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C G are a pair of composition series for G, then their respec tive composition factors can be paired off in such a way that paired factors are isomorphic. We have already proved the first statement of Theorem 5.25. Before illustrating Theorem 5.25 we restate its last assertion as follows: There is a permutation 7r of {I,..., k} such that for i = 0,..., k - 1. Example 1: Suppose that 8 3 is the symmetric group on {I, 2, 3}. Then the series {I} (1 2 3) (1 2 3)} C 1, {(I 2 3) is a composition series for 8 3, Notice that the composition factors are of orders 3 and 2. This is actually the only composition series for 8 3, since (1 2 3)} (1 2 3) {(I 2 3) 123' 231' 312 is the only normal subgroup of 8 3 which is neither 8 3 nor the identity subgroup. Problem 5.50. Let n be a positive integer. What relevance does the Jordan-Holder theorem have to the factoriza-· tion of n into a product of primes? (n> 1).) (Hint: Let G be the additive group of integers modulo n Solution: Let be a composition series for G. Each composition factor Gi+ t/Gi (i = 0,..., l - 1) is simple. As Gl is abelian, each factor Gi + tfGi is abelian. Hence if Gi + t/Gi has any proper subgroup, it would not be simple. So Gi + t/Gi has no proper subgroups. In particular it has no cyclic proper subgroups, so it must be cyclic of order a prime. The number l is therefore the total number of primes (allowing for repetitions) dividing n. By Theorem 5.25, l is uniquely determined. So, as we well know, the total number of prime divisors of n is a constant. Moreover, the uniqueness of the composition factors (asserted in Theorem 5.25) simply means that these prime divisors themselves are unique. Putting these two facts together gives the well-known fact that every integer n > 1 can be written uniquely as a product of primes, if the order in which it is written is disregarded. Sec. 5.5] COMPOSITION SERIES AND SIMPLE GR |
OUPS 165 Example 2: Let Sn be the symmetric group on n letters, n "" 5. Then {I} C An <;:; Sn is a composition series for Sn- For we shall show that An is simple (in Section 5.5e, Theorem 5.34). But An <J Sn and Snl An is cyclic of order two. Hence this series is indeed a composition series. b. Proof of Jordan-Holder theorem Suppose G is a finite group and suppose and 1 1 Go C; G1 C; Ho C; Hl C; (5.21) (5.22) are two composition series for G. We have to prove that k = l and that the composition factors Gi+dGi of (5.21) can be paired off with the composition factors Hi+dHi of (5.22) so that paired composition factors are isomorphic. The proof is by induction on the order [G[ of G. If [G[ = 1, then both assertions are clear. Thus we assume that [G[ > 1 and that the theorem holds for all groups of order less It is useful to observe that if k = 1, then G is simple. Hence l = 1 also, and than [G[. again the desired conclusion holds. So we assume, in addition to [G[ > 1, that k> 1 (and hence l> 1). There are two cases to consider: Gk- 1 = H t- 1 and Gk-l =1= H t- 1• Case 1: Gk - 1 = H t - 1• It is then clear that 1 = Go C;..• C; Gk - 1 (5.23) 1 = Ho C; •.• C; H t - 1 = Gk - 1 and (5.24) are composition series for Gk - 1• But [Gk - 1 [ < [G[. Hence by our induction assumption, the composition series (5.23) and (5.24) have the same length, i.e. k -1 = l-1, and so k = l. Furthermore the composition factors of (5.23) can be paired with the composition factors of (5.24) so that paired factors are isomorphic. But the composition factors of (5.21) are those of (5.23) together with G/Gk-l. Similarly the composition factors of (5 |
.22) are those of (5.24) together with G/Gk - 1• Thus it is clear then that the composition factors of (5.21) can be paired off with the composition factors of (5.22) so that paired factors are isomorphic. This concludes the proof of Case 1. Case 2: Gk - 1 =1= H t- 1• Our method of proof is to produce a composition series, (5.26), which has isomorphic composition factors to those of (5.21) (by Case 1) and a composition series, (5.27), which has isomorphic composition factors to those of (5.22) (by CMe 1). We will then show that (5.26) and (5.27) have isomorphic factors and this will be sufficient to prove the result. First we will show that Gk-1Ht - 1 = G. Observe that both Gk - 1 and H t - 1 are normal subgroups of G, and so Gk-1H1- 1 is also a normal subgroup of G. Obviously Gk- 1H I- 1 con tains Gk- 1 properly, so Gk- 1H I-tlGk-l is a non-trivial normal subgroup of G/Gk-l by the correspondence theorem. But G/Gk-l is simple, so Gk-1HI-dGk-l = G/Gk- 1. This means that (5.25) We now put F = Gk -lnH1- 1 and note that F <J G. Let {1} = Fo C; •.• C; Fm = F be a composition series for F. Then we claim that 166 and FINITE GROUPS {I} {I} Fo C Fo C [CHAP. 5 (5.26) (5.27) are both composition series for G. The only facts that need be verified are that Gk-dF m and Hz-dFm are simple. Now Fm = Gk-inHz- i and by (5.25), G = Gk-iHz- i. Therefore by the subgroup isomorphism theorem (Theorem 4.23, page 125), GIGk-i Gk-iHz-dGk- i """ Hz-d(HI-inGk- i) Hz- |
dFm and similarly GIHz- i (5.28) (5.29) Since both GIGk-i and GIHz- i are simple, it follows that HI-dFm and Gk-dF", are also both simple. Let us compare the composition series (5.21) and (5.26). By Case 1 it follows that they have the same length and their composition factors can be paired off so that paired factors are isomorphic. Similarly for the composition series (5.22) and (5.27). Let us now compare the composition series (5.26) and (5.27). They obviously have the same length, m + 2. Thus the series (5.21) and (5.22) have the same length. What are the composition factors of the series (5.26) and (5.27)? The composition factors of (5.26) are FdFo,..., FmIFm-i, Gk-dFm, GIGk-i while those of (5.27) are FdFo,.. -, Now by (5.28), Hz-dFm ~ GIGk- i; and by (5.29), Gk-dFm ~ GIHI- i. Thus the composition factors of (5.26) and (5.27) can be paired off so that paired factors are isomorphic. It fol lows that the composition factors of (5.21) and (5.22) can be paired off so that paired factors are isomorphic, since the factors of (5.21) can be so paired off with those of (5.26) and the factors of (5.22) can similarly be paired off with those of (5.27). This completes the proof of the Jordan-HOlder theorem. From the Jordan-HOlder theorem we know that the length of a composition series and its factors are uniquely determined for the group. This suggests the following scheme for studying groups which have a composition series. First, find the structure of all groups with composition series of length 1. These are all the simple groups. Assuming now that we know all about groups with a composition series of length n, let G be a group with com position series of length n + 1. Let {I} = Go C Gi C... C Gn C Gn + i = G be a composition series for G. If F is a group with |
a normal subgroup N and FIN ~ H, we say that F is an extension of N by H. In this language then, G is an extension of a group with a composition series of length n, viz. Gn, by a simple group. Hence what we must know is how the structure of a group which is an extension of one group by another is determined. In the next few sections we shall prove that if n > 5, An is simple. The groups An are not all the simple groups and indeed there is no classification of simple groups as yet. This is one of the basic problems of finite group theory. The question of how a group G is built from Hand K if G is an extension of H by K, is considered in Chapter 7. Here too our knowledge is far from complete. Problem 5.51. If two groups have the same composition factors, are they isomorphic? Solution: No. There is a cyclic group G of order 6 and a non-abelian group H of order 6. They have the same composition factors, but they are not isomorphic. Sec. 5.5] COMPOSITION SERIES AND SIMPLE GROUPS 167 c. Cycles and products of cycles We begin with an example of a new notation. Let us consider 8 6• Let () E 8 6 be defined by ( ) Note that the effect of () is to take 3 --'> 5, 5 --'> 4, 4 --'> 6 and 6 --'> 3, and to leave the elements 1,2 unaltered. () is called a cycle. We denote it by (3,5,4,6). More generally, consider 8 n • If al,..., am are distinct integers in {1,2,..., n}, (ai, a2,..., am) stands for the permutation that maps. each integer in {1,2,..., n} {ai,..., am} to itself, and maps We call such a permutation a cycle of length m. As a convention take a cycle of length 1 to denote the identity element. A cycle of length 2 is called a transposition. The inverse of the cycle a = (ai,..., am) is the cycle {3 = (am, am-I,.••, a2, al), since aJa(3) = (aia){3 = ai + l {3 = ai if i =1= m while If j E {1, |
2,...,n} - {al,...,am }, Hence a{3 = L. Similarly (3a = t. j(a{3) = (ja){3 = j{3 = j It is clear that not every permutation is a cycle. Nor is the product of two cycles necessarily a cycle; for example, in 84, (1,2)(3,4) = (~ i : :) which is not a cycle. An obvious question is: can we express every element of 8 n as a product of cycles? The following theorem answers this question. Theorem 5.26: Every element of 8 n can be written as the product of disjoint cycles. (Cycles (ai,..., am) and (b l, •••, bk ) are disjoint if the ai and bi are distinct, i.e. if {al,..., am} n {bl,..., bk } = \Z>.) Proof: Let 7T E 8 n • We say that i E {1,..., n} is fixed by 7T if i7T = i, and we say it is moved if i7T =1= i. We shall argue by induction on the number of integers moved by the permutation 7T. If 7T moves none of the integers {1,2,..., n}, then 7T is the identity permutation. But then 7T = (1), as the 1-cycles are all the identity permutation. Hence we have a basis for induction. So let 7T =1= L and suppose that every element of 8 n which moves fewer integers than 7T, can be written as the product of disjoint cycles. Now suppose a l E {1,...,n} and that a l7T =1= al' Let us define a2,a3 a3 = a27T, • • •, as = as_ I 7T, • • • • Let m be the first integer such that am7T = ai for some integer i with 1 ~ i < m. (As the images of 7T belong to {1,2,..., n}, the terms of the, • • • cannot all be different.) We shall prove, using the minimality of m, sequence ai' a2 that i = 1. Suppose to the contrary that i =1= 1. Then ai = ai _ |
l 7T and am7T = ai = ai As 7T is a one-to-one mapping, ai-I = am' But this contradicts the choice of the integer m. Hence i = 1 and am 7T = al' We consider now the cycle (at'..., am) and note that m> 1. Let, • • • by a2 = a l 7T, 7T. _ I - - T ( al,..., am )-1 7T 168 FINITE GROUPS [CHAP. 5 for T is a permutation that leaves al,..., a implies i'# 1, while a/al,..., am jT oF j )-I7r = j7r '# j; for if j E {a l,...,am }, jT = j. Hence j ~ {a p •••,am} and so jT = j(al,...,a j7r '# j. It follows that T moves fewer integers than 7r. Therefore inductively T can be written as the product of disjoint cycles, say )-l7r = am7r = al. Also if j E {1,2,..., n}, fixed, since a.(al,..., a )-17r = a. l7r = a. then m m 1.- m t t Now if Tj involves an ai' ai = aiT = a/T IT2... TZ) = aiTj" Hence Tj = (ai) = L. Let be those T j which do not involve any of the integers all..., am. Clearly Til' Ti2,...,Tik (il <i2 <··· <ik) If T = L, 7r = (a l, •••, am) and we have proved that 7r is actually a cycle. Otherwise 7r = (ai'..., amhl... Tik and 7r has been expressed as the product of disjoint cycles. Hence the result. Corollary 5.27: If 7r E Sn and ai' a2,..., am are chosen as in the proof of the theorem (i.e. a2 = a l 7r, •••, and am7r = al and ai'..., am distinct), then 7r = (al,..., am)T where a |
T = a7r if a ~ {a l,...,am}, while ajT = aj for j = 1,...,m. Proof: This is precisely what we showed in the proof of Theorem 5.26. This corollary provides us with a method of computing the decomposition of an element 7r E Sn into the product of disjoint cycles. For example, let us write 1 2 3 3 42 4 1 ( 5 6 7 8 9 10 11 12) 8 7 9 11 12 10 5 6 as the product of disjoint cycles. Since h oF 1, we may take 1 for al. Then al = 1, a2 = 3, aa = 2, a4 = 4, as = 1. So m = 4. Hence by the corollary, 7r 1 2 3 4 5 6 7 8 9 10 11 12) (1,3,2,4) 1 2 3 4 8 7 9 11 12 10 5 6 ( Here the second factor T on the right is obtained from 7r by letting it leave 1,2, 3, 4 unchanged, and letting it act as 7r does on the remaining integers. Applying the same technique to T, we find T = (5,8,11)(6,7,9,12). Hence 7r = (1,3,2,4)(5,8,11)(6,7,9,12) In practice it is not necessary to use the corollary rigorously. We need only find the disjoint cycles by choosing al such that a l 7r oF ai' and then taking the cycle (al,..., am) where a;+l = ai7r, am7r = ai' and the al'...,am are distinct. Then we choose b1 E {1,2,...,n} {al,...,am } where bl7roFbl' and find the cycle (bl,...,bk) with bi+l=bi7r, bk7r=b l and b l, •••, bk distinct, and so on. Not only is every element of Sn a product of cycles, but each element can also be ex pressed as the product of transpositions, as can be seen from the following proposition. Proposition 5.28: Every cycle is a product of transpositions. Proof: We assert (al,a |
2,...,ak) = (ai, a2)(al, a3)... (al,ak) For if a is an integer not in {ai,..., ak}, both the left-hand side and the right-hand side leave it unchanged. Sec. 5.5] COMPOSITION SERIES AND SIMPLE GROUPS 169 If i =1= k, then ai(a1,..., ak) = ai+ 1 while ai(a1,a2)(al,aa)... (a1,ak) ai(a1,ai)(a1,ai+l)... (a1,ak) If i = k, then ak(al,..., ak) = a1 while ak(a1, a2)... (ai, ak) = ak(a1, ak) = a1 Thus the effect of the left-hand side and the right-hand side is the same, and so the permuta tions are equal. Problems 5.52. In 8 6 compute a = (1,2,3,4)(2,5) and f1 = (1,2,3,4)-1(2,5). Solution: 1a = (1(1,2,3,4))(2,5) = 2(2,5) = 5. ia = i + 1 if i = 2,3; 4a = 1; 5a = 2; 6a = 6. Hence a = (~ (1,2,3,4)-1 = (4,3,2,1). Then 1f1 = 4, 2f1 = 1, 3f1 = 5, 4f1 = 3, 5f1 = 2, 6f1 = 6. Hence f1 = (4,3,2,1)(2,5). f1 = 2 345 3 4 1 2!). 1 2 3 4 5 6) 1 5 3 2 6 (4. 5.53. Express a and f1 of Problem 5.52 as products of disjoint cycles. Solution: We note that 1a = 5, 5a = 2, 2a = 3, 3a = 4, 4a = 1. Hence a = (1,5,2,3,4). Since 1f1 = 4, 4f1 = 3, 3f1 = 5 and 5f1 = 2, 2f |
