND SUBGROUPS [CHAP. 3 where a3 = al0-2 + b 2cl, b 3 = a 2b 1 + b 2d 1, C3 = alC2 + d 2cl, and da = b 1c2 + d 1d 2. But by the def inition of multiplication, This does not mean the group of 2 x 2 matrices is identical with the group of Mobius transformations. For we have seen in Problem 3.47 that u(a, b, c, d) = u(ka, kb, kc, kd) for any complex number k =F O. But (~ ~) =F (~~ ~~), e.g. if k = -1. The precise rela tionship between the two groups will be given in Problem 4.81, page 120. Problems 3.50. (i) Multiply (a)(_~ 1)(: -~), (b)(~ ~)G ~), (C)G ~)(: ~). (d)(: ~)(~ ~). (ii) Find the inverse of |
(a) G _~), (b) G ~), (c) (~ ~), (d) (~ ~). Solution: (i) () a 8i 20 - 2i ( -3 + 2i) 10 1 2C) 0 1 (b) ( (ii) (a) 14!3 3i ( 14 + 3i 14 : 3i) -7 14 + 3i (b) G ~) -i 0) 0 -i (c) ( d) (l/a -clad) ( lid 0 3.51. Show that matrix multiplication is an associative binary operation. Solution: Let A = (:~ ~: ), B = (:: ~: ), C = (:: ~:). Then (a 1a2 + c1b2)ca + (a1c2 + C1d2)da) a 1az + c1b2)aa + (alc2 + c1d2)ba (b1a2 + d 1b2)aa + (b1C2 + d 1d 2)ba (b1a2 + d1b2)ca + (b 1C2 + d 1d 2)da ( A(BC) = (a1(a2aa + C2ba) + cl(b2aa + d 2ba) al(a2Ca + C2da) + cl(b2Ca + d 2da») b1(a2aa + C2ba) + d 1(b 2aa + d 2ba) b1(a2ca + C.p3) + d 1(b 2Ca + d 2d a) A check of entries shows (AB)C = A(BC). 3.52. Prove D(A) D(B) = D(AB) = D(B) D(A), for any two matrices A and B. I Solution: (a' C') (a C) Let A = b d,B = b' d' aa' + cb' ac' + Cd') Then AB = ba' + db' bc' + dd' (. and r s".3.61, SYMMETRIES OF AN ALGEBRAIC STRUCTURE 83 D(A)D(B) (ad - bc)(a'd' - b'c') = aa'dd' + bb'cc' - |
adb'c' - bca'd' (aa' + cb')(bc' + dd') - (ba' + db')(ac' + cd') D(AB) Because multiplication of complex numbers is commutative, D(.4) D(B) = D(B) D(A). 3.53. Let A ab dC) be a matrix. ( Prove: (i) If D(A) ¥ 0, ( d -c ) D(A) D(A) is the inverse of A. -b - - - D(A) D(A) a (ii) If D(A) = 0, A has no inverse. Solution: (~ ~)(n7:) D~~)) (i) D(A) D(A) ad- bc D(A) ( bd-db D(A) -ac + ac) D(A) -bc + da D(A) (Dt~) D~~))(: ~) D(A) D(A) (ii) If A' is an inverse of A, then D(A') D(A) = D(l) where J is the identity matrix. But D(J) = 1 and D(A) = O. Since zero times any number is zero, A cannot have an inverse. 3.54. Show that the following sets of matrices are subgroups of the group eM of 2 X 2 matrices. (i) 'l{ = {( a b (ii) U {( a dC) I a, b, c, d real numbers, ad - bc ¥ 0 } b dC) I a, b, c, d complex numbers, ad - bc = I} (iii) cP = {( ~ ~) I a, d, c complex numbers, ad ¥ 0 } Solution: (i) (1 'l{C;eM and J = 0 ~)E'l{. If A = (; ~)E'l{ then, as D~A)' D~)' D~A)' D~) are all real numbers, A -1 E 'l{ and 'l{ is a subgroup of eM, as it is easy to check that 'l{ is closed with respect to products. (ii) 1J C; eM and D(l) = 1, implies D(A -1 |
) = 1. Hence U is a subgroup of ~~f, as it is easy to check that U is closed with respect to products. so J E U. Let A E U. Then D(A) D(A -1) = D(J) = 1 (iii) cP C; eM and J E CPo The inverse of A = (~ ~) is (l~a -::;d). Hence A -1 E cP if A E CPo Thus cP is a subgroup of oJIf, as it is easily checked that cP is closed with respect to products. 3.6 SYMMETRIES OF AN ALGEBRAIC STRUCTURE a. Automorphisms of groupoids We have discussed isometries of plane figures. The corresponding one-to-one onto mapping of a groupoid is defined as follows. Definition: Let G be a groupoid. Then an automorphism a of G is a one-to-one mapping of G onto G such that (ab)a = aaba for all a, bEG. 84 GROUPS AND SUBGROUPS [CHAP. 3 Note that isometries preserve distance whereas automorphisms of groupoids preserve groupoid multiplication. As the analog to Theorem 3.5, page 67, we have Theorem 3.13: The set A of all automorphisms of a groupoid G is a subgroup of SG, the symmetric group on G. Proof: I. t, the identity mapping, belongs to A; hence A =F ~. II. If a, f3 E A and a, bEG then (ab)(af3) = ((ab)a)f3 = [(aa)(ba)]f3 = [a(af3)][b(af3)] Thus the composition of mappings is a binary operation in A. III. The identity mapping is an automorphism, and so A contains an identity. IV. (A,') is associative. V. If a E A, let a- 1 be the inverse of a; since a E SG, a- 1 makes sense. Let a, bEG and choose a', b' E A so that a'a = a, b'a = b. Then (a'b')a = abo Hence (ab)a- 1 = a'b' = (aa-1)(ba- 1). Thus A |
is a group, and hence a subgroup of SG. We call A the automorphism group of the groupoid G and sometimes denote it by aut (G). Problems 3.55. Find the automorphism group of (G,o) where G = {a, b} and 0 is defined by the multiplication table (a) a b (b) b a :Effij :8±B Solution: (a) t, the identity mapping, is the only automorphism for the only other possibility is the mapping a defined by aa = band ba = a. But (bb)a = aa = band baba = aa = a; hence (bb)a =F (ba) • (ba). Thus a is not an automorphism. (b) Define a by aa = band ba = a. Note that xy = y. Hence (xy)a = ya = (xa)(ya). The auto morphism group therefore contains the two elements t and a. Notice aa:= t. 3.56. Find the automorphism group of A 3. Solution: (For table of A3 see Problem 3.23(iii).) Theorem 2.6, page 44, showed that for any homomorphism a of a groupoid G into a groupoid G', i.e. a mapping of G into G' such that (glg2)a = glag2a for all gl, g2 E G, the image of an identity in G is an identity in G' and the image of an inverse of g EGis an inverse of ga, i.e. 1a = l' (1 an identity of G and l' an identity of G') and if gh = 1 = hg, gaha = l' = haga. Now an automorphism of a group G is a one-to-one homomorphism of G onto G. Therefore if ta = t. Also ula is either U2 or Ul, as a is a one-to-one onto mapping. a is an automorphism of A 3, Hence there are at most two automorphisms of Aa. Let I be the identity mapping, i.e. tA = t, u1A = U2' U2A = Ul' On checking the homomorphism property, we see that I and A are automor phisms. Note that A2 = I. Thus |
the multiplication table is tl = t, u1I = Ul, u2I = U2' Let A be the mapping I A I~ A~ Sec. 3.6] SYMMETRIES OF AN ALGEBRAIC STRUCTURE 85 3.57. Let a be any element of a group G. Define the mapping Pa of G into G by Pa: g --. a - I gao Prove that Pa is an automorphism of G and that PaPb = Pab where PaPb is the usual multiplication of mappings. Solution: Pa is clearly a mapping. If glPa = g2Pa' then a-Igla = U-lg2a and hence gl = gz. There fore Pa is one-to-one. Also Pa is onto, for if g E G, g has as pre-image aga -I. and so Pa is a homomorphism. Hence Pa is an automorphism. Note that we have used the associative law, so that our argument would not apply to a groupoid which is not a semigroup. If g E G, and thus Pab = PaPb' 3.58. Find the automorphism group of S3' prove that there are no other automorphisms. This problem is difficult.) (Hint: Use Problem 3.57 to find six automorphisms. Then Solution: Refer to the multiplication table of S3 given in Section 3.3(a). By Problem 3.57, P, Pc; I'PT, I P T2'PTa are all automorphisms of Sa· We use the notation of Section 2.4(c), page 37, to denote the effect of these mappings. We use the multiplication table of Section 3.3(a) to calculate the images under the automorphisms., Pu 2, P, C UI U2 (T2 (TI 'TI 'T2 'T3 ) 'TI 'T2 'T3 PUI = (: (TI (TI (T2 'TI 'T2 'T3 ) (Tz 'T2 'T3 'TI PU2 P TI (: UI U2 (TI (T2 'TI 'T2 'T3 ) 'Ta 'TI 'T2 = C (TI (T2 (T2 'TI 'T2 'T3 ) (TI 'TI 'T3 'T |
2 P T2 P T3 C UI U2 (T2 (TI 'TI 'T2 'Ta) 'T3 'T2 'TI = C (TI (T2 (T2 'TI 'T2 'T3 ) (TI 'T2 'TI 'Ta If P were another automorphism, then (TIP (TIP· Now (TIP must be either (TI or (T2' for if say (TIP = 'TI' then contradicts P a one-to-one mapping, since 'P -,. Once (TIP is given, (T2P is known as 'P = L. Hence there are two possible choices for (TIP' (T2P = (TI(TI)P = (Tzp = (TIP(TIP = 'TI'TI =,; but this Now 'TIP must be one of 'TI' 'T2 or 'Ta for if for example, 'TIP = (TI, then Lp = ('TI'TI)P = (TI(TI = (T2 = L, a contradiction. Hence there are 3 possible choices for 'TIP. But once (TIP and 'TIP are known, the effect of P on all the elements of S3 is known, since So this means that there are at most six possible automorphisms. To find the multiplication table we use the result of Problem 3.57, that PaPb = Pab' and PL p, p, PUI PU2 PTI PT2 PT3 PUI PU2 p, PT2 PT3 PTI PU2 p, PUI PTa PTI PT2 PT I PT3 PT2 p, PU2 PUI PT2 PTI PT3 PUI p, PU2 PTa PT2 PTI PU2 PUI p, 86 GROUPS AND SUBGROUPS [CHAP. 3 h. Fields of complex numbers The complex numbers C have customarily two binary operations, addition and multi plication. Notice that if a, b E C, then a - b E C; and if b =1= 0, ab- 1 E C. We are often interested in a subset of C that satisfies the same conditions. This leads us to a field of complex numbers. Definition: A subset F of C is called a field of complex numbers if (a) 1 E F. (b) Whenever a, b E F, then also |
a - b E F. (c) Whenever a, b E F and b =1= 0, then ab- 1 E F. (The definition of field can be extended to sets which are not contained in the complex num bers. See, for example, Birkhoff and MacLane, A Survey of Modern Algebra, Macmillan, 1953.) Of course the complex numbers themselves form a field. Let F be a field. Recall that the set of complex numbers is a group under the usual binary operation of addition, denoted by (C, +), and that C* = C - {O} is a group (C*, x) under the usual multi plication of complex numbers (see Example 5, page 51). Therefore, using Lemma 3.1, page 55, for a subset of a group to be a subgroup, parts (a) and (b) of the definition of a field imply (F, +) is a subgroup of (C, +), and parts (a) and (c) imply (F*, x), F* = F - {O}, is a subgroup of (C*, x). In view of these remarks the definition of a field is equivalent to: Lemma 3.14: A subset F of C is a field of complex numbers if (1) (F, +) is a subgroup of (C, +), (2) (F*, x) is a subgroup of (C*, x) where F* = F - {O}, C* = C - {O}. Problems 3.59. Show that Rand Q are fields of complex numbers. Solution: 1 E R. applies for Q. If a, b E R, then a - b E Rand ab -1 E R whenever b # O. The same argument 3.60. Which of the following sets are fields? F = {a + by'2 I a, b E Q} (i) (ii) F == {a + bi I a, bE Q}, i = R (iii) F == {a + bi I a, bE Z}, i == R Solution: (i) 1 == 1 + 0y'2 E F. Let a + by'2 and a' + b'y'2 be two elements in F. (a + by'2) - (a' + b'y'2) = (a - a') + (b - b') |