1 = 1, f1 ~ (1,4,3,5,2). 5.54. Write a = G 2 4 Solution: 3 4 5 6 6 8 10 12 14 1 7 8 9 10 11 12 13 14) 3 5 7 9 11 13 as the product of disjoint cycles. 1a = 2, 2a = 4, 4a = 8, 8a = 1. 3a = 6, 6a = 12, 12a = 9, 9a = 3. 5a = 10, lOa = 5. 7a = 14, 14a = 13, 13a = 11, 11a = 7. Hence a = (1,2,4,8)(3,6,12,9)(5,10)(7,14,13,11) 5.55. If a, f1 E 8 n are such that (if1)a = if1, then a and f1 commute. Hence prove that disjoint cycles commute. Solution: implies both ill' =1= i (ia)f1 = ill' and if1 = i, and if1 =1= i implies i(af1) = (ia)f1 = ia while i(f1a) = (if1)a = ia. If ia = i, Let i E {1,2,..., n}. If ia =1= i, then iaf1 = if1 while then (if1)a if1 = ill' if if1 =1= i if if1 = i Hence af1 = f1a. Now if a and f1 are disjoint cycles, then ia =1= i f1 does not move either i or ia, and so if1, and so (if1)a = if1. Thus a and f1 commute. if1 = i and (ia)f1 = U;. Similarly if if1 =1= i, implies a moves i and ia and hence f1 moves i and 5.56. Express a of Problem 5.54 as the product of transpositions. Is the expression of an element of 8 n as a product of transpositions unique? Solution: a = (1,2,4,8)(3,6,12,9)(5,10)(7,14,13,11) = (1,2)(1,4)(1,8)(3 |
,6)(3,12)(3,9)(5,10)(7,14)(7,13)(7,11) As (1,2)(1,2) =!, if a = a1a2... am then a = (a1a2... an)(l, 2)(1, 2) the expression as a product of transpositions is not unique. is the expression of a as a product of transpositions, is another expression of a as a product of transpositions. Thus 170 FINITE GROUPS [CHAP. 5 5.57. Find the order of the cycle (av..., am)' (Hard.) Solution: If we write av..., am in a circle, as shown on the right, it is clear that a == (a v..., am) moves each ai one place clockwise, a 2 moves each ai two places further, and in general a i moves each ai j-places further. Hence am moves each ai m-places further, i.e. am moves each ai back to itself. To put this more formally, define am+l == ai' a m +2 == a2,..., a m +m == am' We will prove by i == 1,2,..., m. For induction on, j that for 0 "" j "" m, a i maps ai to ai + i' j == r, consider j == 0, ai == aiar+l == (aiar)a == ai+r"" al == am+l == ai+r+l' If i+r> m, then ai+r == ai+r-m and i+r-m < m r < m). Hence ai+ra == ai+r-ma == ai+r-m+l == ai+r+l' a.; j == r + 1 "" m. Then If i+1' < m, ai+r'" == ai+r+l by the action of a. If i+r == m, ama == (as 1"" i "" m and t and the result is true. If the result is true for • • u", al a2 • • • O • • • In particular, am == t. On the other hand, alas == al+ s # al for 1"" 8 < m. Therefore m is the smallest powE |
.!r of a that yields the identity. Accordingly m is the order of a. d. Transpositions, and even and odd permutations In this section we will produce another way of deciding whether a permutation is even (See the definition in Section 3.3b, page 60.) By Proposition 5.28 and Theorem or odd. 5.26 every permutation is the product of transpositions. However, this decomposition is not unique (Problem 5.56). What we shall show is that a permutation 7r is a product of an even number of transpositions if 7r is even, and the product of an odd number of trans positions if 7r is odd. Our first task is to show that any transposition is odd. We have already noted that (1,2) is odd (Section 3.3d, page 64). We will use the following lemma. Lemma 5.29: Let () E Sn and let (ai,..., am) be a cycle. Then (}-l(al,...,am )(} = (al(},...,am (}) Proof; Let x E {I, 2,..., n}. If x = a;(} for some a;, aj(}(}-l (ai,..., am )(} ai(al,..., am)(} ai+l(} al(} if i =F m if i = m If x =F ai(} for some ai, x(} -1 fl {ai,..., am}. Then X(}-l (ai,..., am )(} = (X(}-l)(} x Thus (}-l(al,..., am)(} = (al(), •••, am(}). Theorem 5.30: All transpositions are odd. If () is an even permutation and is written as the product of transpositions, the number of transpositions is even. If () is odd and is written as the product of transpositions, the number of trans positions is odd. Proof; Let (a, b) be any transposition. We know that (1,2) is odd. Let () be any per mutation such that 1(} = a, 2(} = b. Then, by Lemma 5.29, (a,b) = (}-1(1,2)( |
} If () is even, (}-l is even, (}-1(1,2) is odd, and so (}-1(1,2)(} is odd (Lemma 3.2, page 62). If 8 is odd, (}-l is odd, 8-1(1,2) is even, and hence (}-1(1,2)(} is odd (again by Lemma 3.2). Therefore (a, b) is odd. Now let () be a permutation and let 8 = a • • • am be the product of transpositions. Then by Lemma 3.2, 8 is even if and only if m is even, while () is odd if and only if m is odd. This proves the theorem. l Sec. 5.5] COMPOSITION SERIES AND SIMPLE GROUPS 171 Problems 5.58. Determine whether a and f3 of Problem 5.52 are odd or even, using Theorem 5.30. Solution: a = (1,5,2,3,4) = (1,5)(1,2)(1,3)(1,4). Since a is expressed as the product of an even number of transpositions, a is even. f3 = (1,4,3,5,2) = (1,4)(1,3)(1,5)(1,2), and so f3 is even. 5.59. Determine whether a of Problem 5.54 is even by using Theorem 5.30. Solution: a = (1,2,4,8)(3,6,12,9)(5,10)(7,14,13,11) {(I, 2)(1, 4)(1, 8)}{(3, 6)(3, 12)(3, 9)}{(5, 10)}{(7, 14)(7, 13)(7, 11)} Thus a is even. 5.60. Determine whether (ai'..., am) is even or odd. Solution: (ai'...,~) = (ai' a2)(aV a3)... (ai' am), so (av..., am) is the product of m - 1 transpositions. Thus (ai'..., am) is even or odd according as m is odd or even. 5.61. Let (al"'" ~), (bl |
,..., bm ) be two cycles of 8 n • Prove that there exists 9 E 8 n 9- I (av...,am )9 = (b l,...,bm )· Solution: such that Let 9 be defined by: (1) ai9 = bi for i = 1,...,m; (2) if {CI"'" cn - m } = {1,2,..., n} - {ai'..., am} and if {dl,..., dn - m } = {1,2,..., n}{bl,...,bm }, put ci9 = di for i = 1,...,n-m. Then 9 E 8 n and by Lemma 5.29, 9- I (al"'" a m )9 = (b l,..., bm ). 5.62. If G is a group, the set of all elements conjugate to x is called the conjugacy class containing x. Write down the conjugacy classes of 8 4 using cycle notation. (Hard.) Solution: The conjugacy class containing an element 9 is the set of all conjugates of 9. Our first task is to express each element of 8 4 as the product of disjoint cycles, and then to take all distinct con jugates of each element. We obtain the following classes: {(I)}. (As (1) = I, there are no other conjugates of (1). (i) (ii) What conjugates of (1,2) are there? We know that rl(I,2)9 = (19,29). As 9 runs through the conjugacy class con (a, b) = (b, a), 8 n, 19,29 run through all distinct pairs (a, b). As taining (1,2) is {(1,2), (1,3), (1,4), (2,3), (2,4), (3,4)} (iii) What conjugates of (1,2,3) are there? As 9- 1(1,2,3)9 = (19,29,39), we will get all possible 3-cycles. Hence the conjugacy class containing (1,2,3) is {(I, 2, 3), (1,2,4), (1,3,2), |
(1,3,4), (1,4,2), (1,4,3), (2,3,4), (2,4, 3)} (iv) What is the conjugacy class containing (1,2)(3,4)? r l (I,2)(3,4)9 = 9- 1 (1,2)99- 1 (3,4)9 = (19,29)(39,49) As 9 runs through 8 n, we will obtain the product of all pairs of disjoint cycles. We recall that disjoint cycles commute; for example, (1,2)(3,4) = (3,4)(1,2). Thus the distinct elements in the required conjugacy class are {(I, 2)(3, 4), (1,3)(2,4), (1,4)(2, 3)} (v) What is the conjugacy class containing (1,2,3, 4)? Again we see that we shall obtain all 4-cycles. Hence the required conjugacy class will be {(I, 2, 3, 4), (1,2,4,3), (1,3,2,4), (1,3,4,2), (1,4,2,3), (1,4,3, 2)} 172 FINITE GROUPS [CHAP. 5 e. The simplicity of An, n ~ 5. In this section we aim to prove that An is simple for n ~ 5. What we must do is to show that if H <J An and H # {L}, then H = An. Although the proof involves much calculation, the ideas are easy. Lemma 5.31: Every element of An is the product of 3-cycles, n ~ 5. Proof: Every transposition (al, a2) = (e, al)(e, a2)(e, al) where e, al, a2 are distinct. Now every element of An is the product of an even number of transpositions, and therefore the product of products of pairs of transpositions. So it is enough to prove that a product of two transpositions is a product of 3-cycles. Consider (al' a2)(b l, b2). We may assume that the four integers al, a2, bl, b2 are distinct, for otherwise (al, az)(b l, b2 |
) is either the identity or is itself a 3-cycle. As n ~ 5, there exists an integer e such that 1 """ e """ nand al, a2, bl, b2, e are all distinct. Hence (al, a2)(h b2) = (~, al)(e, a2)(e, al)(e, bl)(e, b2)(e, bl) Lemma 5.32: = {( e, al)( e, az)}{ (e, al)( e, bl)} {( e, b2)( e, bl)} = (e, al, az)( e, al, bl)( e, b2, bl) Let H <J An. If H contains a 3-cycle, then H = An. Proof; Let (a, b, c) E H. Then if x, y, z are distinct elements of {1,2,..., n} and () is an element of Sn that sends a to x, b to y and c to z, we have (}-l(a, b, c)(} = (x, y, z) by Lemma 5.29, page 170. If () is even, (x, y, z) E H as H <J An. If () is odd, then there exists e, f distinct from a, b, c as n ~ 5. Therefore 'lr = (e, f)(} is even. 'lr-l(a,b,c)'lr = = (}-l(e,f)(a,b,c)(e,f)(} (}-l (a, b, c)(} = (x, y, z) Hence by normality, (x, y, z) E H. Thus H contains all 3-cycles of S,,; and by Lemma 5.31, H=An. Lemma 5.33: Let H <J An. If H contains the product of two disjoint transpositions, then H=An. Proof; Let a = (a1, a2)(aa, a4) E H. Since n ~ 5, there exists e E {I,..., n} distinct from a1, a2, aa, a4. Let () = (a1, a2, e); then () E An. (}-la() = (a2, e)(aa, a4) |
a-1(}-la(} = (al, a2)(aa, a4)(a2, e)(aa, a4) = (a1, a2)(a2, e) = (al, e, a2) Thus H contains a 3-cycle, and the result follows from Lemma 5.32. Theorem 5.34: An is simple for n ~ 5. Proof: Let H <J An' H # {L}. Then there exists a E H, a # L. Let a = a 1... ak where k are disjoint cycles. We may suppose without loss of generality that a 1, • • •, a k 1 for i = 1,..., k -1, as the a 1, • • •, a k + al'..., a are arranged so that the length of a i ~ length of a commute by Problem 5.55, page 169. Case 1: i Suppose a 1 = (a1,..., am) with m> 3. Let (J' = (a1, a2, aa)' Clearly, as they move different integers. As (J' E An' Also (1 (1-1 E An and H <J An' commutes with a2 f3 = a- 1(1-1a(J' E H. Note that, •••, ak and so f3 (J'-la(J' = (a2, aa' al' a4,..., am)az.·.. ak a;l... a;-1(a2, aa' al' a4,..., am )a2... ak a;-1(a2, aa' a 1, a4,..., am) (am' am - 1,..., a1)(a2, aa' a 1, a4,..., am) (a1, a4, a2 ) Since f3 E H, the result follows from Lemma 5.32. Sec. 5.5] COMPOSITION SERIES AND SIMPLE GROUPS 173 Case 2: Suppose m = 3, and a2 is also a 3-cycle. Let al = (at' a2, aa) and a2 = (a4, as' as). Let (] = (a2, aa' a4 ) E An. Then H contains Thus H contains k a |
-la-I... a-l(u-la u)(u-la u)a... a k 12 a~la;-1(u-lalu)(u-la2u) = a1 la;-1(al, aa' a4)(a2 (aa' a2, al)(aS' as' a4)(al, aa' a4)(a2, as' as) = (ai' a4, a2, aa' as), as' as) 2a 1 and the result follows from Case 1. Case 3: Suppose m = 3 and a 2, • • •, a k are transpositions. Let al = (ai' a2, aa). Then a2 = (ai' a2, aa)ala2... ak(al, a2, aa)ala2... ak (ai' az' aa)2a~a~... a~ = (ai' a2, aa)2 = (ai' aa' a2) and the result follows from Lemma 5.32. Case 4: Suppose all the ai are of length 2. Then m is even, and a l = (at' az), a2 = (aa' a4). Put u = (a2, aa' a4 ). Then H contains u-1au = (ai' aa)(a4, aZ)aaa4... ak Thus H contains and the result follows by Lemma 5.33. Corollary 5.35: The symmetric group Sn is not solvable for n::=" 5. Proof: An <J Sn and {t} CAn C Sn is a composition series of Sn for n::=" 5 (Example 2, page 165). By the Jordan-Holder theorem this is, up to isomorphism, the only possible com position series. Now IAnl = n!/2, which is even for n::=" 4. Hence [An: {t}] is not a prime. But if any subnormal series of Sn with factors of prime order existed, it would be a com position series. Therefore Sn is not solvable. Problems 5.63. Prove that if G = An' n:;" 5, then the derived group G' of G is G. • Solution: We know that G' <J G. Hence G' = G or else G |
' = {I} as G is simple by Theorem 5.34. If G' = {I}, G is abelian. But An is not abelian for n:;" 5; for example, :) :) 1,2,3)(3,4,5) (3,4,5)(1,2,3) but Therefore G'=G. 5.64. If G = An and n:;" 5, prove that Z(G) = {I}. Solution: As Z(G) <J G, Z(G) = G or Z(G) = {I}. As Z(G) is abelian but G is not (see Problem 5.63), Z(G) # G. 174 FINITE GROUPS [CHAP. 5 5.65. Without using Theorem 5.34, prove that G' = G = An for n "" 5, where G = Sn- ~ Solution: We use Lemma 5.33. Let", = (2,3,4), u = (1,2,3). as G' <J G, the result follows from Lemma 5.33. (3 = ",-l u -1",u = (1,4)(2,3) E G'. Hence 5.66. Are any of AI' A 2, A a, A4 simple? Solution: Both Al and A2 are of order 1, so they are not simple. Aa is of order 3, so Aa is cyclic of prime order and therefore simple since groups of prime order are simple. Finally A4 is of order 12. We have already seen that a group of order 12 is solvable. Indeed we saw in Problem 5.45 that its com position factors have orders 3,2,2 respectively. Hence A4 is not simple. 5.67. Prove that Sn has no non-trivial element in its center for n "" 3. (Hard.) Solution: Let '" E Sn' '" "'" t. Let '" = "'1 ••. "'k be the decomposition of '" into disjoint non-identity cycles. Let "'1 = (aI'..., am)' If m"" 3, let (3 = (aI, a2)' Then 01- 1(301 = (a101, a2(1) = (a2' aa) "'" (3. Hence |