y'2 E F and if a' + b'y'2 "# 0, (a + by'2)(a' + b'y'2)-l aa' - 2bb' a,2 _ 2b,2 + a,2 _ 2b,2 y'2 E F a'b - ab' Therefore F is a field. (ii) F is a field. 1 + Oi == 1 E F. (a + bi) - (a' + b'i) == (a - a') + (b - b')i E F and, if a' + b'i # 0, (a + bi)(a' + b'i)-l == aa' + bb' a' + b' 2 2+ a'b - ab' a' + b' 2 2i E F (iii) F is not a field, since 1 + i "# 0, 1 + i E F but (1 + i)-l == 1/2 - 1/2i El F. Sec. 3.6] SYMMETRIES OF AN ALGEBRAIC STRUCTURE 8'i' c. Automorphisms of fields We have discussed isometries of the plane and automorphisms of groupoids. The cor responding one-to-one onto mapping of a field is defined as follows. Definition: A one-to-one mapping a: of a field F onto itself is termed an automot'phism if (i) (ii) (ab)a = (aa)(ba) for all a, b E F. (a + b)ll' = all' + bll' for all a, b E F. Note that the automorphisms of fields preserve both the operations of addition and multiplication. Theorem 3.15: The set A of automorphisms of a field F forms a subgroup of the sym metric group SF. Proof: We must prove I. A 7'=~; this is true since the identity mapping tEA. II. If a, (3 E A, then (a + b)(a(3) «a + b)a)(3 = (all' + ba)(3 = (aa)(3 + (ba)(3 = a(a(3) + b(a(3) «ab)a)(3 = [(aa)(ba)](3 = [(aa)(3][(ba)(3] = |
[a(a(3)][b(a(3)] and for all a, b E F. Thus composition of mappings is a binary operation in A. (ab)(a(3) III. The identity mapping is in A and is an identity element. IV. (A,') is a semigroup, since composition of mappings is associative. V. If a E A, then a E SF the symmetric group on F. Let 0'-1 be the inverse of a. We claim 0'-1 E A and so a will have an inverse in A as desired. Let a, b E F. Then as a is onto, we can find a', b' E F such that a = a'a, b = b'a. Then ab = (a'b')a and a + b = (a' + b')a. Consequently (ab)a- 1 = a'b' = (aa- 1)(ba- 1) and (a + b)a- 1 = a' + b' = aa- 1 + ba- 1• Thus 0'-1 E A as desired. We have proved that the automorphisms of a field form a group. This group is ex tremely useful. For additional pertinent remarks and references, see Section 5.4a, page 158. Problems 3.61. Find the automorphism group of Q. Solution: We will use the fact that (Q, +) is a group and (Q*, x), Q* = Q - {O}, is a group. Notice that (Q,X) is a groupoid (not a group, since 0 has no inverse) and, because of part (ii) of the definition, the automorphism of the field Q is also an epimorphism (see Section 2.5b, page 42) of groupoid (Q, X) onto (Q, X). Hence by Theorem 2.6, page 44, 1, the multiplicative identity of (Q, x), is mapped onto 1 by any automorohism of Q. Let a be an auotmorphism of Q; then la = 1. Using mathematical induction we show na = n (k + l)a = for all positive la = 1. Assume ka = k for all positive integers n. ka + la = k + 1, by the automorphism property of a. We conclude that na = n integers n. |
for some integer k "'" 1. Then Now (Q, +) is a group and, by definition, any automorphism of Q is an epimorphism of the group (Q, +). Hence by Theorem 2.6, inverses are mapped onto inverses and the identity, 0, of (Q, +) is mapped onto O. Therefore (-n)a = -n for all positive integers n, since na = nand -n is the additive inverse of n. Furthermore, Oa = O. Hence ra = r for all integers. But the automorphism a is also an epimorphism of the group (Q*, X) onto itself so that (±r)-la = 1.. a = 1.... for all positive integers r, because l is the inverse of r. Collecting these facts we see that if m (n "" 0) is any element in Q, then m a r= (m.!') a = mala = m' 1:.. = m. Therefore a is the i~entity mapping and is the only possible automorphism of Q. The automorphism group of Q is of order one. ±r ±r n n n n n 88 3.62. GROUPS AND SUBGROUPS [CHAP. 3 Find the automorphism group of F = {a + bV2 I a, b E Q}. Solution: Any rational number q is an element of F, since q + OV2 = q. If a is an automorphism of F, then for any q E Q, qa = q arguing as in Problem 3.61. Now V2 E F and (V2 V2)a = 2a = 2, since 2 is an element of Q. But V2 a V2 a = (V2 V2)a = 2a = 2, so that (V2 a )2 = 2 or V2 a = ±V2. We conclude that V2 has only two possible images under an automorphism of F. Hence (a + bV2)a = aa + (bV2)a = aa + baV2 a = a + b(V2 a). There are two possibilities: (1) (a + bV2)a = a + bV2, in which case a is the identity automorphism,; (2) (a + bV2) |
a = a + b(-V2), in which case we must check to see whether a is an automorphism. If a: a + bV2 -"> a - bV2, then {(a + bV2 )(a' + b'V2)}a (a + bV2 )a(a' + b'V2)a (a'b + b'a)V2 {aa' + 2bb' + (a'b + b'a)V2}a aa' + 2bb' (a - bV2 )(a' - b'V2) aa' + 2bb' - (a'b + b'a)V2 {(a + bV2)(a' + b'V2)}a (a + bV2)a(a' + b'V2)a and Hence Also, {(a + bV2) + (a' + b'V2)}a (a + bV2)a + (a' + b'V2)a {(a + a') + (b + b')V2}a a + a' - (b + b')V2 (a-bV2) + (a'-b'V2) (a + a') - (b + b')V2 {(a+bV2) + (a'+b'V2)}a (a + bV2)a + (a' + b'V2)a and Hence is an automorphism of F. The automorphism group of F has two elements Thus a a: a + bV2 -"> a - bV2. Notice aa =,., and d. Vector spaces In Physics we represent a force x by a straight line pointing in the direction the force is acting and of length proportional to the magnitude of the force. We shall assume for the moment that all the forces act on a fixed point 0 and act in the Euclidean plane E. It is then possible to represent a force by its endpoint, as we know it begins at O. Any point of course can be represented by its coordinates, so a force can be represented by the coordinates of its endpoint. We can talk of increasing the force x in magnitude by a factor 3, say. The resultant force is written as 3x. If x = (h 12), then 3x = ( |
3h 3/2). Similarly if I is any real num ber, we define Ix to be the force x increased in magnitude by a factor I and we can prove Ix = (fh 112). The sum of two forces x = (h 12) and y = (gl, g2) is a third force z computed by the parallelogram law. Again it can be shown that z = (/1 + gl, 12 + g2). We write z = x + y. The set of all 2-tuples (h 12) is called a vector space of dimension 2 over the field of real numbers (because we can multiply the 2-tuples by real numbers). We shall generalize the concept of two dimensional physical forces in two ways: (i) We shall deal with arbitrary dimensions and not only 2 or 3. (We must therefore relinquish our contact with the real world.) (ii) We shall consider vectors that involve fields other than the real numbers. Let F be any field. Let V = Fn be the cartesian product of n copies of F. Then V consists of the n-tuples (h fz,..., In) where Ii E F. If x = (h..., In) and y = (gl,..., gn) are two elements of V, and I E F, we define p.: V X V ~ V (i.e. p. is a binary operation in V) and w: F x V ~ V by Sec. 3.6] SYMMETRIES OF AN ALGEBRAIC STRUCTURE 89 (x, y)p. (f, x)w (f1+u1,h+u2,...,fn+Un) (f/1...., ffn) We denote (x, y)p. by x + y, and (f, x)w by fx. V together with p. and w is called the vector space of dimension n over the field F. The elements of V are called vectors. Problems 3.63. Find (i) (1,2,3) + (6,7,8), (ii) 4(6, -2, 0, 3). (1+6,2+7,3+8) = (7,9,11) Solution: (i) (ii) (4·6, 4· (-2), 4·0, 4·3) = |
(24, -8, 0, 12) 3.64. Prove that if x, y and z are elements of a vector space of dimension n, then x + y = y + x and (x + y) + z = x + (y + z). Solution: If x = (/v"''/n)'y = (U1""'U n) and z = (h 1,...,hn), then x+y = y+x (11 + Uv..., In + Un) as addition is commutative in any field. (x + y) + z = ((11 + U1) + hI'..., (In + Un) + hn) = X + (y + z) by associativity of addition. 3.65. Prove that if V is a vector space of dimension n, then the elements of V form an abelian group under the operation p.. Solution: V is an abelian groupoid by the preceding problem. (0,0,...,0) is the identity element. The inverse of (fI,/2,··.,In) is (-/1,-/2"..,-In)' 3.66. Prove that if e1 = (1,0,...,0), e2 = (0,1,0,...,0),..., en = (0,0,...,1), of V can be represented uniquely in the form then every element x x = 11e1 + 12e2 +... + Inen Solution: Suppose x = (Iv... '/n). Then indeed x = 11e1 + 12e2+'" + Inen. If x = U1 e1 + U2e2 +.,. +Unen, In = Un and the representation is unique. then (Iv·· ·,In) = (U1,...,Un)· Hence 11 = U1, 12 = U2,..., e. Linear transformations. The full linear group Let V be a vector space and a: V ~ V. Then a is said to be a linear transformation of V if (i) (x + y)a = Xa + ya (fx)a For example, let (/1. h)a = (12, f1)' Then (ii) f(xa), for all x, y E V and f E F {(/1. f2) + ( |
U1, U2)}a = (12 + U2, f1 + U1) = (f2, f1) + (U2, U1) = (/1. h)a + (U1, U2)a Also, Note that linear transformations preserve both the additive and the multiplicative structures of V. Now we have the analog of Theorems 3.13 and 3.15. First let us define Ln(V, F) to con sist of all one-to-one linear transformations of V, the vector space of dimension n over F. Ln(V,F) C. Sv, the symmetric group of V, clearly. 90 GROUPS AND SUBGROUPS [CHAP. 8 Theorem 3.16: Ln(V,F) is a subgroup of Sv. Proof: L E Ln(V, F) as L preserves both addition and multiplication. Hence Ln(V, F) =1= ~. If a E Ln (V, F), we ask whether a-I E Ln (V, F). a-I is one-to-one onto. Is it a linear transformation? Let x, y E V and t E F. Since a is onto, there exists Xl and Y l such that (Xl + Yl)a = Xla + Yla = X + y. Xla = X and Yla = y. Of course Xl = Xa-t, Y l = Ya- l ; and Hence (Xl + Y l) = (Xl + yl)aa- l = (X + Y)a- l and so Xa- l + Ya- l = (x + y)a- l. Also, (tXl)a = t(Xla) = tx; so ((txl)a)a-l = tXl = (tX)a- l, i.e. t(Xa- l) = (tX)a- l. Accordingly a-I E Ln(V,F). Thus if a,[3 E Ln(V, F), [3-1 E Ln(V, F) and we ask whether a[3-1 E Ln (V, F). We have (X + y)a[3-1 ((X + y)a)[3-1 = (Xa + |
Ya)[3-1 (Xa)[3-1 + (ya)[3-1 = X(a[3-1) + y(a[3-1) and and thus Ln (V, F) is a subgroup of Sv' Ln (V, F) is called the full linear group of dimension n. Problems 3.67. Show that if a is a linear transformation of V, a vector space of dimension n, then the effect of a is uniquely determined by its effect on the elements el>..., en of Problem 3.66. Solution: By Problem 3.66 each element of V is of the form x = 11 el +... + In en' Then Xa = 11 (ela) +... + In (ena). Hence the effect of a is known once its effect on the elements el,"" en is known. 3.68. Show that if a is any mapping of {el"'" en} ~ V, a: V ~ V such that eja = eja, j = 1,..., n. then there exists a linear transformation Solution: Each element of V is uniquely of the form Ilel +... + Inen. Define a: V ~ V by (flel +... + Inen) a = Il(ela) +... + In(ena) Then a is a linear transformation, since (fl + Ul)(ela) +... + (fn + Un}(ena) (flel +... + Inen) a + (Ulel +... + Un en) a (ffl}(ela) +... + (ffn)(ena) [{(fl el +... + Inen)a} if a is a linear transformation and ela = ez, e2a = e3"'" en-la = en and and 3.69. Is a E Ln(V,F) ena = el? Solution: Yes. All we must prove is that a is one-to-one and onto. An arbitrary element 11 el +... + In en has 12el + 13e2 +... + Inen-l + 11 en as a pre-image. Also, implies 11=Ul,/2=U2'...,In=Un by Problem 3.66. Hence a is one-to-one. Thus aELn(V,F). CHAP. 3] SUPPLEMENTARY PROBLEMS 91 A |