01 ~ Z(Sn)' If m = 2, let (3 = (aI' a2, aa). Then 01- 1(301 = (a101, a2"', aa"') = (a2, av b) where b = aaOl and b is an integer different from a1 and a2' No matter what b is, 01- 1(301 "'" f3 since a2(3 = as but a201-1(301 = a1" Henc!! 01 ~ Z(Sn)' Thus no non-trivial element of Sn belongs to Z(Sn)' 5.68. Prove that An' Sn and {t} are the only normal subgroups of Sn for n"" 5. (Hard.) Solution: If AnnH = {t}, suppose H"", {t}. If (J E Hand (J "'" t, Let H <J Sn- Then HnA n <l An. Hence AnnH = An or else AnnH = {t}, as An is simple by Theorem 5.34. If AnnH = An, then An C; H. If H"", An' as An is of index 2 in Sn' H = Sn. (J2 E HnA n (J2 = t. Let T E H, T"'" t. Then T is odd, and as (JT is even, T = (J-1 = (J. Hence H consists and so of only two elements, t and (J. As Sn has no center (Problem 5.67), there exists {L E Sn such that {L-1(J{L "'" (J. But {L-1(J{L E H, as H <l Sn; and since H = {t, (J}, this is impossible. This contradicts the assumption that H "'" {t}. (J is odd. As (J2 is even, 5.69. Prove that S~ = An for n "" 5. Solution: As SnlAn is abelian (cyclic of order 2 in fact), An d S~ (Problem 4.68(iii), page 116). As Sn is not abelian, S:, "'" {t}. But S~ <l Sn. Hence S~ = An by Problem 5.68. Alternatively, S~ d A~ = An by Problem 5.63 or 5.65. A look back |
at Chapter 5. In this chapter, we proved the three Sylow theorems. The first gives the existence of Sylow p-subgroups; the second states that a subgroup of prime power order is a subset of one of the Sylow p-subgroups; and the third states that all the Sylow p-subgroups are conjugate. The proofs of the Sylow theorems used a standard technique of finite group theory: induction on the order of the group. It is also worth noting our counting arguments, e.g. in the proof of the class equation. Using the class equation we proved that a group of prime power order has a non-trivial center. Then we proved as a consequence that a group of order pr (where p is a prime) has a normal subgroup of order pr-1. By repeatedly taking the center of factor groups, we defined the upper central series of a group. A group is nilpotent if a term of the upper central series is the group itself. AA Next we gave a method of constructing a group from the cartesian product of two given groups Hand K. This group had isomorphic copies H, K of Hand K respectively which satisfied HK = G and HnK = {I}. Reversing this analysis we showed that if a group G had normal subgroups Hand K with HK = G and HnK = {I}, then G "'" H x K. Using this result and the Sylow theorems, we classified groups of orders 1, 2,..., 15 up to isomorphism. A A A A CHAP. 5] SUPPLEMENTARY PROBLEMS 175 Then we defined solvable groups, so called because they led to a criterion for the solv ability of equations. We noted that our first definition involving a subnormal series with factors of prime order did not extend to infinite groups. So we chose a criterion involving subnormal series with abelian factors for our definition of solvability. We showed that sub groups and factor groups of solvable groups were solvable, and an extension of a solvable group by a solvable group was solvable. We next considered composition series (subnormal series with simple factors) and proved that every finite group has a composition series. In the Jordan-Holder theorem we showed that a composition series has a unique length and unique factors up to isomorphism. In our final section we proved that the groups An for n ~ 5 are simple |
. To do this we needed to express permutations as products of disjoint cycles. This led to a method of determining whether a permutation was even or odd. As a consequence of the fact that An is simple, we concluded that Sn is not solvable for n ~ 5. Supplementary Problems SYLOW THEOREMS 5.70. Prove that the Syl6lw 17-subgroup is normal in a group of order 255 = 3' 5 '17. 5.71. Prove that the Sylow 13-subgroup is normal in a group of order 2· 5 • 13. 5.72. Let A = {O,l} be the set of integers modulo 2, B = {O, 1,..., 64} be the set of integers modulo 65, and G be the set of all pairs (a, b), a E A and bE B. Define the mUltiplication (aI' bl )(a2' b2) = (al + a2, (-1)a2b l + b2) for (at> bl ), (a2' b2) E G. Prov.e that G is a group of order 2' 5· 13 with respect to this multiplication. Is a Sylow 2-subgroup normal? 5.73. Verify the class equation (i.e. equation (5.2), page 135) for the group in Problem 5.72. 5.74. Prove that the Sylow p-subgroup is always normal in a group of order 4p, where p is a prime '= 5. Is this true when p = 3? 5.75. Prove that the normalizer of a Sylow p-subgroup coincides with its own normalizer. 5.76. [G[ = nand [H[ = m where H is a subgroup of G. If Hng-1Hg = {1} for all g E G - H, Let then there are precisely nlm - 1 elements in G which are not in any conjugate of H. (Hint: Examine N G(H).) THEORY OF p-GROUPS 5.77. Show that every subgroup of index p in a finite p-group is a normal subgroup. 5.78. Let G be a finite p-group and H be a proper subgroup of G. Show that N G(H) # H |
. 5.79. Prove that if G is nilpotent and GIG' is cyclic, then G' = {1}. (G' is the derived group of G). 5.80. Use Problem 5.79 to show that elements of co-prime order commute in a nilpotent group. 5.81. Find an example of a group G with a normal subgroup N such that GIN and N are nilpotent but G is not. 176 5.82. FINITE GROUPS [CHAP. 5 Let G(l) = G and G(i+l) = up({[u, xli u E G(i) and x E G}). The sequence of subgroups is called the lower central series of G. Prove that a group G is G(1) d G(2) d... d G(i) d... nilpotent if and only if G(n) = {1} for some positive integer n. 5.83. Find the lower central series for D 2n for positive integers n. (See Problem 5.82 for the definition of lower central series.) DIRECT PRODUCTS AND GROUPS OF LOW ORDER 5.84. Prove that a finite niipotent group is isomorphic to a direct product of anyone of its Sylow p-sub groups and some other subgroup. 5.85. Employ the results of Problem 5.70 to prove that a group of order 255 is isomorphic to a direct product of its Sylow 17-subgroup and another of its subgroups. Thereby show that a group of order 255 is cyclic. 5.86. Show that the direct product of two nilpotent groups is a nilpotent group. 5.87. Let G to G? UP ( (_ ~ ~), G ~)). Show that IGI = 8. Which group of order 8 is isomorphic 5.88. Show that Dn is isomorphic to G = UP ((~ ~), (~ f~l)) where f = e21Tiln. 5.89. Suppose G is a finite group with all its Sylow p-subgroups normal. Show that G is nilpotent. SOLVABLE GROUPS 5.90. Prove that a group of order pqr is solvable when p, q, r are primes and r> pq. 5.91. Show that the direct product of two solvable groups is sol |
vable. 5.92. Prove that a group of order 4p, p a prime, is solvable. 5.93. In Problem 5.49, page 163, we showed that a group G is solvable if and only if G(n) = {1} for some integer n. Use this fact to give alternate proofs of Theorems 5.23, page 161, and 5.24, page 162. 5.94. Show that Dn is solvable for all positive integers n. COMPOSITION SERIES AND SIMPLE GROUPS 5.95. Find the composition factors of a finite p-group. 5.96. Show that there is no simple group of order prm, where p is a prime and m < p. 5.97. Prove Sn is not solvable for n"" 5, without using the fact that An is simple. all 3-cycles of Sn.) (Hint. Consider 5.98. Find a composition series for the quaternion group. 5.99. Show that, except for groups of prime order, there are no simple groups of order < 60. difficult case occurs for order 36. See Problem 5.43, page 158.) (Hint. A Chapter 6 Abelian Groups Preview of Chapter 6 A group G with binary operation. is abelian if, for all g, h E G, g' h = h· g. It is customary to use "+" for the binary operation in abelian groups. We will begin this chapter with a preliminary section in which we restate some of our results and defini tions in additive notation. One of the concepts which we will reformulate additively and generalize is the concept of "direct product" considered in Chapter 5. In abelian groups it is customary to talk about "direct sum" instead of "direct product". We note that the direct sum of abelian groups is again abelian. We call direct sums of infinite cyclic groups free abelian groups. Direct sums satisfy an important homomorphism property which gives rise to the important fact that every abelian group is a homomorphic image of a free abelian group. We then consider classifying abelian groups according to the orders of their elements. An abelian group G which has every element of finite order can be expressed as a direct sum of p-groups, i.e. groups every element of which is of order a power of the prime |
p. An important p-group that we shall introduce here is the p-Priifer group. At this point we have three types of abelian groups: (a) the cyclic groups, (b) the additive group Q of rationals, (c) the p-Priifer groups. The rest of the chapter is devoted to showing that many abelian groups are direct sums of these groups. Recall that a group G is finitely generated if it contains a finite subset X with G = gp(X). We show that every finitely generated abelian group is a direct sum of cyclic groups. Furthermore we associate a set of integers to each finitely generated abelian group. This set of integers, which we call the type of the abelian group, completely classifies finitely generated abelian groups; in other words, two finitely generated abelian groups are isomorphic if and only if they have the same type. This theorem is of great importance in many branches of mathematics. The additive group of rationals Q has the property that if g E Q and n is any nonzero integer, then there exists f E Q such that nf = g. We express this by saying that Q is divisible. The p-Priifer groups are also divisible. Note that these are not the only divisible groups, e.g. the additive group of reals is also divisible. We obtain the pleasing result that if A is any divisible abelian group, then A is a direct sum of p-Priifer groups and groups isomorphic to the additive group of rationals. Note. Any reader who would like a briefer account of abelian groups may refer to Sec tions 6.la, 6.lc and 6.3. This will bring him quickly to the fundamental theorem of abelian groups, i.e. every finitely generated abelian group is the direct sum of cyclic groups. 177 178 ABELIAN GROUPS [CHAP. 6 6.1 PRELIMIN ARIES Here we will practice expressing our ideas in additive notation. a. Additive notation and finite direct sums In this chapter aU groups will be abelian, and we will use additive notation throughout. In terms of additive notation an abelian group is a non-empty set G together with a binary operation + such that (a + b) + c = a + (b + c) for all a, b |
, c E G. (i) (ii) a + b = b + a (iii) There exists an identity element, denoted by 0, such that a + 0 = a for all a E G. The identity element 0 is often termed the zero of G. (iv) Corresponding to each a E G there exists an element b such that a + b = O. This b is unique and is denoted by -a. The element -a is often termed the negative of a. Standard abbreviations are as follows: (i) g + (-h) is written as g - h. (ii) If n is a positive integer we write ng for g +... + g (n times). If n = 0 we write ng for O. If n < 0 we write ng for -g +... + -g (-n times). If G is an abelian group and H is a subgroup, then automatically H <l G and we may talk of the factor group G/H. (Warning: Some authors write G - H for G/H.) Note that, in additive notation, a coset is simply a set of the form g + H.. Instead of talking of multi plication of cosets, we talk of addition of cosets. Thus the sum of two co sets gl + Hand g2 + H is, by definition, The following table is useful in "translating" multiplicative notation into additive nota tion. a and b are elements of a group G, and Hand K are subgroups of G. Multiplicative ab Additive a+b a- 1 -a 1 0 aft na ab- 1 HK aH a-b H+K a+H In Section 5.3a, page 146, we defined the internal direct product of two groups Hand K. Now, in additive terminology, we speak of a direct sum rather than a direct product. Instead of writing H ® K, we write H E9 K. If G = H E9 K, H is called a direct summand of G. Here we are interested in extending the concept of direct sum from two subgroups to a finite number of subgroups. Definition: An abelian group G is said to be the direct sum of its subgroups G1, •••, Gn if each g E G can be expressed uniquely in the form where gi E Gi, i = 1,..., n. n G= ~Gi. i= |
l g = gl +... + gn In this case, we write G = G1 E9... E9 G n or If n = 2 we obtain the same definition of internal direct product that we gave in (The condition (i) of Proposition 5.19, page 146, falls away, as all groups Chapter 5. studied here are abelian.) Sec. 6.1] PRELIMINARIES 179 If G = G1 EB... EB Gn, then Gin Gj = {O} for i # j. For otherwise if x E Gin Gj and x # 0, then x = g1 +... + gn with g1 =... = gi-1 = gi+1 =... = gn = ° and gi = x, and also with g1 =... = gj-1 = gj+ 1 =... = gn = ° and gj = x. But this contradicts the definition of direct sum. Note that if G = H EB K and H = L EB M, then G = L EB M EB K (see Problem 6.15). The following theorem provides a simple criterion for determining when a group is the direct sum of its subgroups. Theorem 6.1: Let G1, • • •, Gn be subgroups of a group G and suppose each element of G can be expressed as the sum of elements from the subgroups G 1, •••, Gn • Suppose also that an equation ° = g1 + '" + gn with gi E Gi for i = 1,...,n, holds only if g1 = g2 = '" = gn = 0. Then G is the direct sum of the subgroups G 1, • • •, Gn • Proof: If g E G, then g = g1 +... + gn with gi E Gi, i = 1,..., n. We need only show that this expression is unique. Suppose is another such expression. Then ° = (g1 - gi) +... + (gn - g~ ) g=li+···+g~ By our hypothesis, i = 1,..., n and the two expressions for g are identical. (g1 - gI) = (g2 - g~) = '" = (gn - g!) = 0. Hence gi = if for The question arises: if G1, •••, Gn are abelian groups, does there exist a group G |