look back at Chapter 3 We have met many important groups, including groups of real and complex numbers, the symmetric group Sn, symmetry groups, the dihedral groups, the automorphism groups of groupoids and fields, and the full linear group. Groups thus arise in many different branches of mathematics, and hence general theo rems about groups can be useful in apparently unrelated topics. In subsequent chapters we will derive general theorems for groups. Supplementary Problems GROUPS 3.70. 3.71. Let n be any positive integer and let Gn = {a + bVn I a, bE Z} where Z is the set of integers. Prove that with respect to addition Gn is a group. When does Gn = Z? Let n be any positive integer. Let Gn = {a+ibVn I a,b E Z} where i = v'=i and Z is the set of integers. Is G n a group with respect to addition? Is G n a group with respect to multiplication of complex numbers? 3.72. Let D = Z X Z, Z the set of integers. Define a group with respect to this operation o. (a, b) 0 (c, d) (a+c,(-l)cb+d). Prove that Dis 3.73. Prove that the group D of Problem 3.72 is not abelian. 3.74. Let G = Z X Q, where Z is the set of integers and Q the set of rationals. Define (a, b) * (c, d) = (a + c, 2cb + d). Prove G is a group with respect to this operation *. 3.75. Is the group of Problem 3.74 abelian? 3.76. 3.77. If we define Is G a group with respect to the operation· defined by (a, b) • (c, d) = (a + c, 2cb - d)? is G (of Problem 3.74) a group with respect to o? (a, b) 0 (c, d) = (a + c, 2- cb + d), Let B = {o I 0: Z ~ Z}. Let W = Z X B. We define a multiplication on W by (m + n, >/;) where for each z E Z, z>/; = (z - n)o + z</>. Pro |
ve that W is a group. (Hard.) (m, o)(n, </» = SUBGROUPS 3.78. Let G be a group and G1 C; G2 C; ••• be subgroups of G. Show that G1 uG2 u··· is a subgroup of G. Find a group G and two subgroups G1 and G2 of G such that G1 u G2 is not a subgroup of G. 3.79. Let Gv G2, ••• be subgroups of G. Prove G1 nG2 n··· is a subgroup of G. 3.80. 3.81. 3.82. Let G be an abelian group. Let H be a subgroup of G. Let S(H) = {x I x E G and xx E H}. Prove that S(H) is a subgroup of G. Let D be the group of Problem 3.72. Determine whether H = {(a,O) I aEZ} and K = {(O,a) I aEZ} are subgroups of D. Let G be the group of Problem 3.74. Determine whether H = {(a,O) I aEZ} and K = {(O,q) I q E Q} are subgroups of G. 92 3.83. GROUPS AND SUBGROUPS [CHAP. 3 Let B be as in Problem 3.77. Let C = {o I 0: Z --> Z, zo = z for all but a finite number of integers z}. Let B' = {x I x = (0, b), bE B}, C' {x I x = (0, c), c E C}. Prove that B' is a subgroup of Wand C' is a subgroup of B'. (Hard.) 3.84. Using the notation of the preceding problem, let W = {x I x = (m, c), where mE Z and c E C}. Prove W is a subgroup of W. (Hard.) SYMMETRIC GROUPS AND ALTERNATING GROUPS 3.85. 3.86. 3.87. 3.88. Let a: Z --> Z be defined by z", = z + 1 for all z E Z. Let (3: Z --> Z be defined by 2n(3 = |
2n, (2n + 1)(3 = 2n + 3 for all integers n. Let y: Z --> Z be defined by 2ny = 2(n + 1), (2n + 1)y = 2n + 1 for all integers n. Prove that a, (3, y E 8 z and show that a", = (3y = y(3. Let G = 8 p, where P is the set of positive integers. Let Sp = {o I 0 E 8 p and zo = z for all but a finite number of z E P}. Prove that 8 p is a subgroup of 8 p • A A Let G = 8 p • Let Gn = {o I 0 E 8 p G = G1 UG2 U···. and zO = z for all z E P such that z > n}. Prove that Let H = {o I 0 E 8 5, 10 = I}. Prove that H K = {o I 0 E 8 5, 10 = 1 or 10 = 2}. Prove that K is not a subgroup of 8 5, is a subgroup of 8 5, What is its order? Let 3.89. Let nand r be positive integers. Let H = {o I io E {I, 2,..., r} for all i E {I, 2,..., r} and 0 E 8 n } 3.90. Prove that H is a subgroup of 8 n and find IHI. Let X be a set and Y a proper subset of X. Let H = {o I 0 E 8 x and yo = y for all y E Y}. Let K = {o I 0 E 8 x and yo E Y for all y E Y}. Prove that Hand K are subgroups of 8 x and that K"dH. Prove that if IYI "'" 2, H =1= K. 3.91. Let H = {o I 0 E A 5, 10 = I}. Prove that H is a subgroup of A5 and find its order. 3.92. Let A, B be sets with IA I = 1. Prove that 8 A x B =0 8 B' (Hard.) 3.93. Prove that if IXI = IYI, 8 x =0 8 y. GROUPS OF ISOMETRIES 3.94. Let 8 be the even integers and ( |
I: 8) subgroup of I(R). {o I 0 E I(R), 80 E 8 for all 8 E 8}. Prove (I: 8) is a 3.95. Let (I: Q) = {o I 0 E I(R) and qo E Q for all q E Q}. Prove (I: Q) is a subgroup of I(R). 3.96. Find the symmetry group of the figure W. 3.97. Find the symmetry group of the figure 8. 3.98. What is the symmetry group of the graph of y = sin x? 3.99. Determine the symmetry group of the circle. 3.100. Prove that if 8 is any subspace of the plane and 8' is a congruent figure, i.e. there is an isometry o such that 80 = 8', then Is =0 Is.. (Hard.) THE GROUP OF MOBIUS TRANSFORMATIONS 3.101. Prove that if M is the group of Mobius transformations, then the only element m E M for which m:n = nm for all n E M is m = t. 3.102. Let u(a, b, c, d) be the Mobius transformation defined by u(a, b, e, d) : z --> cz + d' Prove that if and only if either a = a', b = b', e = c', d = d' or a = -a', b = -b', u(a, b, e, d) = u(a', b', e', d') e = -e', d = -d', given ad - be = a'd' - b'e' = 1. (Hard.) az+ b CHAP. 3] SUPPLEMENTARY PROBLEMS 93 3.103. Let N be the set of Mobius transformations u(a, b, e, d) with b = 0. Prove that N is a subgroup of M, the group of all Mobius transformations. 3.104. Prove that if mEN (N defined as in Problem 3.103), then there exists sEN such that ss = m. 3.105. Let 'U be the set of all matrices (: :), where a, b, e, d are integers such that ad - be = 1. Find the set of all matrices such that SYMMETRI |
ES OF AN ALGEBRAIC STRUCTURE 3.106. Prove that the automorphism group of a finite group is finite. 3.107. Find a finite groupoid G with IGI > 2, whose automorphism groupoid is of order 1. 3.108. Let G be a non-abelian group. Prove that the automorphism group of G is not of order 1. 3.109. Prove that the subset K of the symmetric group 8 4 defined by K = {G ~ : :), G ~! :), G 2 3 I 4 is a subgroup of 8 4, Find the automorphism group of K. (Hard.) 3.110. Let F = {a + ibV171 a, b rational numbers, where i = H}. Verify that F is a field under the usual operations of addition and multiplication of complex numbers. Determine the automorphism group of F. 3.111. Let V be the vector space over the rationals of dimension n + 1. Let 8 = {al aELn+dV,F), (I,O,...,O)a=(I,O,...,O)} Prove that 8 is a subgroup of Ln+ dV, F) and that 8 =0 Ln(V, F). Chapter 4 Isomorphism Theorems Preview of Chapter 4 We say that two groups are isomorphic if they are isomorphic groupoids. Here we shall prove three theorems which provide a means of determining whether two groups are isomorphic. The main concepts that arise are those of subgroups generated by a set, cosets, and normal subgroups. We find a structure theorem for cyclic groups. The contents of this chapter are indispensable for any further understanding of group theory. 4.1 FUNDAMENT ALS a. Preliminary remarks We begin by reminding the reader of our previous results. A group is a semigroup in which every element has an inverse. Consequently we have the following. (1) The identity is unique. (2) The inverse of an element is unique (Theorem 2.2, page 33), and if G is any group and (Theorem 2.1, page 31.) g, hE G, then (gh)-l = h-lg- l. (3) The product of n elements al,..., an, in that order, is independent of the bracketing (Theorem 2 |
.5, page 39). (4) Homomorphisms, monomorphisms and isomorphisms for groups are defined as they are for groupoids. (Section 2.5, page 40.) (5) If G is a group, and 6 any homomorphism of G into a groupoid, then G6 = {x I x = g6, g E G} is a group. For by Theorem 2.6, page 44, G6 has an identity, is associative, and each element has an inverse. Note that 6 maps the identity of G to the identity of G6 and that (g-l)6 = (g6)-l for each g in G. (6) If 6: G ~ K is a homomorphism from the group G to the group K, and if H is a subgroup of G, then H6 is a subgroup of K. For 61H is a homomorphism of H into K and, by (5), H6 is a group. (7) Isomorphic groups are roughly the same except for the names of their elements. (See Section 2.5d, page 45.) The following theorem is useful. Theorem 4.1: If a and b are two elements of a group G, then there exist unique elements x and Y such that ax = band ya = b. Proof: We consider first the solution of the equation ax = b. then a(a-lb) = (aa-l)b = b. Hence the equation ax = b has a solution. If we put x = a-lb, Suppose aXl = band aX2 = b; then aXl = aX2. Multiplying both sides of the equation on the left by a-I, we have a-l(axl) = a- l (ax2), (a-la)xl = (a- l a)x2 or Xl = X2 The argument for solving ya = b is similar; in fact y = ba- l is a solution. Also, if Yla = band Y2a = b, then Yla = Y2a. Multiplying both sides by a-l on the right, we get (Yla)a- l = Yl = (Y2a)a- l = Y2 94 Sec. 4.1] FUNDAMENTALS 95 Problems U. Prove that the groups given by the following |
multiplication tables are isomorphic. 1 -1 G: -:rn o 1 H: O~ 1 liE] Solution: Let 8: G.... H be defined by 18 = 0, -18 = 1; then 8 is a one-to-one onto mapping. If it is also a homomorphism it will be an isomorphism. We must check that (glg2)8 = g18g28 for all pos sible choices of gl and g2 in G, i.e. we must check (1, 1)8 = 1818 (i) (ii) (-1'1)8 = (-18)(19) (iii) (-1·-1)8 = (-18)(-18) (iv) (1' -1)8 = (18)' (-18) (i) to (iv) hold. Hence (i) holds.) Therefore G == H. (Thus for (i): 1· 1 = 1 by the multiplication table. 18 = O. 18' 18 = 0 • 0 = O. Of course 8 had to be some mapping of G to H. How did we know which was the right mapping to choose? Examining the multiplication table for G, it is obvious that 1 is the identity for G. o is the identity for H. We remarked that any homomorphism must map an identity to an identity. Thus the choice for 8 was quite clear. 4.2. Prove that 8 2, the symmetric group of degree 2, is isomorphic to G, where G is the group of Problem 3.5, page 53, with m = 2. Prove 8 3 == Da the dihedral group of degree 3, i.e. the symmetry group of the equilateral triangle. (Difficult.) Solution: The multiplication table for G is o 1 :8iE while the multiplication table for 8 2 is (Problem 3.20, page 58): {3.c=0 {3~ Let 8: 8 2.... G be defined by As it is one-to-one and onto, 8 2 == G..8 = 0, {38 = 1. Then it can be checked that 8 is a homomorphism. The multiplication table for 8 3 is on page 57, that of Da is in Problem 3.40, page 76. As we have used the same Greek symbols for 8 a and Da, we face the risk of not knowing whether u, for example, refers to an |
element of 8 a or to an element of Da. To avoid such ambiguities, we will rename the elements of D a, replacing a u by an 8 and a 'T by a t. The multiplication table then becomes 81 82 8a t t t82 Ua 82 8a 81 t8a t t82 83 81 82 t82 t8a t t t82 t8a 81 82 8a t82 t8a t 8a 81 82 t8a t t82 82 83 81 96 ISOMORPHISM THEOREMS [CHAP. 4 j == 1,2,3, satisfies TjTj t. If (J is an isomorphism from Sa to Da, then Note that an element Tj, Tj(J Tj(J == (TjT)(J == t(J == 81' So (J can only map the Tj, j == 1,2,3, among the elements t, t82, t8a since these are the only non-identity elements of Da which have the property that their squares are 81' As (J must map t to 81' it maps 0"1> 0"2 onto the elements 82,8a. If we know the effect of (J on 0"1> since 0"10"1 == 0"2' we know the effect of (J on 0"2' Also if we know the effect of (J on Tl, then, because T2 == T10"2 and Ta == T10"1' we know the effect of (J on all the elements of Sa. So we have to experiment. A suitable mapping (J: Sa --> Da must satisfy 0"1(J == 82. or 8a while Tl (J == t, t82 or t8a. We try the following definition. Let t(J == 81> 0"1(J == 82 and Tl(J == t. Then we must have 0"2(J == 83, T2(J == t8 3, Ta(J == t82 if (J is to be an isomorphism. To check whether this mapping is an isomorphism, we must check whether this mapping is a homomorphism. As a mechanical procedure of doing this we use the following table. The entry in the second row and third column, for example, is calculated as follows: In the bottom corner we place 0"1(J· 0" |
2(J. is a homomorphism, (0"1(J)(0"2(J) == (0"10"2)(J. Hence these two entries should be the same in each square of the table. Checking through this table, we see that the entries in each square are equal. Hence (J is a homo morphism and as it is one-to-one onto, (J is an isomorphism. In the top corner we place (0"10"2)(J. If (J 4.3. Prove that if (J: F --> G and <f>: G --> F are two homomorphisms such that (J<f> == identity mapping on F and <f>(J == identity mapping on G, then (J and <f> are isomorphisms of F onto G and of G onto F respectively. Solution: (J is one-to-one, for if x(J == y(J, then x(J<f> == y(J<f>. But (J<f> is the identity on F. Hence x == y. Similarly <f> is one-to-one. Next let U E G; then U<f> E F. U¢(J == U; hence U is the image of an element of F under (J and so (J is an onto mapping. Thus (J is an isomorphism. Similarly <f> is an isomorphism. 4.4. Prove that if Ul' U2, Ua are elements of a group G, then the equation UlxU2 == Ua has a unique solution. Solution: If we put x == U;lUaU;l, we find U1XU2 == Ua. If U1X1U2 == Ulx2U2 == U3, then on multiplying by U;l on the left and U;l on the right we have U;1(U1X1U2)U;1 == U;1(U1X2U2)U;1 or Xl == X2' Sec. 4.1] FUNDAMENT ALS 97 4.S. Prove that if G is a finite group and H is an infinite group, then G and H are not isomorphic. Solution: If G == H, there is a one-to-one mapping from G onto H. But this is not possible since G |
is finite and H is infinite. 4.6. Prove that Sn == Sm if and only if n = m. Solution: Sn has order n! and Sm has order m! Now if Sn == Sm' then there is a one-to-one mapping of Sn onto Sm. So Sn and Sm have the same order, i.e. n! = m!, and this implies n = m. On the other hand every group G is isomorphic to itself. In fact the identity mapping of G onto G is an iso morphism. Hence n = m implies Sn == Sm. 4.7. Prove that if G == H, then H == G. Solution: Let (J be an isomorphism from G onto H. Then (J-1 is an isomorphism from H to G, and so H == G. (See Problem 2.38, page 42.) 4.8. Prove that if G == Hand H == K, then G == K. Solution: Suppose (J is an isomorphism from G to Hand ¢ an isomorphism from H to K. Then (J¢ is an isomorphism from G to K, i.e. G == K. (See Problem 2.38, page 42.) 4.9. Prove that there are infinitely many groups, no two of which are isomorphic. Solution: Consider the symmetric groups S1> S2'.... Then by Problem 4.6, no pair of these groups is isomorphic. 4.10. Prove that if G is a finite group and H is a subgroup of G, H ¥- G, then G and H are not isomorphic. Solution: We observed in the solution of Problem 4.6 that if two finite groups are isomorphic, they have the same order. Since the order of H is less than that of G, it follows that G and H are not isomorphic. b. More about subgroups Let G = 8 4 and let X = {l1s' T 7 } where I1s = (~ 2 3 ~) 1 4 and -e 2 3!) 2 3 1 T7 - (See Problem 3.21, page 59.) Suppose we wish to refer to a product such as I1ST7 or T711SI1S or I1ST71181T;lT;11181. It will be convenient to have some general notation. We will write X'l • |
•• X'n n'where £i = ±1, Xi E X 1 to represent the product of n elements chosen from X or the set of inverses of the elements 1 will mean the inverse of Xi' For example, if £1 = £2 = 1, of X, where xI will mean Xi' and X i£5 = £6 = -, n ta d th en Xl X 2 Xa X4 X5 X6 S n s I1s' x 2 - X 4 - X5 - 1 a d X xa - X6 - T7, 1 - 'a '5 '4 '6 '2 '1 - - - £4 - £a for I1ST71181T;lT;11181. Example 1: If g = X;l... x:n, then g-l = h where h = x;;-'n... X;'l. Proof: gh = x'l •.• x'n X -'n •.. X -'1 1 n n 1 Similarly hg = 1. 98 ISOMORPHISM THEOREMS [CHAP. 4 We proved in Lemma 3.1, page 55, that if H is a subgroup of G and xl' x 2 E H, then XIX;1 E H. We now generalize this and prove Lemma 4.2: If H is a subgroup of G and Xc H, then H d {X~l... x:n I Xi EX, fi = ±1, n a positive integer} Proof: Recall that as H is a group, L E H and x, y E H implies y-l E H, xy-l E H, and xy E H. We prove the lemma by induction on n. Let n = 1. Then xII E H since Xl E H. Hence X~l E H. Assume, by induction, that X = X~l... x:n E H for n = k. Let Xk+1 EX. Since x, X~k// E H where f k+1 = ±1, XX'k+l = X'l... X'k+l E H Hence X'l... x'n E H for all n. This proves the lemma. Now if X is "large enough", e.g. X = H, we may have k+l k+ |
l n I I H = {X;l... x:n I Xi EX, fi = ±1, n a positive integer} We ask what happens if X is not "large enough", i.e. if X is a subset of Hand S = {X;l... x:n I Xi E X, fi = ±1, n a positive integer} is S a subgroup of G? Lemma 4.3: Let G be a group and let X be a non-empty subset of G. Let S = {X;l... <n I Xi EX, fi = ±1, n a positive integer} Then S is a subgroup of G. If H is any subgroup containing X, H d S. Proof: We must prove that: (i) S oF Y'>; this is true because there exists Xl E X as X is non-empty. (ii) If f, g E S, then fg-l E S (Lemma 3.1, page 55). f,g E S means f = X;l... x:n (fi = ±1) and g = y~l... y:m (7]i = ±1) where Xi and Yi are elements of X. Hence g-l = y;;'T/m... y;T/l and fg - l = X'l... x'n y-T/m... y-T/I = X'l... x'n x'n+l... x'n+m n+m n+ I m n n I I I where xn+ l = Ym'..., xn+m = YI and fn+l = -7]m'..., fn+m = -7]1. Therefore fg-l E Sand S is a subgroup of G. If H d X, we use the previous lemma to conclude H d S. We denote S by gp(X) and call S the subgroup generated by X. If a group can be generated by a finite set, we call it a finitely generated group. Example 2: What is gp({l}) in the group of Problem 3.5, page 53, where m = 3? Recall that the multiplication table is = gp({l}) = {X;l... x:n I Xi E {1}, Ei = ±1, n a positive integer} Now 1 E Sand 2 |
E S. Also, 1· 2 = 0 E S. Hence gp({l}) is the whole group. We remind the reader that, for example, in the multiplicative group of nonzero rationals the inverse of a, which we have denoted in this section by a-I, is 1/a, i.e. in this case a - I has the meaning usually associated with it when a is a number. But in the additive group of rationals a-I is -a. Sec. 4.1] Problems FUNDAMENT ALS 99 4.11. Let G = Z, the additive group of integers. What is gp({l})? Solution: gp( {I}) d 1 + 1 + 1 +... + 1 (n"" 1). Hence gp( {I}) contains all positive integers. v n times gp({I}) contains 1-1 + 1-1 +.., + 1-1 = -1 + (-1) +... + (-1) (n"" 1). Hence gp({I}) con- tains all negative integers. \ Y n times I \ V n times J Also, gp({I}) contains 1'1- 1 = 1 + (-1) = 0. Thus gp({I}) = Z. 4.12. Let G = Q Solution: the additive group of rationals. Find gp({I}). Exactly as in the last problem, gp({I}) = Z. Since no other elements can arise as sums or differences of 1, gp( {I}) =F Q. 4.13. Find the subgroup of the multiplicative group of rationals generated by {2}. Solution: 2- 1 = l The elements of gp({2}) are either of the form 2n or 2- n, n a positive integer. 4.14. Determine the subgroup H of 8 3 generated by 0'1 and 7"1 of Section 3.3a, page 57. Solution: We use the multiplication table for 8 3 shown in page 57. 0';-1 = 0'2; hence H contains 0'2' As H contains 7"1' it contains 7"3 = 7"10'1 and 7"2 = 7"10'2' Thus H contains all the elements of 8 3, and so H = 8 3, 4.15. Determine the subgroup of the symmetry group of a square generated by those isometries that leave two vertices fixed |
. (Hard.) (Hint: To see what is happening, cut out a square from a piece of card board and label the four vertices. Perform the isometries on the figure.) Solution: We refer to Problem 3.39(ii), page 73. 81 leaves G and I fixed; 82 leaves Hand J fixed; 85 leaves all vertices fixed. Hence we require 8 = gp({85, 82' 81}), and this must contain 87, since 8182 = 87' It is easy to prove 8182 = 87, for the effect of 87 and 8182 is the same on three points not on a single straight line and this is sufficient by Lemma 3.7, page 71. Note that the inverses of 82,81,87 are 82,81,87 respectively. Let T = {85' 87' 82, 81}' We assert that T is a subgroup. All we must check is that t 1, t2 E T implies t- 1 = t, all we must check is that tlt2 E T for t 1, t2 E T. implies tIt;;I E T. Since t E T tlt2 = 85 E T. Therefore if 85 is either tl or t 2. If tl = t 2, then As 85 is th~ identity, tlt2 E T 8182 = 8281 = 87 E T; 8187 = 8781 = 82 E T. Finally, we need only consider the following cases: 8287 = 8782 = 81 E T. Then T is a subgroup of the symmetry group of the square. But 8 d T, and T d {85' 82' 81}' Hence T d gp(85' 82, 81) = 8, by Lemma 4.3. Thus T = 8. 4.16. Find the subgroup of M, the group of Mobius transformations (see Section 3.5a, page 78), generated by 'I: z ~ -z (z =F "'), (In the notation of Section 3.5a, 'I = 0'(-1,0,0,1) and 7" = 0'(1,1,0,1). This is a difficult problem.) Solution: Let O'(n,E) be the mapping defined by z..... eZ + n for z =F "', and "'..... "', where e = ±1 and n is any integer. In other words |