which is the direct sum of isomorphic copies of the groups Gi? This question is answered in the following theorem. Theorem 6.2: Let G1, • • •, Gn be abelian groups. Then there exists a group G which is the direct sum of isomorphic copies of G1,..., Gn. Proof: The proof follows closely the argument of Problem 5.30, page 146, so we will be brief. Here we will use additive notation. Let G be the cartesian product G1 x G2 X •.• x Gn • If (g1,..., gn) and (h1,..., hn) E G, define (g1,..., gn) + (h1,..., hn) = (g1 + h1,..., gn + hn). n components Then G is a group. AFurthermore, let Gi = {(O,..., g:, 0,..., 0) I gi E Gi}' Then Gi is a subgroup of G, and Gi ~ Gi for i = 1,..., n. It is clear that every element (g1,..., gn) of (g1,0,..., 0) + (0, g2, 0,...,0) +... + (0,0,...,0, gn). Hence G is uniquely of the form G = G1 EB... EB Gn and the result follows. A A An important result which we will prove in Section 6.1c states that if G is the direct sum of G1,..., Gn and H is the direct sum of H 1,..., Hn and Gi ~ Hi for i = 1,..., n, then G~H. In Section 6.1b we will define the concept of an indexed family Gi, i E I. The reader who studies Section 6.1c without reading Section 6.1b may take I = {I,..., n} and Gi, i E I as shorthand for G1,..., Gn. Then ~ Gi is simply ~ Gi. n iEI i=1 180 ABELIAN GROUPS [CHAP. 6 Problems 6.1. Prove that the negative of a + b is -a - b. Solution: (a+b) + (-a-b) (a + b) + [(-a) + (-b)] [a + (b + (-a» |
] + (-b) [(a + (-a» + b] + (-b) [(a + b) + (-a)] + (-b) [a + «-a) + b)] + (-b) (0 + b) + (-b) = b + (-b) o 6.2. 6.3. 6.4. Prove that if G is abelian and H is a subgroup of G, then G/H is abelian. Solution: (I + H) + (g + H) = (f + g) + H = (g + f) + H = (g + H) + (I + H) Let H be a subset of an abelian group G. Prove that H is a subgroup of G if and only if I, g E H implies 1 - g E H. Solution: This is exactly the same argument as in Lemma 3.1, page 55. Let n be an integer and G an abelian group. Prove that if g, hE G, then n(g + h) = ng + nh. Solution: If n = 0, n(g + h) = 0 by definition. Furthermore ng + nh = 0 + 0 = o. Thus n(g + h) ng + nh when n = O. If n > 0, let n = m + 1. Then m "" O. Inductively we may assume that m(g + h) = mg + mho Keeping this in mind, n(g + h) (m + l)(g + h) = m(g + h) + (g + h) = mg + mh + g + h mg + g + mh + h =:= (m + l)g + (m + l)h = ng + nh Finally if n < 0, n = -m for m > 0, and so n(g + h) = m(- (g + h» = m«-g) + (-h» = m(-g) + m(-h) = ng + nh 6.5. Let n be any integer and G an abelian group. Prove that the mapping e which sends g to ng for each g in G is a homomorphism. Solution: Now, in additive notation, to say that e is a homomorphism means that (g + h)e = ge + he |
for all g, h E G. But, by Problem 6.4, (g + h)e = n(g + h) = ng + nh = ge + he. 6.6. Let G be any abelian group and let Xc: G (X # 0). Prove that gp(X) = {y I y = rixi + r2xZ +... + rnxn, ri E Z, xi E X} (6.1) In particular then gp(x) = {rx IrE Z}. Solution: Let H be the right-hand side of (6.1). Then H is a subgroup of G, for H # 0. Moreover if then h = rlxl +... + rnxn and k = SIYI +... + spYp belong to H clearly and Xi' Yj E X), (ri' s; E Z Thus H is a subgroup of G. It follows from the definition of H that Xc: H. Since gp(X) is the smallest subgroup of G containing X, we have then gp(X) c: H. But if Xl'...'Xn E X, then rix i +... + rnXn E gp(X) for every choice of ri E Z. Hence gp(X):2 H, and so gp(X) = H. When X = {x}, then gp(X) is, by (6.1), the set of all multiples of x. 6.7. Let Hand K be subgroups of G. Prove that H + K G = H + K, prove that G = H $ K. (H + K = {h + k I h E Hand k E K}.) Solution: is a subgroup of G. If HnK = {O} and H + K # 0 since both H # 0 and K # 0. So we have to prove that if u, v E H + K, then (h, h' E H, k, k' E K). Thus u - v = (h - h') + u - v E H + K. Now u = h + k, v = h' + k' (k - k') E H + K since Hand K are subgroups of G. If H n K = {O} and if we consider two expres s |
ions hI + kl = hz + kz where hI' hz E Hand kl' kz E K, then X = hI - hz = kz - kl belongs to both Hand K. Therefore X = 0 and hI = hz, kl = kz. Hence the expression of an element in the form h + k is unique. Since G = H + K, it follows that G = H $ K. Sec. 6.1] PRELIMINARIES 181 6.S. Let G have a subgroup H and suppose that G/H is infinite cyclic. Prove that H is a direct summand of G. Solution: Let G/H = gp(g + H) where g E G and let S = gp(g). Consider x E SnH. Then x = ng for some n E Z. As x E H, n(g + H) = ng + H = H. But g + H is of infinite order in G/H. Thus n = 0 and so SnH = {O}. If x E G, x E ng + H, i.e. x = ng + h for some n E Z and hE H. Hence G = S + H. Therefore, by Problem 6.7, G = S EEl Hand H is a direct summand of G. 6.9. Let G = A EEl B and let H be a subgroup containing A. Prove that H A EEl (BnH). Solution: As every element of G is uniquely of the form a + b where a E A and bE B, A + (BnH) = A EEl (BnH). What we must prove then is that A + (BnH) = H. If hE H, h = a+ b where a E A and bE B. Hence b = h - a. But, as A ~ H, h - a E H. Thus bE BnH and He A + (BnH). But H"dA and H"dBnH. Hence H = A EEl (BnH). 6.10. Let G be abelian and let a E G be of order n, bEG of order m. Show that the order of a + b divides the least common multiple 1 of m and n. Solution: Since both m and n divide I, let 1 |
= qm = rn for some integers q and r. Then Ib = qma + rnb = 0 + 0 = O. Thus the order of a + b divides I. I(a + b) = la + 6.11. Let G = H EEl K. Prove that G/H ~ K. Solution: G = H + K. Then by the subgroup isomorphism theorem (Theorem 4.23, page 125), G/H = (H + K)/H ~ K/(KnH). Now HnK == {O}, and so G/H ~ K as required. 6.12. Prove that if G is an abelian group, then the set S of elements of G of order a power of a fixed prime p is a subgroup. Deduce that a finite abelian group has one Sylow p-subgroup for each prime p dividing IGI. Solution: If a and b are of order a power of p, then -b is of order a power of p. So, by Problem 6.10, a - b is of order a power of p since the least common multiple of two powers of a prime p is a power of p. Hence S is a subgroup. If G is finite, then S is the Sylow p-subgroup of G. For if P is any subgroup of G of order a power of p, by the definition of S, P ~ S. So every Sylow p-subgroup of G is contained in S. S itself is of order a power of p (Problem 5.6, page 132). Since the order of a Sylow p-subgroup is the maximal power of p dividing the order of G, every Sylow p-subgroup of G must coincide with S. Thus, for each prime p, there is precisely one Sylow p-subgroup of G. 6.13. If g E G, then g = gl + g2 where the order of gl divides 4 and the order of g2 divides 9; Let G be an abelian group of order 36. Prove that: (i) (ii) G = A EEl B where A is the Sylow 2-subgroup and B is the Sylow 3-subgroup of G. Solution: (i) Let g be of order 2t3s• Put 2t = m and 3s = n. Then, since (m,n |
) = 1, there exist integers a, b such that am + bn = 1. Hence g = (am + bn)g = (am)g + (bn)g = gl + g2' say Now since ngl = n(am)g = a(nm)g = 0, and similarly mg2 = 0, (i) is proved. (ii) Clearly A nB = {O}, so A + B = A EEl B. Now if g E G, then, by (i), g = 91 + g2 where 91 is of order dividing 4 and g2 is of order dividing 9. By the preceding problem, the set of all elements of order a power of 2 is the Sylow 2-subgroup A, and so gl EA. Similarly g2 E B. Hence g E A + B and we conclude that G = A EEl B. 182 6.14. ABELIAN GROUPS [CHAP. 6 Show that the group C of complex numbers is with respect to addition the direct sum of the sub group consisting of all the reals and the subgroup consisting of all the pure imaginary numbers. Solution: Let R = {a + iO I a an arbitrary real number}, and let 1= {O + ib I b an arbitrary real num i2 = -1). Then clearly R and I are subgroups of C, RnI = {O}, and, if a+ib E C, ber} a + ib = (a + iO) + (0 + ib) belongs to R + I. Thus C = R EB I. (here 6.15. Let G = H EB K and H = L EB M. Prove that G = L EB M EB K. Solution: Every element g of G can be expressed in the form g = h + k where hE Hand k E K. But h = l + m where l ELand mE M. Hence g = l + m + k. Now if g = II + ml + kl with II E L, m 1 E M and kl E K, put II + m 1 = hi E H. Then g = h + k = hi + kl and consequently h = hi and k = k 1• As h = l + m = II + ml' l = II and m = mi' Hence the result. h. (See Note on page 177.) Infinite direct sums It is convenient to |
label the subsets of a set X with the elements of a second set. We are already familiar with such a device, e.g. in labeling a collection of sets AI, A 2, • •• we have labeled with 1= Z. In general, if I is an arbitrary set we shall denote by Ai, i E [, such a collection of labeled sets. A collection of labeled sets Ai, i E [, is called an indexed family. More formally, let (): I ~ X be an onto mapping. Then () is said to be an indexing with the elements of I of the set X. We will denote the image of i by Xi, i.e. Xi = i(). The collection Xi, i E I, is then called a family of indexed sets. We generalize our definition of direct sum to apply to the direct sum of an infinite number of subgroups. Definition: An abelian group G is said to be the direct sum of its subgroups Gi, i E I, if there is a unique expression (but for order) for g for each g E G, g =/= 0, of the form g = gl +.., + gk where gj E Gj', j = 1,..., k, with 1',2',..., k' distinct elements of I and no gi is zero. Note that as G is abelian, gl +... + gk = gk + gk-l +... + gl for example; hence the uniqueness of the expression is understood to be without regard to the order of the elements g 1, •••, gk. We write G = ~ Gi • iEI If I is finite, it is easy to see that a group which is the direct sum in the sense of the above definition is also the direct sum in the sense of the definition of Section 6.la, and conversely. Usually we will use the definition of Section 6.la whenever I is finite. We note that if G = ~ Gi and i,j E I, i =/= j, then GinGj = {OJ. For if x E GinGj and x =/= 0, then x is expressible as i E I and x = gl where gl E Gi x = g2 where g2 E Gj But this contradicts the definition of direct sum. Sec. 6.1] INFINITE DIRECT SUMS 183 The analog of Theorem 6.1 |
is the following: Theorem 6.1': Let Gi, i E I be subgroups of the abelian group G. Suppose each element of G can be expressed as the sum of elements of the subgroups Gi • Suppose also that if 1',2',..., k' are distinct elements of I and that the equation o = gl +.., + g1c where gj E Gj' holds if and only if gl = g2 =... = gk = O. Then G = ~ Gi • iEI As the proof is similar to that of Theorem 6.1, we omit it. Again, as in Section 6.1a, the question arises: is an indexed family of abelian groups, does there exist a group G which is the direct sum of isomorphic copies of the groups Gi? This question is answered in the following theorem: if Gi, i E I Theorem 6.3: Let Gi, i E I be an indexed family of abelian groups. Then there exists an abelian group G which is the direct sum of groups isomorphic to Gi • Proof: Let G = {B I B: 1-7 U Gi, iB E Gi for all i E I, and iB is the zero element of 'l' = B + cp by i'l' = iB + icp. Gi for all but a finite number of i E I}. If B, cp E G, define i E I We assert that G is a group. First, if B, cp E G, then B + cp is clearly a mapping of I into i ~ I Gi and B + cp maps all but a finite number of the elements of I onto zero elements. Hence B + cp E G. Note that B + cp = cp + B so that G is abelian. Next, G is associative. For if CPI' CP2' CP3 E G, and if i E I, i(( CPI + CP2) + CP3) = i( CPI + CP2) + iCP3 Hence (CPI + CP2) + CP3 = CPI + (CP2 + CP3)' iCPI + iCP2 + iCP3 = iCPI + (iCP2 + iCP3) iCPI + i( CP2 + CP3) = i( CPI + (CP2 + CP3)) The mapping 'rj: i -70 E Gi for |
all i in I is the identity of G. For if BEG, then for all i in I, i('rj + B) = i'rj + iB = 0 + iB = iB so that 'rj + B = B. Finally, if BEG, define cP: i -7 -(iB). Then Thus B + cP = 'rj and cP is the inverse of G. i(B + cp) = iB + icp = iB + (-(iB)) = 0 = i'rj' A A A Now we prove that if Gi = {B I BEG and jB is the zero of Gj for all j in I with perhaps the exception of iB}, then Gi ~ Gi • Note that Gi is a subgroup of G; for if B, cP E G, then j'l' = jB - jcp = 0 - 0 = O. Thus 'l' E {J.j. on putting 'l' = B - cP we note that if j oF i, j E I, Next, let v.: G. -7 G. be defined by Bv. = iB, BEG.. Clearly v. is a mapping of G. into Gi • To see that Vi is one-to-one, suppose B, cP E Gi and BVi = cpvi• This means that iB = icp. But jB = jcp = 0 for every j E I, j oF i and so B = cpo Next we prove v. is onto. Let a E Gj and define B: 1-7 i ~ I Gi by iB = a and jB = 0 if j oF i. Then B E Gi and Bv; = a, and so Vi is one-to-one and onto. Vi is a homomorphism; for if B, cP E Gi and 'l' = B + cP, tA t then Therefore Vi defines an isomorphism. (B + cp)vi = 'l'vi = i'l' = iB + icp = BVi + cpvi Finally we show that G = ~ Gi. We already noted that G is abelian. We need to show that if BEG, B = BI +... + Bk where each Bi belongs to one of the subgroups Gj • If BEG, then iB |
is not the zero of Gi for only a finite number of elements of I, say iI,..., ik • Let BI E Gil be such that ilBI = ilB, B2 E Gi2 be such that i2B2 = i2B,.•., Bk E Gik be such that ikBk = ikB. It is clear that iEI A A 184 ABELIAN GROUPS [CHAP. 6 o = 01 +.., + Ok Now suppose that 7], the zero of G, is of the form 7] = 01 +... + Ok " where OJ E Gr and 1',2',..., k' are distinct elements of I. Then for 1 ~ j ~ k, (since 1',2',..., k' are distinct) from which OJ = 7] for every j = 1,..., k. Thus j'7] = 0 = j'(01 +... + Ok) = j'Oj i E I (Remark: Problem 6.17 shows how the mappings of this proof link up with cartesian G = ~Gi products.) The existence of direct sums is a very powerful result, and we will use it repeatedly. We note the important result that: given Gi ~ Hi for each i E I, then if G is the direct sum of its subgroups Gi and H is the direct sum of its subgroups Hi, we can conclude G Eo< H (Section 6.1c). Problems 6.16. Let G be an abelian group with subgroups G;, Gj n UP U Gi) = {O} for each ( i E I i".; i E I. Suppose that G = UP ( U Gi ) iE I and that j E I. Prove that G is the direct sum of its subgroups G;, i E I. Solution: We need only show that if o = U1 +... + Uk where Ui E Gi, and 1',2',..., k' are distinct elements of I, then U1 = U2 =... = Uk = 0 SUppose, if possible, that some Ui is not zero, say U1 oF O. Then -U1 = U2 +... + Uk U1 = U2 = '" = Uk = 0 and the result follows. and U2 +... + Uk E up( U Gi)' i E I i".1' |
By hypothesis then, Hence ~.17. Compare the construction of a direct product involving cartesian products with the construction involving mappings by showing that an ordered pair can be thought of as a mapping. Hence show that the two groups obtained are isomorphic. Solution: An ordered pair can be thought of as a mapping from {1,2}. The image of 1 gives the entry in the first position, the image of 2 gives the entry in the second position. Let G = G 1 X G2 and let G be the group as constructed in Theorem 6.3 with I = {1,2}. Let P: G ~ G be defined by (lp = (1(1,2(1) for any (I E G. Then p is clearly a one-to-one and onto mapping. Also p is a homomorphism. For if 'I' = (I + </>, «(I + </»p =,yp = (1'1',2'1') = (1(1 + 1</>,2(1 + 2</» = (1(1,2(1) + (1</>,2</» = (lp + </>P 6.18. Let 11' be the ratio of the circumference of a circle to its diameter. Let G be the subgroup of the additive group of the real numbers generated by the numbers 11',11'2,11'3,.... Let Gi = UP(11'i), i = 1,2,.... Prove that G = ~ Gi where P is the set of positive integers. Use the fact that 11' is not the root of any polynomial with integer coefficients. iEP Solution: Clearly each element of G is of the form U1 +... + Un where each Ui belongs to one of the implies that U1 = U2 =... = groups G1, G2,. •.• Then we need only show that 0 = U1 +... + Un Sec. 6.1] HOMOMORPHISMS OF DIRECT SUMS; FREE ABELIAN GROUPS 185 gp = 0 where the gi E Gi" with 1',...,n' distinct elements of {1,2,... }. Now gi = Zi".i' where Zi E Z and i = 1,...,n. Then o = Zl".l' + Z2".2' +... + zn".n' If not |
all zi are zero, ". is the root of a polynomial with integer coefficients, contrary to the statement in the problem. Hence the result. c. The homomorphic property of direct sums and free abelian groups Let G = A EEl B and let H be a group which contains isomorphic copies A and B of A (but not necessarily that H = A EEl B). and B respectively. Suppose that H = A + B What connection, if any, is there between G and H? It turns out that H is a homomorphic image of G. This follows from Theorem 6.4. This theorem when applied to particular cases leads also to important results: (1) Theorem 6.5 and (2) the concept of a free abelian group. Theorem 6.4: Let G = A EEl B and let H be any group. Let 0, cp be homomorphisms of A into Hand B into H respectively. Then there exists a homomorphism ~: G ~ H such that ~ IA = 0, ~ IB = cp. Proof: If g E G, then g = a + b uniquely where a E A, b E B. Define g~ = aO + bcp. ~ is uniquely defined and so is a mapping of G to H. Note that if gi = ai + bi where ai E A and bi E B (i = 1,2), then alO + a20 + blCP + b2CP (a l + a2)O + (b l + b2)cp alO + blCP + a20 + b2CP = (a l + bl)~ + (a2 + b2)~ Hence ~ is the required homomorphism as ~ IA = 0, ~ IB = cp. In exactly the same way we can prove that if G = ~ Gi and if for each i E I, Oi: Gi ~ H is a homomorphism of Gi to H, then there exists a homomorphism 0: G ~ H such that 8 IGi = Oi. We shall often say that 0 extends the mappings 8i, or that 8 is an extension of the mappings Oi. i E I We use this result to prove: Theorem 6.5: Let Gi """ Hi, i E I. If G is the direct sum of its subgroups Gi and H is the direct sum of its subgroups Hi, then |
G """ H. Proof: Let Oi: Gi ~ Hi be an isomorphism for all i E I. Then by Theorem 6.4 there i = Oi for each i E I. To prove 0 is an exists a homomorphism 8: G ~ H such that 0 IG isomorphism, we need only show that its kernel is trivial since 0 is clearly onto. If g E G, then g = gl +... + gn where gj belongs to the subgroup Gj, with 1',2',..., n' distinct ele ments of I. Thus gO = gl8l' +... + gn8n' = hI +... + hn, where hi E Hi' Then g8 = 0 only if each hj = 0, since H is the direct sum of its subgroups H;. Since h j = gjOj' and OJ' is an isomorphism, we have gj = 0 and g = O. Thus Ker 0 = {O}, and so G"""H. We will now apply Theorem 6.4 when each Gi is infinite cyclic. To begin with, let C = gp(c) be infinite cyclic and H any abelian group. Note that if cp is a mapping of {c} into H, then there exists a homomorphism 8: C ~ H such that 0 I{c} = cp. The homo morphism 0 is simply defined by putting (rc)O = r(ccp) for each l' E Z. It is readily seen that this does define a homomorphism. 186 ABELIAN GROUPS [CHAP. 6 Now let each Gi be infinite cyclic, Gi = gp(Ci), i E I. Let X = {c; liE I}. If (J: X ~ H where H is any abelian group, then there exists a homomorphism (J*: ~ Gi ~ H such that i E I (J* IX = (J. For, corresponding to each Gi, we know from the remark above that there exists (Ji: Gi ~ H such that Ci(Ji = cd). Hence it follows by Theorem 6.4 that a homomorphism there exists a homomorphism (J*: ~ Gi ~ H which agrees with (Ji on Gi. So we have the required result. i E I Corollary 6.6: The direct sum G = ~ |
Gi satisfies the following condition: for every i E I mapping phism (J*: G ~ H such that (J* IX = (J. (J: X ~ H, H any abelian group, there exists a homomor A group G which contains a subset X such that (i) G = gp(X), (ii) for every mapping (J: X ~ H, H any abelian group, there exists a homomorphism (J*:G~H such that (J*IX = (J, is called a free abelian group. G is said to be freely generated by X and X is called a basis for G. We have shown that the direct sum of infinite cyclic groups is a free abelian group. Conversely we have the following Theorem 6.7: If G is a free abelian group freely generated by a set X = {Xi liE I}, then G is the direct sum of its subgroups Gi = gp(Xi) and each Gi is in finite cyclic for all i E I. Proof: This theorem is proved by showing that G is isomorphic to a direct sum of in finite cyclic groups. To this end let H be the direct sum of its subgroups Hi, H = ~Hi i E I where Hi = gp(hi) is an infinite cyclic group generated by hi. (We know such a direct sum exists by Theorem 6.3.) Let (J: X ~ H be the mapping defined by Xi(J = hi. Then (J can be extended to a homomorphism (J* of G into H, by the definition of a free abelian group. On the other hand H is the direct sum of the infinite cyclic groups Hi. Thus by Corollary 6.6, the mapping cf>: {hi liE I} ~ X defined by hicf> = Xi can be extended to a homomorphism cf>* of H into G. Actually cf>* and (J* are inverse isomorphisms. To see this, suppose g E G. Then g = nIXI' +... + nrXr' where 1',..., r' E I and nl,..., nr E Z. Accordingly, (g(J*)cf>* = [nl(xI,(J*) +... + nr'{xr,(J*)]cf>* = (nIhl' + |
... + nrhr,)cf>* = nl(hl,cf» +... + nr(hr'cf» = nIXI' +... + nrXr' = g and so (J*cf>* is the identity mapping on G. Similarly cf>*(J* is the identity mapping on H. This implies that (J* is a one-to-one mapping of G onto H. For if g, g' E G, then g(J* = g'(J* implies that (g(J*)cf>* = (g'(J*)cf>*. Since (g'(J*)cf>* = g', we have g = g'. Furthermore if h E H, then h = (hcf>*)(J*. Thus (J* is one-to-one and onto. (g(J*)cf>* = g and Note that each Gi is infinite cyclic, since (J* is an isomorphism and Gi(J* = Hi. Finally we show that G is the direct sum of its subgroups Gi • If 1',2',..., r' are distinct elements of I and gl,..., gT are nonzero elements of GI "..., Gr, respectively, then if (6.2) it follows that gl(J* +... + gr(J* = O. But then, as gi(J* E Hi' and gi(J* =1= 0, we have a contradiction as H = ~ Hi. Hence (6.2) does not hold. Finally gp(Gi liE I) = G, and so G = ~ Gi • i E I iEI Sec. 6.1] HOMOMORPHISMS OF DIRECT SUMS; FREE ABELIAN GROUPS 187 Corollary 6.8: Every abelian group is the homomorphic image of some free abelian group. Proof: If G is an arbitrary group whose elements are gi, infinite cyclic, then as we have seen, F = ~ gp(Xi) B: Xi 4 gi extends to a homomorphism of F onto G. i E I i E I, and gp(Xi), is is free abelian and the mapping i E I, Problems 6.19. If Igp(a)1 = m, Igp(b)1 = nand (m, n) |
= 1, then G = gp(a) EB gp(b) is cyclic of order mn. Solution: We show that G = gp(a + b). If l is the order of a + b, then implies Ia = lb = 0, by the definition of a direct sum. Consequently I is divisible by the order of a and the order of b. Since (m, n) = 1, mn I I and we conclude mn = I, so that G = gp(a + b). l(a + b) = la + Ib = 0 6.20. Find IGI where G is the direct sum of n cyclic groups of order 3. Solution: Let GI,..., Gn be subgroups of G and suppose G = GI EB ••• EB Gn, where each Gi is of order 3. Each expression of the form gl + g2 +... + gn' where gi E Gio gives rise to unique elements of G. There are 3 different possible choices for glo 3 for g2, etc. Hence the total number of possible choices is 3·3····· 3 = 3n. Hence IGI = 3n • 6.21. 6.22. 6.23. 6.24. Prove that A is a direct summand of G if and only if there exists a homomorphism 0 of G onto A (Hint. Use K = Ker o. Also consider g - go to prove such that 0IA is the identity on A. g E A +K.) Solution: Suppose that G = A EB B. Then the identity homomorphism on A and the trivial homomorphism on B extend to a homomorphism 0: G --> A which satisfies the required conditions (Theorem 6.4). Conversely, if such a homomorphism exists, let K = Ker 0 and let x E An K. Then x = XO, since x E Ker o. Hence A nK = {O} and so gp(A, K) = (g - a)o = go - ao = a - a = 0 and as 0 IA = identity on A. But xo = 0, A EB K. Now letting g E G, go = a for some a E A. Then g - a E K. Thus g E A EB K and G = A EB K. implies nx = 0 and n is the smallest positive A group G is said to |
be of exponent n if x E G integer with this property. Let H be the direct sum of two cyclic groups of order n, generated by Xl and X2 respectively. Put X = {Xl' X2}' Prove that if G is any group of exponent nand 0 is a mapping of X into G, then there exists a homomorphism 0* of H into G such that 0* Ix = O •• Solution: Let HI =.qp(XI) and H2 = gp(X2)' There satisfying X10I = xlo, for gp(xlo) is cyclic and is of order dividing n. Similarly there is a homomorphism O2 : H2 --> G satisfying x202 = x20. Thus there is a homomorphism 0*: HI EB H2 --> G such that 0* IHI = 0 1 and 0* IH2 = 92, by Theorem 6.4. The result follows. is a homomorphism 0 1 : HI --> G Let G be freely generated by a finite set X, IXI = n. Prove that every element of G is uniquely of the form mixi +... + mnxn, where mj E Z and Xi EX. Solution: Let Gj = gp(Xi)' Now by the theorem on free abelian groups we know that G = GI EB..• EB Gn, and each G; is infinite cyclic. Then each element of G is uniquely of the form gl +... + gn where gi E Gi. But we know from the theory of infinite cyclic groups that gi = mixi uniquely. The result follows. Let G be the direct sum of cyclic groups Gi of order 2 where Let E denote the set of even positive integers. Then H = gp(G i liE E) group of G. Prove that H "" G. i E P, the set of positive integers. is clearly a proper sub Solution: Let 0i: Gi --> G2i for all i E P be an isomorphism. (Such an isomorphism exists, since all the Gi are cyclic of order 2.) We apply Theorem 6.5 to obtain the result. 188 6.25. ABELIAN GROUPS [CHAP. 6 Let G = A EB B where A is cyclic of order 32 and B is cyclic of order 52. Prove that aut (G) ~ aut (A |
) EB aut (B) (hard) and hence compute laut (G)!. (aut (G) is the automorphism group of G (see Section 3.6a, page 83).) Solution: If a E aut (A), identity on B. is one-to-one and onto.) Similarly for elements [3 of aut (B). We use the symbols a*, [3* represent corresponding automorphisms of G. Note that then a can be used to form an element of aut (G) by letting it act as the (Theorem 6.4 states that a extends to a homomorphism. We must check that it to (a + b)a*[3* = (aa + b)[3* = aa + b[3 = (a + b)[3*a* from which a* [3* = [3*a*. Note that a* = [3* implies a* = [3* = " the identity automorphism. The mapping a --> a* is an isomorphism of aut (A) into a subgroup of aut (G). The mapping [3 --> [3* is an isomorphism of aut (B) into a subgroup of aut (G). As (aut (A))* n (aut (B))* = {,}, and as the elements of (aut (A))* commute with the elements of (aut (B))*, we have (aut (A))* + (aut (B))* = (aut (A))* EB (aut (B))*, by Theorem 5.16, page 144. Now let ()IA induces an auto morphism ()A on A, for A() must go into a subgroup of order 9 and by Sylow's theorem there is only one subgroup of order 9 (as G is abelian). Similarly ()IB induces an automorphism ()B on B. Then () E aut (G). (a+b)()1()~ = (a()+b)()~ = a()+b() = (a+b)() from which ()1()~ = (). Thus aut (G) = (aut (A))* EB (aut (B))* and, by Theorem 6.5, aut (G) ~ aut (A) EB aut (B) To compute laut (G)I we must compute a E aut (A) and let A = gp( |