, O'(n,e) = O'(e, n, 0, 1). 100 ISOMORPHISM THEOREMS [CHAP. 4 We will show that the subgroup generated by '1 and r consists of all O'(n. El' n any integer, ::s = {O'(n.E) I n any integer, • = ±1}. We claim gp('1,r) =::s. Any element of ::s is • Now zr- I = z-l. Also, Zr"'r = z+n and zr- I... r - I = z-n for any • = ±l, Let of the form 0' ( n. E) '------v--' n '-----v---' n positive integer n. Hence for any arbitrary integer n, O'(n. I) E gp('1, r). (We must check what happens to 00, but this presents no difficulty.) Also, O'(n. I) '1 = 0'( -no _I) E gp('1, r) and so O'(n. E)1 • = ±1, n any integer, belongs to gp('1, T). Thus gp('1, r) d ::So Note that '1 = 0'(0. _I) and r = 0'(1. I) belongs to::S. So we need only show that ::s is a subgroup of the group of Mobius transformations to conclude that ::s d gp('1, r). To show ::s is a subgroup, we Il = ±l, But O'(~.O) = 0'(-6m.ol' since zO'(m.o) O'(-om.o) = need only show that O'(n.E) 0'(~.6) E::S, Ilm = 1l2Z + 8m (Ilz + m)1l Ilm, and we conclude that O'(n.E) O'(~,O) = O'(no-om.Eo) E::S. zO'(n. El O'(~. a) = (.z + n) 0'( -am. a) =.Ilz + nil - Ilm = 1l2Z = z. Then c. Exponents We have seen in the previous section that we |
often are forced to consider the product of (Note that as we are dealing with groups, it is not necessary m a's (m> 0), e.g. a'...• a. '-----r----' m to indicate in which order the multiplication is performed. See Section 4.1a.) It is convenient to introduce the notation am for the product of m a's, m > O. Then am. an is the product of m a's followed by n a's, i.e. am. an = am+n (Section 2.4d, page 39). Our idea is to extend the exponent notation in a sensible way to zero and negative exponents. We would naturally like the law (4.1) to be true when m and n are arbitrary integers. Now if it were true that aDam = am, then multiplication by aO leaves am unchanged. Hence we have only one choice in extending the exponent notation and retaining the law (4.1), namely putting aO = 1, the identity. Now if m = -n where n > 0, m + n = O. Because we want (4.1) to be satisfied, we must have am+n = aO = 1, i.e. we must put am = (an)-I. Note that (an)-I = (a-I)n = a-no Thus we have defined am to be the product of m a's if m > 0, (i) (ii) 1 if m = 0, (iii) the product of -m a-l's if m < 0, hoping thus to satisfy (4.1) for all m and n. (4.1) is true if m, n are both positive. If both are nonnegative, again by running through the possible cases (4.1) holds. If both m and n are negative, then aman = (a-I)(-m). (a-I)(-n) = (a-I)(-m + -n) = (a-I)-(m+n) = am+n If m and n arenonpositive, again the result is easily verified. If m> 0 and n < 0, then by checking the various possibilities m > -n, m = -n and m < -n, we find aman = am(a-I)(-n) = am+n. Another result which holds for exponents is (am |
)n = amn (4.2) We already know that (4.2) holds when n = -1. If m,n are positive, (am)n is the product of n elements, each of which is the product of m a's. Hence (am)n is the product of mn a's.'If m is negative, n positive, If now n is negative, Hence (4.2) is proved. (a-I)-mn as -m> 0, n> 0 amn as mn < 0 (am)(-I)(-n) = ((am)-I)-n = (a-m)<-n) a(-m).(-n) (by our previous remarks) Sec. 4.2] CYCLIC GROUPS 101 In the study of groups there are two main notations for the binary operation. One is the multiplicative notation we have employed up until now. The other is the additive nota tion. We denote the binary composition by + in this case. The identity is denoted by zero, 0, and the inverse of a by -a. The result of performing n > 0 compositions of the same element, i.e. of taking a + a +... + a, we denote by na. The law (4.1) becomes \ v~--~' n na +ma (n+m)a while the law (4.2) becomes n(ma) (nm)a In other words, translation takes place according to the following dictionary: Multiplicative notation ab 1 Additive notation a+b o -a na It is immaterial which notation one uses. But additive notation is most often used for a group in which the order of the composition of two elements is irrelevant, i.e. in which a + b = b + a for all a, b in the group. Such a group is called abelian, after the Norwegian mathematician Niels Henrik Abel, or commutative (Section 2.2, page 29). Problems 4.17. Find 13, 1-4 where 1 E Q, the additive group of rationals. Solution: In the additive group of rationals the binary operation is the usual addition. Then 13 means 1 0 1 0 1 where 0 is the binary operation in Q. Hence 13 = 1 + 1 + 1 = 3. Also 1-4 means (1"1)4, i.e. 1- 1 0 |
1 -1 0 1 -1 0 1 -1 where 0 in (Q, +). Thus 1-4 = (-1) + (-1) + (-1) + (-1) = -4. is the binary operation under discussion. Now 1 -1 = -,1 4.18. Find 22, 2- 3 where 2 E Q*, the multiplicative group of nonzero rationals. Solution: 22 = 2·2 = 4 and 2- 3 = (2- 1)3 = (t)3 =!- F' d 3 m a, a - _ (1 2 3 4) 2 1 4 3 4.19.,an element of 8 4, Solution: u2 = 1 and so uS = la = a. 4.20. Find Tn, where T is as defined in Problem 4.16. Solution: Tn = a(n,1)' See Problem 4.16. 4.2 CYCLIC GROUPS a. Fundamentals of cyclic groups If gp(X) = H, we say H is generated by X. To get an understanding of groups, a good plan is to investigate the simpler groups first. So we begin by considering groups which can be generated by a single element. We call such groups cyclic. Thus a group H is cyclic if we can find an element x E H such that H = gp({x}). We will usually write gp(x) instead of gp({x}). 102 ISOMORPHISM THEOREMS [CHAP. 4 Lemma 4.4: gp(x) Proof: {t I t = X T gp(x), r an integer}. Cyclic groups are abelian. { XEI ••• x En I x. E {x} €. = ±1 n > O} 1 n 1't'{x EI • • • X En I €i = ± 1, n > O} {xct <,)I €i=±I, n>o} {XT I r any integer} If a, b E gp(x), then a = xT, b = x', ab = xTx· = xT+', ba = x'xT = xT+.. Hence ab = ba for any two elements of a cyclic group. Thus we have shown that cyclic groups are abelian. Suppose now that H = gp(x) and IHI = m |
(m < 00). Then we know that the elements of H are of the form XT for various integers r. The Xi cannot be distinct for all integers i. Con sider XO = 1, x,..., Xl-I and suppose these are distinct but that Xl = x k for some k < l, k ~ 0; then XI(Xk)-1 = Xl- k = 1. If k 7'= 0, m = l- k < land xm is equal to xO. But we assumed this was not so. Hence k = 0 and Xl = XO = 1. We will show that S = {1,x,x 2 is actually H. This is easy. First notice that as Xl = 1, every positive power of x is in S. Furthermore, X-I = Xl-I. Hence every negative power of x lies in S. But H = {XT I r any integer}. Therefore H ~ S and so S = H as stated., •••,Xl } - I Thus we have proved I~emma 4.5: Let G be cyclic of order m generated by the element x. Then G = {XO, xl,..., xm-I}. Furthermore xm = 1, and xm is the least positive power of x that is 1. We ask a simple question: do cyclic groups of order m exist for all finite integers m > O? Yes! Let us consider in the symmetric group Sm of degree m the element Then u m m 1 2 1 2 ( m-l m m-2 m :) m-l m) 1 2 and so the elements £, Um' •• •,0-;::-1 are distinct and H = gp(um ) is cyclic of order m. Hence there exist cyclic groups of order m for each m > o. And now we ask another question: are there two essentially different cyclic groups of order m? Rephrasing the question, we ask: are two cyclic groups of order m isomorphic? Lemma 4.6: Let G = gp(x), H = gp(y) be each of order m. Then G 5!! H. Proof: G = {XO, x, x2, ••., xm- I }, H = {yO, yl,..., ym-l}. Let 0: G ~ H be defined |
by xiO = yi (i = 0,1,..., m -1) Then 0 is one-to-one onto H. To prove it is an isomorphism we must show it is a homomorphism. Consider Sec. 4.2J CYCLIC GROUPS 103 Now 0 ~ i, j ~ m -1. Then 0 ~ i + j ~ 2(m -1) = 2m - 2 and so o ~ r ~ m - 1 and f = 0 or 1. Hence i + j = fm + r where (xi+j)8 = (x.m+r)8 = (xEmxr)8 = (xr)8 = yr But Hence ((xi)(xj))8 = (xilJ)(xjlJ). Thus IJ is a homomorphism and, as it is one-to-one onto, it is an isomorphism. We now ask the obvious question: are there any infinite cyclic groups and are two infinite cyclic groups isomorphic? Consider the element a in the symmetric group Sz on Z, the set of integers, defined by Za = Z + 1, Z E Z As Zan = Z + n, am = an elements and so H is an infinite cyclic group. implies m = n. Then gp(a) = H, say, has an infinite number of Recall that G = gp(x) = {xn I n any integer}. If there exists an integer m> 0 such that xm = 1, then G will consist of only a finite number of elements (see the remarks pre ceding Lemma 4.5). Consequently if G is infinite, there exists no m =1= 0 for which xm = 1. For we have already shown that there can exist no m> 0 for which xm = 1; while if xm = 1 for m < 0, then X C- m ) = 1 and (-m) > O. If Xl = x n, n =1= l, then x n - l = 1. But this contradicts the condition that there exists no m such that xm = 1. Hence the elements of G are simply the powers xn of x, and two such powers xm and xn are equal if and only if m=n. Now we can easily prove that two infinite cyclic groups are isomorphic. Let G = gp(x), H = gp(y) |
both be infinite cyclic groups. Then each element of G is uniquely of the form xn, n an integer, and each element of H is uniquely of the form yn, n an integer. Define (xn)8 = yn. 8 is a one-to-one onto mapping. Furthermore, (xnxm)8 = (xn+m)8 = yn+m and (xn8)(xm8) = ynym = yn+m. Hence xn8xm8 = (xnxm)8. Therefore 8 is an isomorphism and G and H are isomorphic groups. Collecting our results, we have proved Theorem 4.7: There exist cyclic groups of all orders, finite and infinite. Any two cyclic (We therefore often talk about groups of the same order are isomorphic. the cyclic group of order m, or the infinite cyclic group, or sometimes the infinite cycle.) If x is an element of a group G, then we define the order of x as the order of gp(x). Note that if x is of order m < 00, then xm = 1 and m is the first positive integer r for which x, = 1. If x is of infinite order, then xm = 1 implies m = O. If x is of order m, m < 00, we say x is of finite order. Lemma 4.8: Let x be of order m < 00. If x' = 1, then m divides r. Proof: Put r = qm + s where 0 ~ s < m. Then 1 = x, = xqmxs = xs. As m is the first integer greater than 0 for which xm = 1, s = O. Hence m divides r. Problems 4.21. Prove that the additive group of integers is infinite cyclic. Solution: Z = gp(l). As Z is infinite, it is infinite cyclic. 4.22. Prove that the group of Problem 3.5, page 53, is cyclic of order m. Solution: The group is gp(l), and its order is m. 104 4.23. 4.24. ISOMORPHISM THEOREMS [CHAP. 4 Prove that (Z, +) and the subgroup of M, the group of Mobius transformations, generated by the mapping 1J: Z --> Z + 1, |