a). a must take a onto an element of order 9; so if aa = aT, (r,3) = 1. Hence the pos sibilities are r = 1,2,4,5,7,8. Each of these gives rise to an automorphism of A, as can be checked. Thus laut (A)I = 6. Similarly laut (B)I = 20. Accordingly, laut (G)I = 6 X 20 = 120. laut (B)I. Let laut (A)I and 6.2 SIMPLE CLASSIFICATION OF ABELIAN GROUPS, AND STRUCTURE OF TORSION GROUPS a. Tentative classifications: torsion, torsion-free, and mixed Consider the following three examples of abelian groups: Q, the additive group of rationals; QIZ the factor group of the additive group of the rationals by the integers; C, the multiplicative group of complex numbers. Each of these groups is not isomorphic to the others, but how would we prove that? One way is to examine the orders of the elements of the groups. N ow every element of Q except 0 is of infinite order and every element of QIZ is of finite order. For if r E Q, r = min where m, n are two integers. Thus n(r + Z) = nr + Z = m + Z = Z. Let us, to avoid confusion, continue to use the multi plicative notation for C. We assert that C has elements of infinite order and also elements of finite order. Recall that the identity of C is 1. Note that (-1)2 = 1 implies that -1 is of order 2 and 3T = 1 if and only if r = O. Hence -1 is of finite order and 3 is of infinite order. Summarizing, we have (i) Q has every element but the identity of infinite order. (ii) QIZ has every element of finite order. (iii) C has elements of finite order and elements of infinite order. It is then easy to see that the three groups are not isomorphic. If G is a group in which every element other than the identity is of infinite order, G is said to be torsion-free. If G is a group in which every element is of finite order, G is said to be a torsion group. If G has both an element of |
infinite order and an element (not equal to the identity) of finite order, G is said to be mixed. These three concepts provide us with a rough classification of abelian groups and, as we have seen above, distinguish between Q, Q/Z and C. Sec. 6.2] SIMPLE CLASSIFICATION. STRUCTURE OF TORSION GROUPS 189 Problems 6.26. Let G be the direct sum of torsion groups. Prove that G is a torsion group. Solution: Let G = l. Gi • If the order of Xj is Pj' then i E I If g E G, g = Xl +... + Xn for some integer nand Xj belongs to some G j.. (Pl'...• Pn)g = (Pl····· Pn)Xl + '" + (Pl'...• Pn)Xn = 0 + '" + 0 = 0 6.27. Prove that QIZ and G, the direct sum of groups Gi (i E Z) where each Gi is cyclic of order 2, are not isomorphic. Solution: Following the method of Problem 6.26, it is easy to prove that every nonzero element of G is of order 2. Since! + Z is of order 3, QIZ and G are not isomorphic. 6.28. Prove that QIZ and G, the direct sum of groups Gi (i E Z) where each Gi is cyclic of order 3i, are not isomorphic. Solution: G is easily shown to have every element of order some power of 3 by following the method of Problem 6.26. Since -l- + Z is of order 4 in QIZ, G and QIZ are not isomorphic. b. The torsion subgroup In Section 6.2a we introduced a tentative classification of abelian groups into torsion free, torsion and mixed groups. In this section we consider the question of whether it would not be possible to split a mixed group into a torsion-free group and a torsion group. This would provide the following program for investigating abelian groups: (1) Investigate torsion-free groups. (2) Investigate torsion groups. (3) Investigate how they may be put together to form mixed groups. Such a program is found to be too difficult to accomplish completely, but it does lead to |
some significant results. Let T(G) be the set of all elements of G of finite order. Then T(G) is a subgroup of G, as we show in the following Theorem 6.9: T(G) is a subgroup of G (termed the torsion subgroup of G). G/T(G) is torsion-free. Proof; Let a, bEG be of order m, n respectively. Then mn(a - b) = mna - mnb = 0 - 0 = 0 Thus if a, b E T(G), a - b E T(G) and T(G) is a subgroup of G. Now consider G/T(G). Assume g + T(G) is of finite order n, i.e. n(g + T(G)) = ng + T(G) = T(G). It follows that ng E T(G). As T(G) consists of all the elements of G of finite order, there exists m such that m(ng) = O. Then g is of finite order and g E T(G); hence g + T(G) = T(G). Therefore the only element of finite order in G/T(G) is the zero T(G). Thus G/T(G) is torsion-free. Problems 6.29. Prove that if G is a group and H a subgroup of G such that GIH is torsion-free, then H contains the torsion subgroup of G. Solution: Let g in G be of finite order. Then g + H is of finite order in GIH. Since GIH is torsion-free, g + H = H. This means g E H, and so every element of finite order in G is contained in H, i.e. the torsion subgroup of G is contained in H. 190 ABELIAN GROUPS [CHAP. 6 6.30. Is the set consisting of 0 and of all elements of infinite order of a group automatically a subgroup? Solution: No. For example, let G = H EEl K where H = gp (h) is infinite cyclic, and K = gp (k) order 2. Now h + k and -h are both of infinite order. However (h + k) + (-h) = k order. is of is of |
finite 6.31. Prove that if the set consisting of 0 and the elements of infinite order of a group G constitutes a subgroup, then G is either a torsion-free group or a torsion group. Solution: Suppose G is mixed and H = {h I hE G and h is either 0 or of infinite order}. Since G is mixed, we have g E G of infinite order and g'("# 0) E G of finite order. Now g' - g is of infinite order so that (g' - g) E H. But as H is a subgroup, (g' - g) + g = g' E H. Therefore g' is 0 or an element of infinite order, contradicting the choice of g'. Hence G is not mixed. 6.32. Prove that if T is the torsion subgroup of G, then TnH is the torsion subgroup of any given subgroup H of G. Solution: If a E H is of finite order, then a E T. Hence a E TnH, and so the torsion subgroup of H is contained in TnH. Conversely TnH consists of elements of finite order, and so TnH is contained in the torsion subgroup of H. Thus we have proved that TnH is the torsion subgroup of H. 6.33. Find the torsion subgroup of R/Z where R is the group of real numbers under addition and Z is the subgroup of integers. Solution: Suppose r + Z is of finite order (r E R). Then for some nonzero integer n, n(r + Z) = Z. But n(r + Z) = nr + Z, and so nr E Z. This means that r is a rational number. Thus T(R/Z) ~ Q/Z, then a = m/n where Q is the subgroup of rational numbers. On the other hand, if a + Z E Q/Z, where m,n E Z and n"# O. So Hence a + Z is of finite order and Q/Z ~ T(R/Z). Thus we have proved that Q/Z = T(R/Z). n(a+Z) = n(m/n+Z) = n(m/n. Structure of torsion groups. Prufer groups A group G is called |
a p-group or a p-primary group for some prime p if every element of G is of order a power of p. (If G is finite, it follows that the order of G is a power of p; see Problem 5.6, page 132. The definition of p-group given here thus coincides with that In this section we show that a torsion group is of Chapter 5 when the p-group is finite.) built out of p-groups. Thus the study of torsion groups becomes essentially the study of p-groups. Theorem 6.10: Let G be any torsion group and let Gp = {g I g has order a power of p}, p any prime. Then if II is the set of all primes, G = ~ Gp p E II Proof: We let the reader show that each Gp is a subgroup of G. Suppose that g E G and is of order p;lp;2... p:n, PI,..., Pn distinct primes and rl,..., rn positive integers. Let q = p~l... p:~-/; then (q,p:n) = 1. Thus there exist integers a and b such that aq + bp:n = 1. It follows that g = aqg + bp:ng. Now aqg is of order p:n, and so aqg E Gpn. bp:ng is of order q. Since q is less then the order of g, we may assume inductively that bp:ng is the sum of elements belonging to GPn-t> G pn - 2,..., GPI' Thus every element of G can be expressed as the sum of elements belonging to the Gp, i.e. G is generated by the subgroups Gp • Sec. 6.2] SIMPLE CLASSIFICATION. STRUCTURE OF TORSION GROUPS 191 To show that G = ~ Gp, we must prove that gl +... + gn = 0 (where gi E GPi and PI,..., Pn are distinct primes) occurs only for gl = g2 =... = gn = O. pEn We proceed by induction on n. For n = 1 it is certainly true. If true for n, consider gl+... +gn+l = 0 and let gn+l be of order P~+l' Then and By |
the inductive hypothesis, we have Now P~+lgl = 0 implies gl = 0 as gl is of order some power of PI and PI =1= Pn+l. Arguing similarly for g2,..., gn, we have gl = g2 =... = gn = O. Hence also gn+ I = 0 and we con clude that G = ~ Gp • pEn The subgroups Gp are called the p-components of G. Example 1: Let us apply this theorem to Q/Z, i.e. the additive group of rationals modulo the integers. Q/Z is clearly a torsion group. {x + Z I x + Z of order a power of p} = {x + Z I prx E Z} {m/pT + Z I for various integers rand 0 """ m < pr~} By the theorem, Q/Z = ~ (Q/Z)p pEn (Q/Z)p is called the p-Priifer group (also called a group of type poo). In Section 6.4 the p-Priifer groups will be fundamental. Note that the p-Priifer group (Q/Z)p = U Cr where Cr = gp(l/pr + Z), since (Q/Z)p = {m/pr + Z I for various integers rand 0""" m < pr-l}. Clearly (Q/Z)p d Cr and (Q/Z)p C U Cr. The result follows. r=1 r=l 00 00 We now have at our disposal cyclic groups of all orders, the additive group of rationals, and the p-Priifer groups, together with all their direct sums. These, as we shall prove, constitute a large class of abelian groups. Problems 6.34. Use Theorem 6.10 to prove that an abelian group of order pq, where p and q are different primes, is the direct sum of a cyclic group of order p and a cyclic group of order q. Solution: Let G be of order pq. Then by Theorem 6.10, G = Gp EB Gq ; for if r is any prime other than p or q, Gr = {O}. Why does Gr = {O}? If g E G" then, as G is of order pq, pqg = O. Hence r divides pq, which |
is not the case. Note that Gp ¥- {O}, Gq ¥- {O} by Proposition 5.9, page 137. IGpl divides pq. Since Gq ¥- {O}, IGpl = p or q. As the elements of Gp are of order a power of p, IGql = q. Hence, as the only group of prime order is cyclic, it follows that IGpl = p. Similarly we have the result. 6.35. Show that a direct sum of p-groups is again a p-group. Solution: Let G = ~ Gi where each Gi is a p-group. Let g E G. Then g = gl +... + gn where each gi E Gi" 1',2',..., n' E I. Let pr be the maximum order of the gi' Then prg = O. Thus the order of g is a power of p, and so G is a p-group. i E I 192 6.36. 6.37. ABELIAN GROUPS [CHAP. 6 Show that if G is a p-Priifer group, for each integer k"" 0 and each element g E G there exists an element hE G such that kh = g. (We shall call a group with this property divisible.) Solution: Let g E G. Then g = 1n/pr + Z where 0 =:: m < pro Let k = pSl where p and l are co-prime, and let h! = m/pr+s + Z. Then pSh! = g. As land pr+s are co-prime, there exist integers a and b such that al + bpr + S = 1. Therefore h! = (al+ bpr+s)h! = alh! + bpr+sh! = alh! Put h = ah1• Then '" i=l Let G = ~ Gi where Gi is a cyclic group of order Pi and P1' P2. " are the primes in ascending order of magnitude. Let H be a p-Priifer group for any prime p. Prove that G is not isomorphic to H. Solution: Let Pj be a prime different from p. Then the Pjth component Gpj "" {O} but Hpj = {O}. Thus G is not isomorphic to H. d. Independence and rank We introduce here an important concept in abel |
ian group theory, the concept of rank. Let G be an abelian group. A subset X of G is called independent if whenever Xl,..., Xn are distinct elements of X and n1,..., nr are integers such that then n1X1 =... = nrXr = O. n1X1 +... + nrXr = 0 (6.3) Note that if G is torsion-free and X is independent, then equation (6.3) implies n1 = nz =... = nr = O. Suppose X = {Xi liE l} and Xi "1= Xj for i "1= j. Then if X is an in dependent set, it follows readily that gp (X) = ~ gp(Xi). i E I We need one further definition. An element X in a group G is dependent on a subset X of G if nx + n1X1 +... + nrXr = 0 for some choice of Xl,..., XT E X, nand nj E Z and nx "1= O. In other words, X is dependent on the subset X if there is an integer n with nx"I= 0 and nx E gp(X). We say Y C G is dependent on Xc G if every element of Y is dependent on X. Observe also that if G is torsion-free with subsets X, Y and Wand if X is dependent on Y and Y is dependent on W, then X is dependent on W. For if X E X, then for some integer n"l= 0, and integers n1,..., nT, nx = n1Y1 +... + nTYT (Yi E Y). Since Y is dependent on W we can find integers m1 "1= 0,..., m T "1= 0 such that miYi E gp(W) for i = 1,..., r. Put m = ml... m T. Then i = 1,...,r, and consequently mnx E gp(W). Furthermore, clearly mniYi E gp(W) for since G is torsion-free, mnx"I= O. Thus every element of X depends on W. The main result that we shall now prove is called the Steinitz Exchange Theorem. Theorem 6.11: Let G be tors |
ion-free and let A = {a1,...,am} be an independent subset of G. Suppose B = {b 1, • • •, bn } is another subset of G such that A is dependent on B. Then n::=" m and B depends on Au C where C is a subset of Band ICI = n-m. Proof: We will use induction on m. For m = 1 it is clear that n ~ m. Now a1 depends on B means that there exist integers z,..., Zn such that Z "1= 0,, za1 + z l b1 +... + znb n = 0 Sec. 6.2] SIMPLE CLASSIFICATION. STRUCTURE OF TORSION GROUPS 193 and thus for some integer i, Zi =1= 0, 1 ~ i ~ n. Hence B depends on {at} U (B the result holds for m = 1, with C = B - {b i }. {b i }). Thus N ext we assume that the result holds for m = r, and consider the case m = r + 1. Then by the inductive hypothesis, n ==== r. Now {aI,..., a,.} depends on B, and so inductively B depends on {al,...,aT}UD where Dc;;;,B and IDI=n-r. But {aT+t} depends onB and B depends on {aI,..., aT} U D. Then by our remarks above, {aT+ I} depends on {aI,..., aT} U D. Thus we can find integers y =1= 0, YI,..., YT, Zl,..., Zs such that yaT+I = Ylal +... + yraT + zldl +... + zsds, dl,..., ds E D Suppose, if possible, that Zl = Z2 =... = Zs = 0. Then yar+l = Ylal +... + yraT implies that the elements aI,..., aT+! are not independent. So some Zj =1= 0. Let C = D {di }. Then it is clear that Since B depends on {al,...,ar}UD, then B depends on {al,...,ar+I}UC |
. Finally Cc;;;,B and ICI=IDI-1. Thus {al,..., aT + I} U C. {al,..., aT} U D depends on ICI = (n - r) - 1 = n - (r + 1) = n - m as desired, and the proof of Steinitz's theorem is complete. Let us call a subset 8 of a torsion-free abelian group G a maximal independent set if 8 is independent and (i) (ii) if g E G and g tl 8, then 8u {g} is not an independent set. Suppose now that G, a torsion-free abelian group, has a maximal independent set 8 that is finite. Let T be any other finite maximal independent set. By the Steinitz exchange theorem, 181 ~ ITI. Also by the same theorem, ITI ~ 181. Hence we can without ambiguity define the rank of a torsion-free abelian group G which has a finite maximal independent set 8 to be 181. If G does not have a finite maximal independent set 8, we shall say G is of infinite rank. It is easy to see that if G and H are isomorphic groups, then they have equal ranks. As a consequence of these remarks we obtain a result concerning free abelian groups. If F is free abelian with a finite set of free generators X, then X is a maximal in dependent set of F (see Problem 6.41 below). Hence the rank of F is IXI. Similarly if F is also freely generated by a finite set Y, then rank of F = IYI. Hence IYI = IXI. Thus we have Corollary 6.12: If F is a group freely generated by two finite sets X and Y, then IXI = IYI. We have as yet not proved that all abelian groups have maximal independent sets. To do so we need a result called Zorn's lemma. Before stating the lemma, we consider the following examples: (a) Let cP = {A, B, c, D}, where A = {0,1}, B = {1,2}, C = {0,2}, D = {O, 1,2, 3}. We inquire: does cP have a largest element, i.e. one that contains all the elements of CP? Clearly it does, for |
D:dA, D:dB, D:dC and D:dD. Thus D is a largest element. (b) Let cP = {A, B, C, E}, where A, B, C are as in (a), and E = {1, 2, 3}. Now there is no largest element of CPo We search for some concept replacing that of a largest element. Note that although E is not a largest element, no element other than itself contains it. Then E is called a maximal element. Similarly C and A are maximal elements, whereas B is not. (c) If ICPI is finite, then it is clear that cP has maximal elements. For we choose any element Al of CPo If there is an element of cP that contains Al properly, we call it A 2. If there is an element of cP that contains A2 properly, we call it A 3 • Continuing in this way we get a chain of elements of CP, Al C A2 C... C Ai C.. '. As cP has only a finite number of elements, this chain ends at An, say. Clearly An is a maximal element. 194 ABELIAN GROUPS [CHAP. 6 (d) On the other hand, not all sets P of sets have maximal elements. For example, let An = {O,l,..., n} for n = 1,2,.... Let P = {Ai liE P, the positive integers}. Then P has no maximal element. For if X E P, then X = Ai for some i, and X ~ A i + l, but X""" A i + l • Zorn's lemma establishes a criterion for determining whether a set P of sets has a maximal element. What one needs is some condition for handling an ascending sequence of sets such as the Ai in (d). To state the criterion we need some definitions. implies X = A. (1) We define A to be a maximal set in P if for each X E P, X d A (2) Let C be a subset of P with the property that if X, Y E C. then either X ~ Y or Y ~ X. Then C is called a chain in P (in (d) above, P itself is a chain). We are now in a position to state Zorn's lemma: Let P be a set of sets. Suppose that for every chain C in P, |
U X is an element of P. Then P has a maximal element. XEC We will not prove Zorn's lemma. We take it as an axiom. We could assume a more innocent sounding axiom instead, namely the axiom of choice, which says that an element from each set may be chosen from a collection of sets. The proof of Zorn's lemma can be derived from the axiom of choice (see Problem 6.42 for a sketch of the proof). Using Zorn's lemma we prove Theorem 6.13: Let G be any abelian group. Then G has a maximal independent set. Proof: Let P = {X I X ~ G and X an independent set}. Let C be any chain in P. Let C = {Xi liE I}. To apply Zorn's lemma we must show that U = i 'd r Xi E P. Clearly Is U an independent set? If not, U is a dependent set. This means that it is pos U ~ G. sible to find distinct elements Ul,..., Un E U and integers rl,..., rn such that rlUl +... + rnUn = 0 with at least one riU; """ O. As U = U Xi, Ul E Xl', U2 E X 2• where 1', 2' are elements of I. Then since C is a chain, either Xl' ~ X 2• or X 2• ~ Xl" Thus Ul, U2 both belong to some element of C. Continuing the argument in this way we find that Ul,..., Un all belong to some Xi E C. But this is a contradiction, since every element of C is independent. So U is independent and U E P. We conclude, using Zorn's lemma, that P has a maximal ele ment and this is precisely the maximal independent set required. Hence the result follows. i E r From Theorem 6.13 we conclude that if G is of infinite rank, then G has an infinite maximal independent subset. Problems 6.38. Find the rank of the additive group Q of rationals. Solution: If ml/nl and mJ~ are elements of Q, ml> m2, nl, n2 integers, then Thus every set of two elements is dependent. Accordingly the rank of Q is 1. 6.39. Show that the p-Priifer group has no independent set consisting of two |
elements. Solution: Let x, y be elements of G, a p-Priifer group, x -:;6 0, Y -:;6 O. Then x, y E Cr, say, for some r (see Example 1, page 191, for the notation Cr). Let C,. = gp(g) and let gp(x) be of order pt. Since pr-tg is of order pt, gp(pr-tg) = gp(x). Thus gp(x) = gp(pig) and gp(y) = gp(p;g) for some i, j. If i ~ j, it follows that gp(y) ~ gp(x). (If i"" j, we merely reverse the roles of x and y.) Con sequently x = ry for some integer 0 < r < order of y. Thus (-r)y + 1· x = 0 and x, yare not independent. Sec. 6.2] SIMPLE CLASSIFICATION. STRUCTURE OF TORSION GROUPS 195 6.40. Prove that if Hand K are torsion-free groups of finite rank m and n respectively, then G = H EEl K is of rank m + n. Solution: (Difficult.) Let hI'...,hm be a set of independent elements of H, and k1'..., k n a set of independent ele ments of K. Then the set hI'...,hm, kl>".,kn is independent. For if r1h1 +.,. + rmhm + slk1 +.., + snkn = 0, then But HnK = {O}, and so r1h1 +... + rmhm = -Slk1 Thus, by the independence of hl>..., hm and k1'..., kn' '" - snkn = 0 r 1 = r2 =... = rm = Sl = S2 = '" = Sn = 0 Now suppose that {hI'..., hm' k1'..., kn} is not maximal; say, there exists an element h + k where hE Hand k E K such that {hI>..., hm, k1'..., kn' h + k} is independent. Since {hI'..., hm' h} is not independent, there exist integers tl>..., tm, t not all |
zero such that t1 hI +... + tmhm + th = 0 If t = 0 we have a contradiction to {hI,..., hm} being independent, as then at least one of t 1,..., tm is nonzero. Hence t #- O. Next, {k, kl>..., kn} is not independent, i.e. there exist s, sl'..., sn' not all zero, such that sk + slk1 +... + snkn = O. Hence arguing as above, it follows that s #- O. Thus st1h1 +... + stmhm + st(h + k} + ts 1k 1 +... + tsnkn = s(th + t1h1 +... + tmhm} + t(sk + slk1 +... + snkn} = 0 But as st #- 0, {hI' h2'..., hm, k1'..., kn' h + k} is not independent. Thus {hI' hz,..., hm' kl>..., kn} is a maximal independent set and rank H EEl K = m + n. 6.41. (a) If F is free abelian with a finite set of free generators X, prove that the rank of F is IXI. (b) Prove that F cannot be generated by fewer than IXI elements. Solution: (a) Let X = {Xl,..., x n}. By Theorem 6.7, page 186, G = gp(X1) EEl gp(x2) EEl ••• EEl gp(xn) We proceed by induction on n. For n = 1, G is infinite cyclic, and the rank of G is clearly 1. If true for n = r, suppose n = r + 1. Then gp(X1) EEl ••• EEl gp(xr) is of rank r, and G is the direct sum of a torsion-free group of rank r and a group of rank 1. By Problem 6.40, G is thus of rank r + 1, and the result follows by induction for all n. (b) Let G = gp(gl'..., gr)' X is an independent set. Then by the Steinitz exchange theorem (Theorem 6.11), as X is dependent on {gl'... |
, gr}, n"" r. Thus we obtain the result. 6.42. Let X be a set and cP a collection of subsets of X. Suppose that if A E CP, all subsets of A belong to CPo (a) Prove that if A E cP is not a maximal element, there exists a set A'" = {A, x} E cP with xG!:A. (b) Assume that A * = A if A is maximal, otherwise that A'" has been defined equal to {A, x} with X G!: A as stated in (a). Let C be a chain in CPo Suppose that if Ci' i E I, is a family of elements in C, then U Ci E C. Suppose also that if A E C, A'" E C. Prove that cP has a maximal element. i E 1 Solution: (a) If A E cP is not maximal, there exists a set BE cP such that B #- A and B:2 A. Hence there exists x E B - A. Since {A, x} is a subset of B, {A, x} E CPo (b) Let M = U C. Then ME C. and so M* E C. But M* d M. However, M contains every element of C; in particular, it contains M*. Therefore M = M'" and we conclude that M is a maximal element of CPo CEe Remarks. (1) In assuming that A * can be defined we used implicitly the axiom of choice. (2) The proof of Zorn's lemma requires converting the theorem into this problem. For details see P. R. Halmos, Naive Set Theory, Van Nostrand, 1960. 196 6.43. ABELIAN GROUPS [CHAP. 6 Let G be an arbitrary non-abelian group. Prove that G has a maximal abelian subgroup (i.e. one that is not properly contained in an abelian subgroup of G). Solution: We use multiplicative notation for G since it is not abelian. Let P be the set of all abelian subgroups of G. Let C be a chain in P and let U = U X. Then U is a subgroup of G. For if g, hE U and if g belongs to XI E C and h to X 2 E C, then as either |
XI C X 2 or X 2 C XI, it fol lows that g, h belong to some element X of C. Hence gh- I E X as X is a subgroup, and gh- I E U. Also, gh = hg as X is an abelian subgroup of G. Consequently U is abelian. Hence U E P. By Zorn's lemma, P has a maximal element M, say. M is the maximal abelian group sought. XEC 6.3 FINITELY GENERATED ABELIAN GROUPS a. Lemmas for finitely generated free abelian groups In Section 6.3b we will show that all finitely generated abelian groups are direct sums of cyclic groups. We will do this by using a lemma (Lemma 6.15) about subgroups of free abelian groups. The relationship between Lemma 6.15 and finitely generated abelian groups is easily obtained by noting that all abelian groups are factor groups of free abelian groups. Lemma 6.14: Let G = gp(al) EEl... EEl gp(an) be the direct sum of infinite cyclic groups. If bl = al + r2a2 +... + rnan, where r2,..., rn are any integers, then G = gp(b l ) EEl gp(a2) EEl '" EEl gp(a n ) Proof: As gp(b l, a2,..., an) = gp(al,..., an) = G, we must only show that if Sl,..., Sn are any integers, then implies all Si are O. (6.4) Substituting bl = al + r2a2 +... + rna n into (6.4) and collecting terms, we obtain As G = {aJ} EEl... EEl {an}, sIal + (S2 + slr2)a2 + Thus SI = S2 =... = Sn = 0 and the proof is complete. The next lemma is a crucial one. We recall that a basis CI, •.., Cn for a finitely generated (see Section free abelian group G is a set of elements such that G = gp(cI) EEl ••• EEl gp(c n ) 6.lc). Lemma 6.15: Let G be |
free abelian, the direct sum of n cyclic groups. Let H be a subgroup of G. Then there exists a basis CI,..., Cn of G and integers UI,..•, Un such that H = gp(UICI, U2C2,..., Unc n ). Proof: We use a, b, C to denote basis elements of G, h, k, 1 to denote elements of H, q, r, s, t, u, v to denote integers. We prove the result by induction on n. For n = 1, G is cyclic and the result is a consequence of Theorem 4.9, page 105. Assume the result is true for free abelian groups of rank less than n where n > 1. Let G be free abelian of rank n. We assume also that H # {O}. For if H = {O}, we may take an arbitrary basis CI, •.., Cn for G. Then H = gp(UICI,..., UnCn) where UI =... = Un = O. To every basis we associate an integer, called its size (with respect to H). Let {aI,..., an} be a basis for G and let q be the smallest nonnegative integer such that there exists h E H with q2,..., qn integers (6.5) Then q is termed the size of the basis {aI, a2,..., an}. Assume {aI,..., Un} is a basis of smallest size, i.e. if {b 1, •••, bn} is a basis of G, then the size of {b 1, •••, bn } is not less than q. Sec. 6.3] FINITELY GENERATED ABELIAN GROUPS 197 Let h be as in equation (6.5). We show that q divides q2,..., qn. From the division algorithm, if qi is not divisible by q, qi = riq + Si where 0 < Si < q. Hence h = q(al +riai) +... + Siai +... + qna" But if we put bl = ai, b2 = a2,..., bi = al + riai,..., bn = an, we obtain a basis by Lemma 6.14. Furthermore this basis is |