CC1J = 00 are isomorphic. Solution: By Theorem 4.7 all we need prove is that UP(1J) is infinite, as we know from Problem 4.21 that implies n = m. Thus UP(1J) is infinite and (Z, +) is infinite cyclic. But since Z1Jn = Z + n, 1Jn = 1Jm so UP(1J) =0 (Z, +). Find the order of (i) U = G 2 1!) E S3' defined by Z1J = -Z, 001J = 00. (ii) U 2 3 3 4 ~) E S4, (iii) the map 1J of M Solution: (i) u¥ t 2 1 2 1!) = t. Hence u is of order 2. u, u2, u 3 are not t, but u4 = t. Thus u is of order 4. (ii) (iii) 1J2 = t and so 1J is of order 2. 4.25. Let G be abelian. Let x, y E G be of orders r, s respectively. Show that xy is of order rs if rand s are co-prime, i.e. have no common prime divisors. Solution: Note that since G is abelian, the order of xy divides rs, by Lemma 4.8. If (xy)m = 1, i.e. xmym = 1, then xm = y-m and 1 = (xm)T = y-mT. Therefore s, the order of y, divides -mr. Since s does not divide r, s must divide m. Similarly we can show r divides m. Hence rs divides m. So if m is the order of xy, m is divisible by rs and also m divides rs. Thus the order of xy is rs. (xy)n = xnyn for any integer n. Since (xy)TS = xTSyTS = 1, 4.26. Show that if G is a cyclic group of order m < 00 and s is co-prime to m, then as = bs (a, bEG) implies a = b. Find a group G and a nonzero integer n such that there are two elements a, bEG with an = bn but a ¥ b. Solution: Since G is ab |
elian, so (ab-1)S = as(b-1)s = 1. Since G = Up(x) and the order of G is m, then ab- 1 = XT for some r, and (XT)S = 1. Hence x TS = 1 and m divides rs. But sand m are co-prime; then m divides r, say r = qm. Now ab- 1 = x qm = 1 and so a = b. 1 3) b = (1 2 3). Then a2 = b2 = t but a ¥ b. In S3, let a = (~ 2 3'1 3 2 4.27. Show that if G = Up(x) and G is of finite order rand s is co-prime to r, then UP(XS) = G. Solution: The distinct elements of Up(xs) are l,xs,x2s, •••,x(n-l)s where (XS)ft = x ftS = 1 and n is the least such positive integer. Since x ns = 1 and G is of order r, r divides ns. As rand s are co-prime, r divides n. Hence there are at least r distinct elements in Up(xS). But as G:l Up(XS) and G itself has only r elements, Up(xs) = G. 4.28. Find a group which is not abelian. (Hint. Consider S3.) Solution: See Section 3.3a, page 57, where we pointed out that UITI ¥ TIUl. Hence S3 is not abelian. 4.29. Prove that a subgroup H of S3 is cyclic if H ¥ S3. Solution: A survey of the subgroups of S3 shows that if H is a subgroup of S3 and H ¥ S3' then H is either cyclic of order 3 or cyclic of order 2 or cyclic of order 1. To obtain all the subgroups of S3, we refer to the multiplication table for S3 in Section 3.3a, page 57, and list all the subsets of S3. Then we check which subsets are subgroups. Of course since a subgroup must contain the identity, there is no need to go through the process of finding all subgroups quite so crudely. Nevertheless this method will suffice. Sec. 4.2] CY |
CLIC GROUPS 105 4.30. Prove that (Q, +) is not cyclic. Solution: If (Q, +) is cyclic, there exists q = min, m and n integers (n"" 0), such that gp(m/n) = Q. Of course m"" O. Each element (,.., 0) of Q would then be of the form q + q +... + q with some suitable choice of the positive integer r, or else of the form -q-q - '--v---------' r... -q. '--v---------' r But 1/2n E Q. 1/2n = q +... + q implies 1/2n = rm/n, i.e. 1 = 2rm; then 1 - 2rm = O. '------y------J r But rand m are integers; hence the equation 1- 2rm = 0 is not true. If 1/2n = -q -q -q r terms in all, a similar argument leads to a contradiction. Therefore (Q, +) is not cyclic.... -q, 4.31. Prove that an abelian group generated by a finite number of elements of finite order is finite. Solution: Let G = gp( {Xl'..., X n }), n < 00, and suppose G is abelian. Then every element g in G is of the form g = (Xij E X, OJ = ±1) Since G is abelian, we can rewrite g in the form To see this we need only observe that if is = it for 8 < t in the first expression for g, then g =xY l " 'x 1 Yn n ( YvYz,·· ·,Yn III egers't ) (4.3) g = Xi £1 l ES+Et ••. Xis Er ••• Xir i.e. we can always "collect" all occurrences of any X in a product. Now if Xl' X2' •••, Xn are all of finite order, then the number of distinct elements given by (4.3) is finite. For if k i is the order of x;, i = 1,2,..., n, the distinct powers of Xi are 1, Xi' X;,..., X~i-l. Thus the number of distinct ele m |
ents given by (4.3) is at most klkz... k m and so G is finite. b. Subgroups of cyclic groups Before beginning the study of a new section it is a good idea to list the natural ques tions. If we want to know something about the subgroups of cyclic groups, we might ask: (i) Are subgroups of cyclic groups cyclic? (ii) Does there exist a subgroup of any given order? (iii) How many distinct subgroups of a cyclic group (less than or equal to the order of the group) are there? (iv) How many subgroups of a given order are there? We tackle each of these questions. Theorem 4.9: (i) Let H be a subgroup of G = gp(x). Then H is cyclic and either H = gp(XI) where Xl is the least positive power of x which lies in H or else H = {1}. If the order of G is m < 00, then lj m and the order of H is mil. If the order of G is infinite, H is infinite or H = {1}. (ii) Conversely if l is any positive integer dividing m, then S = gp(XI) is of order mil. Consequently there is a subgroup of order q for any q that divides m. (iii) The number of distinct subgroups of G is the same as the number of distinct divisors of m = jGj < 00. (iv) There is at most one subgroup of G of any given order for G finite. 106 ISOMORPHISM THEOREMS [CHAP. 4 (i) Proof: If H # {I}, there exists xn # 1 E H. As H is a subgroup, x- n E H. Now one of n,-n is positive. Hence we can talk meaningfully about the smallest positive power Xl E H. Clearly, H-:1S = gp(XI). Suppose xr E H; then r = ql+s, 0 ~ s < l, and xr(X1q)-1 = XS E H But Xl is the least positive power of x that belongs to H. Thus s = 0 and so r = ql and (xr) = (Xl)q E S. Hence S = H. If the order of G is m < |
00, then m = ql + s, 0 ~ s < l. Now and so XS E H. Then s = 0, as Xl is the least positive power of x that lies in H. Hence l divides m, and m = lq. Clearly (Xl)q = 1, and q is the least positive integer for which this occurs. Then, by Lemma 4.5, writing a = Xl, we have H = gp(a) = {aO,a l } and hence IHI = q = mil. If the order of G is infinite, all the powers of x are distinct, and so Xl, X21, •.• is an infinite set of distinct elements of H. Thus H is infinite., •••,a q - l lim, l> O. Put mil = q and Xl = a. Then S = gp(XI) = {I, a,..., a q - I}, as (ii) Let q q = xm is the least positive power of a which is 1 (for if a · = 1, q' < q, then x lq' = 1 a and lq' < m, contradicting the fact that xm is the least positive power of x which is 1). Thus S is a subgroup of order q. Consequently if we start out with a positive integer q which divides m and we put l = mlq, then S is a subgroup of G of order q. (iii) Let ll' l2,..., In be the distinct divisors of m. Then put HI = gp(xZ,),..., Hn = gp(xln). We know IHil = mil;. These are n distinct subgroups of G (because their orders are different). Are there any more subgroups? By (i) any subgroup H of G will have to be generated by Xl where l is a positive integer dividing m. Hence l = li' say. There fore H = Hi. Thus the subgroups of G are simply HI, H 2, •••, H n, as desired. (iv) If Hand K are two subgroups of G with IHI = IKI, then Hand K are Hi and H; of part (iii) above, for some i and j. But IHil = mlli, IH;I = mil;. Since IHI = IKI, 1, |
; = l; and therefore i = j and H = Hi = H; = K. The reader will perceive that our knowledge of the cyclic groups is in some ways quite comprehensive. We know in the case of finite cyclic groups what the distinct subgroups are, we know they are cyclic and we know which cyclic subgroups appear. In the case of infinite cyclic groups we can easily prove there are an infinite number of subgroups. We will distinguish between them in Theorem 4.24, page 126, using the concept of index which will be introduced in Section 4.3b. The reader might naturally be led to consider now groups generated by two elements, hoping that similar powerful conclusions can be obtained, e.g. that every subgroup of a two generator group is a two generator group. But in going from one to two generators we lose control. It has been shown that every countable group is a subgroup of a two generator group, so we can never hope for a simple account of two generator groups. Problems 4.37. A subgroup H of a group G is called proper if H # G and H # {l}. Let G be a cyclic group of order a prime p. Prove that G has no proper subgroups. Solution: We know from Theorem 4.9 that the number of subgroups of G is the same as the number of distinct divisors of p, which are p and 1. Hence the number of distinct subgroups of G is two. As {l} and G itself are two distinct subgroups, the number of proper subgroups is zero. Sec. 4.3] COSETS 107 4.38. Prove that the only groups which have no proper subgroups are the cyclic groups of order p and the group consisting of the identity alone. Solution: Let G be a group with no proper subgroups, G # {1}. Let g E G, g # 1. Then 8 = gp(g) is a subgroup by Lemma 4.3. Since g E 8, 8 = G as G has no proper subgroups. Hence G is cyclic. If G is cyclic of order mn, m, n # 1, then, by Theorem 4.9, G has a subgroup of order m. But this is a proper subgroup. Hence G is cyclic of prime order or else possibly infinite cyclic, say G = gp(x) = {...,x |
-2,x-l,xO,xl,X2,... }. But H=gp(X2) is a subgroup not equal to {1}, and not equal to G since x ~ H. Hence G can only be cyclic of order p, a prime. 4,39. Find a group with two distinct subgroups both of the same order. rHint. Consider 8 3,] Solution: Let Ta = (1 2 3), T2 = (1 2 3). Then Igph)1 = Igp(T2)1 = 2. 132 213 4.40. Find a group which is of infinite order but has a subgroup of finite order. (Hint. Try the group of Mobius transformations, M, of Section 3.5, page 77.) Solution: Let 71: z ~ lIz, 00 ~ 00. Then gp(71) is of order 2, but M is infinite. 4.41. Let H be a subgroup of G. Let g E G. Prove that the set 8 = {g-lhg I hE H} is a subgroup of G. Prove that fJ: H ~ 8, defined by hfJ = g-lhg, is an isomorphism of H onto 8. If K is a :finite cyclic subgroup of G which contains both Hand 8, prove that H = 8. (Hard.) Solution: Since H # 0, 8 # 0. Let g-lh1g, g-lh2Y E 8. Then (g-lh 1g)(g-lh2g)-1 = g-lh1gg- 1h"2 1g = g-1(h1h"21)g E 8 because H is a subgroup implies h1h"21 E H. Thus 8 is a subgroup. fJ is an onto map, since hfJ = g-lhg. then g-lh1g = g-lh2Y. Pre-multiply by g and post-multiply by g-l: g(g-lh1g)g-1 = g(g-lh2Y)g-l. Hence hI = h2 and so fJ is one-to-one onto. We need only check that fJ is a homomorphism to conclude the proof: If h1 |
fJ = h2fJ, h 1fJh2fJ = g-lhlg' g-lh2g = g-lh1h2g = (h1h2)fJ Thus Hand 8 are isomorphic. If K is a finite cyclic subgroup containing Hand 8, then Hand 8 are both of finite order; and since they are isomorphic, IHI = 181. But, by Theorem 4.9, K has only one subgroup of any given order. Hence H = 8. 4.3 COSETS a. Introduction to the idea of coset In this section we propose a natural question which introduces the idea of a coset. Cosets are important for other reasons: (i) With co sets we can perform useful counting arguments for finite groups. (ii) Cosets of a subgroup sometimes enable us to construct a new group from an old. We can also see how a group G is built up from one of its subgroups H and the group constructed from the cosets of H. (iii) The fundamental idea of a homomorphism can be re-interpreted in terms of the idea of a group constructed from cosets. What is the natural question we ask? In Section 3Ac, page 67, and Section 3.4e, page 73, we defined the group I of isometries of the plane E and the isometry group Is of a given figure S in E. Recall that an isometry 0' of E belongs to Is if for each tEE, to' E S implies t E S, and 8 E S implies 80' E S. Suppose 0' E I - Is. We ask: which ele- 108 ISOMORPHISM THEOREMS [CHAP. 4 ments 0 of I are such that So = Ser? In the case where S is an equilateral triangle we know that Ser is a congruent equilateral triangle by Lemma 3.6, page 71. So the question we are asking is this: which other elements 0 of I send the equilateral triangle S onto the equilateral triangle Ser? Let SO = Ser. How are 0 and er related? If (j were equal to er, then Oer- 1 = L. If 0 and er were widely different, we would expect Oer- 1 to be any thing but L. Let q, = Oer- 1. We will show that q, is an element of Is. |
To do so observe that q, is an isometry of the plane E, since it is a product of isometries of E. Now SO = Ser implies that for each s E S there is atE S such that sO = ter, and conversely. Therefore for each s E S, sq, = sOer- 1 = (ter)er- 1 = t E S Suppose now that x E E and xq, E S. We must show that xES to complete the proof that q, E Is. Suppose xq, = t E S, Le. X(Oer- 1) = t. Now there exists an s E S such that ter = sO. Hence xO = ((xO)er-1)er = ter = sO and so xO = sO. As 0 is one-to-one, x = s. Hence xES. This means q, E Is. We have of course 0 = q,er. If we write Iser = {Terl T E Is}, we may put our deduction in the form 0 Elser. We have thus shown that every isometry 0 of E satisfying SO = Ser lies in the set of isometries Iser. Conversely if 0 Elser, then 0 = q,er for some q, E Is; then SO = Sq,er = Ser. This means that Iser consists of all the isometries 0 of E for which SO = Ser. Such a subset Iser of the group I of all isometries of E associated with the subgroup Is of I is called a right coset of Is in I. More generally we have the following Definition: Let G be a group and H a subgroup of G. Then a right coset of H in G is a subset of the form Hg = {x I x = hg, hE H} for some g in G. We define a left coset of H in G to be a subset of the form gH = {x I x = gh, hE H}. Note that a coset is a right or left coset according as the element g is on the right or. the left of H. In the case where the group G is written additively, Le. + is used to denote the binary operation, a right coset is written H + g. Of course, H + g = {x I x = h + g, hE H}. Problems |
4.42. Let G be the cyclic group of order 4 generated by {a}. Let H = gp(a2 ). Find all right co sets of H in G. Show that two cosets are either equal or else have no elements in common, and prove that the union of these cosets is G. Solution: H -1 = H = {1, a2 } is a right coset. Ha = {a, a3 } is a right coset. Ha 2 = {a2, a4 } = {1, a2 } = H. Ha3 = {a3, a 5} = {a3, a} = Ha. Thus the distinct co sets of H in G are Hand Ha. HnHa = 0, and HuHa = {1, a2, a, a3 } = G. 4.43. Let H be the trivial subgroup of a group G, i.e. H = {1}. Determine the distinct right co sets of H in G. Solution: If g E G, Hg = {lg} = {g}. Thus the co sets are the sets consisting of single elements of G. Sec. 4.3] COSETS 109 4.44. Find the right co sets of H = gp(7"3) the right cosets of K = gp(al) in S3? Solution: in S3 with the notation of Section 3.3a, page 57. What are H = (,,7":1), H, = H, Hal = {al,7"2}, Ha2 = {a2,7"I}' K = {ai' a2',}. The cosets of K are K, = K and K7"1 = {7"I' 7"2, 7"3}' These are all the cosets of H in S3' 4.45. Let A, B, C be subsets of a group G. If X and Yare subsets of G, we define XY = {g I g = xy, x E X and y E Y}. Prove that A(BC) = (AB)C. Hence conclude that if H is a subset of G, I, g E G, then (i) (fg)H = l(gH), (ii) H(fg) = (Hf)g, (iii) (fH)g = I |
(Hg). Solution: Let x E A(BC); then x = ad where a E A and dE BC. But dE BC implies d = be, where bE Band e E C; hence x = a(be) = (ab)e E (AB)C and so A(BC) \: (AB)C. Similarly (AB)C \: A(BC). Therefore A(BC) = (AB)C. (i), (ii) and (iii) follow immediately, e.g. (i) is the case where A = {f}, B = {g} and C = H. 4.46. Let G be a group with a subset H. Show that IH = HI implies!-IH = HI-I. Solution: IH = Hf. Hence 1-I(fH) = f-I(Hf), (f-lf)H = (f-IH)! and H = (f-IH)I Thus Hf- I = ((/-IH)f)I-1 = I-IH. Note that we have used the "associative law" proved in Problem 4.45. h. Cosets form a partition. Lagrange's Theorem In the problems above it is clear that any two right (or left) co sets are either disjoint or exactly the same and that the union of all the right (or left) co sets of H in G is G. We recall that a family of subsets of a set G is a partition of G if they are disjoint and their union is G. The examples above point to the following: Theorem 4.10: Let H be a subgroup of a group G. Then the right (left) co sets of H in G form a partition of H in G, i.e. the union of all the right (left) co sets of H in G is G itself and any pair of distinct cosets has empty intersection. Proof: It is easy to show that each element of G occurs in at least one right coset. (The proof for left cosets is similar and is not included here.) For if g E G, then g E Hg since 1 E Hand 1· g = g. Suppose now that Ha and Hb are two cosets of H in G and that HanHb # ~, i.e. there exists g E Han Hb. |
Then g = h'a = h"b, h', h" E H. Hence a = h'-Ih"b = h"'b, h'" E H, since the product of two elements of H belongs to H. Therefore Ha = {ha I hE H} = {h(h"'b) I hE H} C Hb as hh'''b = hlb, hi E H Similarly Hb C Ha. Thus Ha = Hb and any two cosets are either disjoint or identical. In Section 4.3a we mentioned in (i) that co sets are useful for counting arguments; this follows from Theorem 4.10. If G is of finite order and H a subgroup of G, then, since the cosets of H in G are disjoint, the order of G is the sum of the number of elements in each coset. We use this fact in proving Theorem 4.11 (Lagrange's Theorem): The order of a subgroup H of a finite group G divides the order of G. Proof: Let the distinct co sets of H in G be Hg l, Hg2,..., Hgn. Since these form a partition of G, (4.4) 110 ISOMORPHISM THEOREMS [CHAP. 4 What is IHgl? We will show that the mapping ()g: H ~ Hg defined by h() = hg is a one IHI = IHgl. Clearly {)g is onto by the definition of right to-one onto mapping and hence coset. If hlg = hzg, then multiplying by g-l on the right we conclude that hi = h2 and so IHI = IHgl. Thus hl{)g = h2{)g for each i, implies hi = h2. Therefore ()g is one-to-one and onto and IGI = nlHI by (4.4). IHI = IHgil. Hence Corollary 4.12: Let G be a finite group and g an element of G of order m. Then m divides IGI. Proof: The order of g is the order of gp(g) which is a subgroup of G. Then by Theorem 4.11, Igp(g)1 divides IGI. But m = Igp(g)l. Hence m divides IGI. Corollary 4 |
.13: If G is of finite order n, and g E G, then gn = 1. Proof: Every element of G must be of finite order. Let g E G be of order m. Then, by the preceding corollary, m divides IGI and so IGI = qm. Hence glGI = gqm = (gm)q = 1 and the result follows. Definition: Let the number of right cosets of H in G be called the index of H in G. Denote it by [G:H]. Note that [G: H] is read as "the index of H in G", i.e. in the opposite order to which G and H appear in [G: H]. Corollary 4.14: If G is a finite group, IGI = IHI· [G: H]. Proof: In the proof of Theorem 4.11 we conclude with "Hence IGI = nlHI... ". Since n is the number of cosets of H in G, i.e. n = [G: H], we have IGI = IHI [G: H]. (i) S7 has no subgroup of order 11; (ii) D4 has no subgroup of order 3; (iii) if g E A5 Problems 4.47. Show that: and g7 = t, then g = t. Solution: (i) If H were a subgroup of S7 of order 11, then, by IS71 = 7! = 7·6·5· 4 • 3 • 2 = 7.5.32.24 • Theorem 4.11, 11 divides IS71. But in the prime decomposition of IS71 there is no 11. Hence there is no subgroup of order 11. ID41 = 8. Since 3 does not divide 8, Theorem 4.11 tells us there is no subgroup of order 3. (H) (iii) If g7 = t, then either g = t or g is of order 7, since gm = 1 implies that the order of g divides m. As m = 7 and is a prime, the only possibility if g =j:. t is that the order of g is 7. Now by Corollary 4.12 it would follow that 7 divides IA51, which is not true. Hence g = t. 4.48. Prove that if G is a group of prime order, then G |
is cyclic. Solution: If G = {1}, there is nothing to prove. If 1 =j:. g E G, gp(g) = H is a subgroup of G. Hence its order, by Theorem 4.11, divides the prime IGI. As IHI =j:. 1, IHI = IGI since the only divisors of IGI are 1 and IGI. Thus H = G, as He G and Hand G have the same number of elements. 4.49. Give the right and left cosets of H = gp({'7}) where '7 = u(O, 1, 1, 0) of Mobius transformations, Section 3.5a, page 77. is an element of M, the group Solution: If uta, b, e, d) EM, then Hu(a, b, e, d) = {t' uta, b, e, d), '7u(a, b, e, d)} = {uta, b, e, d), u(b, a, d, e)} by the rule for multiplication which is obtained in Problem 3.46, page 79. Now uta, b, e, d)H = {uta, b, e, d), u(e, d, a, b)}, right and left co sets are in terms of a, b, e, d. as is easily checked. Thus we know what the Sec. 4.3] COSETS 111 c. Normal subgroups We discussed left and right cosets of a subgroup H in G. Each gives rise to a partition of G. How do these partitions compare? In particular, are they the same? Sometimes yes. In Problem 4.43, the right cosets are just the elements of G; the left cosets can similarly be shown to be just the elements of G. Sometimes no. In Problem 4.49 the right coset containing a(a, b, c, d) also contains a(b, a, d, c). But the left coset containing a(a, b, c, d) contains a(C, d, a, b). With a suitable choice of a, b, c, d we can ensure that a(b, a, d, c) >/= a(C, d, a, b), e.g. if a = b = c = 1, d = |
0, then a(I, 1, 0, 1) oF a(I, 0,1,1). Thus the left and right cosets of H in G do not coincide. We ask: when do the right and left cosets of a subgroup H in a group G coincide? Suppose every left coset of H is also a right coset of H in G. Let a E G. aH contains a, as does Ha. Since the right cosets form a partition, the only right coset containing a is Ha. But we have assumed that there is some right coset which is the same as aH. Hence it must be Ha, and so aH = Ha. In other words if every left coset of H in G is a right coset, then for every a E H we must have aH = Ha. Proposition 4.15: A necessary and sufficient condition for the left co sets of H in G to provide the same partition as the right is that for each a E G, aH = H a. Proof: We have proved above that if every left coset is a right coset, Ha = aH. Let Ha = aH for all a E G. {Ha I a E G} is the set of all the right cosets of H in G. There fore every right coset of H in G is a left coset of H in G. Similarly every left coset of H in G is a right coset of H in G. This completes the proof. If Ha = aH for all a E G, then for each hE H, ha = ahl for some hI E H.' Hence a- 1ha E H for each hE H. Conversely if a- 1ha = hI, for some hI belonging to H, ha = ah 1• Hence Ha C aH. If ah E aH and assuming x- 1hx E H for all x E G and all hE H, then ah = ah(a- 1a) = «a- 1)-lha- 1)a E Ha. Accordingly Ha d aH and aH = Ha. Thus we have Proposition 4.16: aH = Ha for all a E G if and only if a- 1ha E H for all h E Hand all a E G. Definition: A subgroup H of a group G is normal (also called invariant) in G if g-lh |
g E H for all g E G and all hE H. We write H ~ G and read it as: "H is a normal subgroup of G". By Proposition 4.16, H is normal in G if and only if Hg = gH for all g E G (equiva lently, g-IHg = H). In Section 4.3a we gave in (ii) and (iii) reasons for the importance of some cosets. The cosets we had in mind are those arising from normal subgroups. Problems 4.50. Prove that every subgroup of an abelian group is a normal subgroup. Solution: Let G be abelian and H any subgroup of G. Then if g E G and hE H, g-lhg = h; for since G is abelian, gh = hg and hence multiplying by g-1 on the left, h = g-lhg. Then if hE H, g-lhg E H. Thus H is a normal subgroup of G. 4.51. Prove that An <J Sn for each positive integer n. Solution: Let U E An' l' E Sn. Is 1'- 1U1' E An? Now l' is either odd or even. If l' is even, then l' and 1'-1 E An and so 1'- 1U1' E An' If l' is odd, then 1'-1 is also odd, and hence 1'- 1 U is odd, for u E An. Since 1'- 1 U and l' are odd, their product 1'- 1U1' is even. Hence 1'- 1 U1' E An. Note that we used Lemma 3.2, page 62. 112 4.52. ISOMORPHISM THEOREMS [CHAP. 4 Let H be a finite cyclic subgroup of G, and let H <J G. Let K be a proper subgroup of H. Prove that K <J G. (Hard.) Solution: Let K = gp(y). Note that if y is of order m, and g E G, then g-lyg is also of order m, since implies g-lyrg = 1 and therefore yr = 1. Hence m (g-lyg)m = g-lymg = 1 and (g-lyg)r = 1 divides r, and the order of |
g-lyg is m. Since H <J G, g-lyg E H. Therefore gp(g-lyg) is a sub group of H of order m. Then by Theorem 4.9(iv), gp(g-lyg) = K. In particular, g-lkg E K for any kEG. Hence K <J G. d. Commutator subgroups, centralizers, normalizers We will now introduce some subgroups which are normal. 1. If G is a group, we define the center of G, denoted by Z(G), to be {z I z E G and for all g E G, gz = zg} Z(G) turns out to be a normal subgroup of G (see problems below). 2. If G is a group and x, y E G, then x-1y-1xy is called the commutator of x and y or, more briefly, a commutator. We often write [x, y] for the commutator x-1y-1xy. The sub group of G generated by all commutators is called the commutator subgroup (also called the derived group) of G and is denoted by G'. Again G' turns out to be normal in G. Proceeding along somewhat different lines, let A be a subset of a group G. 1. The centralizer C(A) of A (in G) is defined by C(A) = {cl cEG and for all aEA, ca=ac} C(A) is a subgroup of G (see problems below). If A is an abelian subgroup, A is normal in C(A) (see problems below). 2. The normalizer N(A) of A in G is defined by N(A) = {n I nEG and An = nA} N(A) is a subgroup of G and, if A is a subgroup of G, A is normal in N(A). Furthermore, if A is a subgroup of G, A is normal in G if and only if N(A) = G. These facts will be proved in the problems below. The details concerning the groups Z(G), G', C(A), N(A) appear in the problems below. In Chapter 5 we will use the concepts we have just introduced. Problems 4.53. Prove that the inverse |
of a commutator is a commutator. Solution: [x, y] = x-1y-1xy = z, say. So z-l = y-1x-1yx = [y, x]. 4.54. Prove that G' is normal in G. Solution: We must show that if g E G and hE G', then g-lhg E G'. If h is a commutator, say h = x-1y-1xy, then g-lhg = g-lx-1gg-1y-lgg-lxgg-1yg = x11Y11X1Y1 = [Xl> Y1] where Xl = g-lxg (consequently x 1 = g-l x -1g ) and Y1 = g-lyg. Now any element h of G' is a product of commutators and their inverses; and as an inverse of a commutato~ is a commutator, every element h of G' is a product c1 ••• ck of commutators. Therefore g-lhg = g-1(C1 ··· Ck)g = g-lc1gg-1c2g··· g-lckg = d1d2 ··· dk where di = g-lCjg. But we have just shown that d j is a commutator. Hence if hE G', g-lhg E G'. 1 Sec. 4.3] COSETS 113 4.55. Show that G is abelian if and only if G' = {I}. Solution: Suppose G is abelian and that x, y E G. As x and y commute (i.e. xy = yx), [x, y] = x- 1y- 1xy = x- 1x = 1. Then G' is the subgroup of G generated by 1, and G' = 1. Now if G' = {I},. then in particular any commutator [x,y] = x- 1y-1 xy = 1. Hence x(X- 1y-1 xy ) = x and y(y-1 xy ) = yx, i.e. xy = xy. Thus G is abelian. 4.56. Show that every element |
in Aa is a commutator of elements in S3. Hence show that S; = A 3. Solution: We use the table of Section 3.3a, page 57. 7"1 = 7"~1. 7"~lU~17"lU1 = 7"lU27"lU1 = U1U1 = U2. t. Thus every element of Aa is a commutator of ele 7"~lU;17"lU2 - 7"lUl7"lU2 - U2U2 - U1. 7"~lt-l7"lt ments in S3 (A3 is listed in Problem 3.23, page 63). If we can show that all commutators belong to A 3, then A3 = S~. This is a matter of trying all possibilities, e.g. 7"~17";17"17"2 = 7"17"27"17"2 = U1. Hence the result. 4.57. Show that the commutator subgroup of M of Section 3.5a, page 77, is infinite. finite cyclic subgroup generated by a commutator.] [Hint. Find an in Solution: [u(2, 0, 0,1), u(l, 1,0,1)] = u(l, -1, 0, 1) = u, say. Now gp(u) is infinite cyclic, as un = (1, -n, 0,1) for each n. Since the commutator subgroup of M contains gp(u), it is infinite. 4.58. Prove that if G is any group, Z(G) is a normal subgroup of G. Solution: 1 E Z(G), since 19 = gl then g(glg;l) = (ggl)g;l = gl(gg;l) = glg; l g since gg2 = g2g for all g E G. Consequently Z(G) ~ 0. If g1> g2 E Z(G) and implies g;lg = gg;l. It g E G, follows that Z(G) is a subgroup of G. If g E G and hE Z(G), then gh = hg and so h = g-lhg. Hence U- 1hU E Z(G) for |
all U E G and all hE Z(G). Thus Z(G) is normal in G. 4.59. Show that C(A) is a subgroup of G and, if A is an abelian subgroup of G, A <J C(A). Solution: 1 E C(A) and so C(A) ~ 0. If gl' g2 E C(A), and a E A, then U2a = ag2 and hence au; 1 = u;la, i.e. u2- 1 E C(A). Now ag1g;1 = U1aU;1 = U1g2-1a and so glg2-1 E C(A) if gl, g2 E C(A). Therefore C(A) is a subgroup of G. If A is an abelian subgroup, then each a E A belongs to C(A). Now if g E C(A), then for each a E A, ga = ag, i.e. g-lag = a E A. Accord ingly A <J C(A). 4.60. Show that if A is a subset of G, then N(A) is a subgroup of G. Show that if A is a subgroup of G, then A <J G if and only if N(A) = G. Solution: N(A) ~ 0, since 1 E N(A). Let I,g E N(A). Using the results of Problems 4.45 and 4.46, (fg-1)A = 1(g-lA) = I(Ag-1) = (IA)g-l = (AI)g-1 = A(fg-1). implies Ig-1 E N(A). Therefore N(A) is a subgroup of G. Clearly, if A is a IA = AI implies UA = Ag and Hence I, U E N(A) subgroup, A C N(A) and A <J N(A). If A is a subgroup and A <J G, then for each g E G, UA = Ag. Hence U E N(A) and so G C N(A). Therefore G = N(A). If A is a subgroup and N(A) = G, then since A <J N(A), A <J G. 4.61. Find all |
normal subgroups of Sa. Solution: We use the notation and multiplication table in page 57. Clearly {t} and S3 are both normal subgroups of S3. There are no normal subgroups of Sa con taining elements of order 2 except Sa. The elements of order 2 in Sa are, as we readily check by using the multiplication table for S3' 7"1,7"2 and 7"a. Suppose for example that a normal subgroup N of S3 contains 7"1; then U~17"lU1 = U27"lU1 = 7"2 E N. Similarly 7"17"2 = U2 E N and u~ = U1 E N, and so it follows in this way that N = S3. 7"3 E N. Hence 114 ISOMORPHISM THEOREMS [CHAP. 4 We have shown that if N is a normal subgroup of S3, then if N contains elements of order 2, N == S3. Therefore if N # {,}, then N must contain elements of order 3 (there are only elements of order 1,2, or 3 in S3). Now 0"1 and 0"2 == O"~ are the only elements of S3 of order 3. In fact {" 0"1, 0"2} is a normal subgroup of S3. For example, 7";-10"17"1 == 7"10"17"1 == 0"2 E {" 0"1> O"z}. Accordingly S3 has precisely three distinct normal subgroups. 4.62. Show that if A is a subgroup of G and B <J G, then AB is a subgroup of G, where AB == {x I x == ab, a E A, b E B} Solution: AB # 0, as 1 == 1'1 E AB. bi E B. Now glg;1 If gl, g2 E AB, then gl == alb l, g2 == a2b2 where ai E A and a l bl b2- l a;1 == a l b3a 21 (where b3 E B) a la;l a2 b3a;1 == ala;l(a;I)-lb3(a;l) ab, say, where a == ala;1 E A, and b == aZb3a;1 E B as B <J G. |
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