of smaller size than the size of {al,...,an}, contrary to our assumption. Thus Si = 0 and q divides qi for i = 2,..., n. Let qi = riq. Then h = q(al + rZa2 +... + r"an) Let Cl = al + r2a2 +... + rnan. Then, by Lemma 6.14, {Cl, a2,..., an} is a basis for G. Also (6.6) If k = t1al +... + tnan E H, it follows that tl is divisible by q. For if t! = uq + v with o =:: v < q, then l = k - uh E H has v as its coefficient of al. As v < q, by the minimality of q, v = O. Therefore l = k - uh E gp(a2,..., an) Hence l E gp(a2,...,an)nH = L, say. From this we conclude that if k E H, then where l E L. k = uh + l (6.7) By the inductive hypothesis there exist a basis C2,..., Cn and integers U2,..., Un such that L is generated by U2C2,..., UnCn. Hence by (6.7) every element of H belongs to gp(h, U2C2,..., UnCn). On the other hand, H contains h, U2C2,..., UnCn. Thus Put Ut = q. By (6.6), H = gp(h, U2C2,..., UnCn) Also, Ct,..., Cn is a basis for G. Hence the result follows. Note that if any 14 is negative, we can replace Ci by its inverse -Ci. In this manner we can assume that the 14 are nonnegative. Lemma 6.16: Suppose G = A EB B. Let At, B! be subgroups with At C A, B! C Band N = A! +B!. Then GIN"" AlA! EB BIBt. Proof: Let K = AIAt EB BIB!, and let (): A ~ AIAt and </>: B ~ BIBt be the natural (), </> extend to a homomorphism 'ifr of G into K. Then Ker |
'ifr:;) Ker () = A! homomorphisms. and Ker'ifr:;)Ker</>=Bt. Thus Ker'ifr:;) At+B!. Now let xEKer'ifr. Then x=a+b, x'ifr = (a + At) + (b + B t) and this is the identity element only if a E A! a E A, b E B. and b E B!. Hence x EAt + B t, and so Ker'ifr = At + B t. By the homomorphism theorem (Theorem 4.18, page 117) GIN"" K and the result follows. Corollary 6.17: Let G be free abelian with basis C!,..., Cn. Let H = gp( UtC!,..., UnCn) where Ut,..., Un are nonnegative integers. Then GIH is the direct sum of cyclic groups of orders u~,..., u~, where u{ = Ui if Ui "1= 0 and u; = 00 if Ui = O. Proof: The result follows by repeated application of Lemma 6.16. b. Fundamental theorem of abelian groups The following theorem is called the fundamental theorem of abelian groups. Theorem 6.18: Let G be a finitely generated abelian group. Then G is the direct sum of a finite number of cyclic groups. 198 ABELIAN GROUPS [CHAP. 6 Proof: G """ FIR where F is a finitely generated free abelian group (Section 6.1c). By Lemma 6.15, F has a basis Cl, •••, Cn such that R = gp(UICl, ••., Uncn) for some nonnegative integers Ul,.••, Un. We now apply Corollary 6.17 to conclude that G """ FIR is the direct sum of cyclic groups. Corollary 6.19: If G is finitely generated, it is the direct sum of a finite number of infinite cyclic groups and cyclic groups of prime power order. Proof: It is only necessary to show that a cyclic group of composite order is the direct sum of cyclic groups of prime power order. This we have already done in Problem 6.19, page 187. Corollary 6.20: If G is a group |
with no elements of finite order and G is finitely generated, then G is free abelian. Proof: G is the direct sum of a finite number of cyclic groups each of which must be infinite cyclic as G has no elements of finite order. Thus the result follows. Problems 6.44. Prove that every finitely generated torsion group is finite. Solution: By the fundamental theorem of finitely generated abelian groups, if G is finitely generated it is the direct sum of a finite number of cyclic groups. If G is a torsion group, then it is the direct sum of a finite number of finite cyclic groups. Hence G is finite. (Compare with Problem 4.31, page 105.) 6.45. Let G = ~ Gi where Gi is a cyclic group of order 2 for finitely generated. '" i=1 i = 1,2,.... Prove that G is not Solution: Every element in G is of finite order, for if U(¥ 1) E G, U = Ul + U2 +... + Un, Ui E Gi, (i' E Z), and 2U = 2Ul +... + 2Un = 0 +... + 0 = O. Thus G is a torsion group. If G were finitely gen erated. G would be finite by the preceding problem. But G is clearly infinite. Therefore G cannot be finitely generated. c. The type of a finitely generated abelian group In Section 6.3b we proved that a finitely generated abelian group is a direct sum of cyclic groups. However, such a decomposition is not unique: first, the direct summands are not unique (see Problem 6.46 below); moreover, the number of direct summands can vary (see Problem 6.19, page 187). We say that two decompositions are of the same kind if they have the same number of summands of each order. For example, two decompositions of a group into the direct sum of three cyclic groups of order 4 and two cyclic groups of infinite order are said to be of the same kind. A concrete example of two decompositions of the same kind is given in Problem 6.46. As we remarked in Corollary 6.19, every finitely generated group can be decomposed into the direct sum of a finite number of cyclic groups of prime power or else infinite order. |
Our aim is to prove Theorem 6.21: Any two decompositions of a group G into the direct sum of a finite num ber of cyclic groups which are either of prime power order (# 1) or of in finite order, are of the same kind. Proof: We shall separate the proof into four cases: (1) both decompositions involve only infinite cyclic groups, (2) both decompositions involve only cyclic groups of order a power of fixed prime p, (3) both decompositions involve no infinite cyclic groups, and (4) the general case. Sec. 6.3] FINITELY GENERATED ABELIAN GROUPS 199 Case 1. where h Ii for j = 1,..., k and i = 1,..., l respectively, are infinite cyclic groups. A From Corollary 6.12, page 193, we conclude that k = l. (Alternatively we may proceed as in Problem 6.52.) Case 2. Both decompositions involve only cyclic groups of order a power of a fixed prime p. We shall write for any integer n, nG = {ng I g E G}. If G is a group, nG is a subgroup (Problem 6.53). To prove case 2 we will need the following lemma. Lemma 6.22: Let G = A EB B. If n is any integer, then nG = nA EB nB. Proof: As nAnnBcAnB={O}, gp(nA,nB)=nAEBnB. If gEnG, there exists hE G such that nh = g. Let h = a + b, a E A and b E B. Then g = nh = na + nb. Accordingly nG C nA EB nB enG and so nG = nA EB nB. Corollary 6.23: Let G = Al EB... EB A k • Let n be an integer. Then nG = nAl EB... EB nA k Proof: We apply Lemma 6.22 to one direct summand at a time. Then the result follows. Corollary 6.24: Let G be expressed as the direct sum of k i cyclic groups of order pi, 1 ~ i ~ r. Then pG is expressible as the direct sum of k i cyclic groups of order pi-l where 2 ~ i ~ r. |
Proof: This is an immediate consequence of Corollary 6.23 and the fact that if A is cyclic of order pi, pA is cyclic of order pi-l. Hence the corollary follows. We are now in a position to prove case 2. We proceed by induction on the order of G. If IGI = 1 or p, then the result is immediate. If the result is assumed true for all groups of order less than n that satisfy the conditions of case 2, then let \G\ = n. Sup pose G is expressed as the direct sum of k i cyclic groups of order pi for 1 ~ i ~ r, and also as lj cyclic groups of order pi for 1 ~ j ~ s. Then pG is expressible (by Corollary 6.24) as the direct sum of k i cyclic groups of order pi-l for 2 ~ i ~ r on the one hand, and as the direct sum of li cyclic groups of order p i - l for 2 ~ i ~ s on the other. As \pG\ < \G\, it follows by the induction assumption that r = sand ki = li for 2 ~ i ~ r. Now we must still prove that kl = ll. But \G\ = pkl(p2)k2... (pr)kr = p!.(p2)!2... (pr)!r, and so II = k l • Thus we have proved both decompositions are of the same kind, as required. Case 3. G is expressed in two ways as the direct sum of a finite number of cyclic groups of prime power order. We have dealt with the case where only one prime is involved. We proceed by in.., duction on the number of primes involved. Let p be one of the primes involved. Let A l, •••, Am be all the direct summands of order a power of p in the one decomposition, Bl,..., Bn the other direct summands involved, so that G = Al EB... EB Am EB Bl EB... EB Bn Putting A = Al EB... EB Am and B = Bl EB... EB Bn, it follows that G = A EB B. Let Xl,..., X k be all the direct summands of order a power of p in the second de composition, Y 1, •••, Y! the remaining direct summands, so that G = Xl EB |
... EB X k EB Y1 EB... EB Yr Put X = Xl EB... EB X k, Y = Yl EB... EB Yr. Then G = X EB Y. We claim that A = X and B = Y. 200 ABELIAN GROUPS [CHAP. 6 Let g EA. Then g = x + y where x E X and y E Y. Now the order of any nonzero element of Y is coprime to p. As g is of order a power of p, y = 0 (Problem 6.54). Hence g E X, and so A ~ X. Similarly X ~ A and we conclude that A = X. By a similar argument B = Y. Thus A1 EB... EB Am = Xl EB... EB X k and B1 EB... EB Bn = Y 1 EB... EB Y!. By the induction hypothesis, A1 EB... EB Am and Xl EB... EB X k on the one hand, and B1 EB... EB Bn and Y 1 EB... EB Y! on the other, are of the same kind. Hence the two decompositions are of the same kind and the result follows. Case 4. Let G be expressed as the direct sum of cyclic groups of prime power order or of infinite order in two ways, say G = 11 EB... EB 1m EB F1 EB... EB Fn = 11 EB... EB Ire EB F1 EB... EB F! where h Ii are infinite cyclic groups and F i, Fi are groups of prime power order. A A A A A A Let T(G) be the set of all elements of finite order (see Theorem 6.9, page 189). Then T(G) is the direct sum of the direct summands of finite order in both cases (Problem 6.55). Thus A A T(G) = F\ EB '" EB Fn = F1 EB... EB F! Hence by case 3, F1 EB... EB F m and P1 EB... EB FI are of the same kind. A A Also G/T(G) "'" 11 EB... EB 1m"'" 11 EB... EB Ire (by Problem 6.11, page 181). By Problem 6.56, 11 EB... EB 1m is the direct sum of k infinite cyclic groups. Then k = m by case 1. Therefore we have proved that 11 EB 12 EB... EB 1m EB F1 EB |
... EB Fn and 11 EB 12 EB... EB Ire EB F1 EB... EB F! are of the same kind. This completes the proof of the theorem. A A A A A If a finitely generated group G is the direct sum of cyclic groups of orders p~l,..., p~k and 8 infinite cyclic groups, where P1,...,Pk are primes, P1 ~ P2 ~... ~ Pk, r1,..., rk posi then the ordered k + 1-tuple (p~l, ••., p~k; 8) is tive integers with ri ~ ri+ 1 called the type of G. (The definition of type differs slightly from book to book. Usually it is applied only to p-groups.) By Theorem 6.21 the type of G is uniquely defined. We can now give a criterion for the isomorphism of two finitely generated abelian groups. if Pi = Pi+ 1, Theorem 6.25: If F and G are two finitely generated groups, then they are isomorphic if and only if they have the same type. Proof: Let F = A1 EB... EB A k. is an isomorphism, then G = A 1cp EB If cp: F ~ G... EB Akcp (Problem 6.56). As Aicp "'" Ai, it follows that F and G have the same type. Conversely, if F and G have the same type they are clearly isomorphic (Theorem 6.5, page 185). Problems 6.46. Let G = A EB B where A and B are cyclic of order 2. Find C and D such that G = C EB D where C and D are cyclic of order 2 and C -# A and C -# B. Solution: Let A = {O, a}, B = {O, b}. Put C = {O, a + b}. Then C is cyclic of order 2. Also put D = B. ThenC+D={O,a+b,b,a+b+b=a},andsoC+D=G. Also CnD={O}. Thus G=CEBD. 6.47. If the type of F is (11)..., fk; f) and that of G is (U1>..., Up; U) where f1 = p~l, f2 = |
p;2,..., fk = p~k, U1 = q~l,..., UI = q:! and Pk < q1> where the Pi and qi are primes, find the type of FEB G. Solution: We have P1 "" P2 ""... "" Pk < q1 ""... "" ql' Hence the type of F EB G is (f1,·· ·,h,U1'··.,UI; f+ U) Sec. 6.3] FINITELY GENERATED ABELIAN GROUPS 201 6.48. If F, G and H are finitely generated abelian groups, show that FEB G 2" F EB H G2"H. implies that Solution: Express F, G and H as direct sums of cyclic groups of prime power and infinite orders. If the type of F is (h..., h; I), and that of G is (gl'..., gl; g) while that of H is (h1'..., hm ; h), then the type of F EB G is (a1'...,·ak + l; I + g) where a1'..., ak + 1 is 11,..., ik, gl,..., gl in some order, while the type of FEBH is (b 1,...,bk +m ; I+h) where b1,·..,bk+ m is 11,...,h,h1,...,hm in some order. For two abelian groups to be isomorphic we require that (by Theorem 6.25) their types are the same. Accordingly the types of G and H are the same and G 2" H. 6.49. Find up to isomorphism all abelian groups cf order 1800. Solution: Observ~ that 1800 = 233252. So an abelian group of order 1800 is a direct sum of a group of order 23, a group of order 32 and a group of order 52. The possible types of!l group of order 23 are (23 ; 0), (2 2,2; 0), (2,2,2; 0). Thus there are precisely 3 groups of order 8. The possible types of a group of order 32 are (3 2 ; 0) and (3,3; 0), so there are 2 non-isomorphic groups of order 9. Similarly |
there are 2 non-isomorphic groups of order 25. Then the total number of non-isomorphic groups of order 1800 is 3 X 2 X 2 = 12. Compare the ease with which we solve this problem with the effort required to find all the groups (non-abelian as well as abelian) of order 8 which we have considered in Chapter 5. 6.50. Let p be any prime and let m be any integer. Prove that the number of groups of order pm is equal to the number of ways of writing m = r1 +... + rk where r j, • • •, rk are positive integers and r1 "" r2 ""... "" rk' Solution: A group of order pm has all its elements of order a power of p. Hence its type will be of the form (pT1, pT2,..., pTk; 0) with r1 "" r2 "".., "" rk' Since G is of order pm, and we conclude that r1 +... + rk = m. IGI = pT1pT2... pTk = pT1 +... +Tk 6.51. Prove that a cyclic group of order pn, where p is a plime, is not expressible as the direct sum of nontrivial subgroups by the following two methods: (1) directly, (2) by using Theorem 6.21. Solution: (1) Suppose G = A EB B where A, B are nontrivial subgroups of G. Clearly IAI "" pn-1 and lGI = pn. Let G = gp(g). Then g = a + b, a E A, and bE B. As pn-1g = IBI "" pn-1 as pn-1 a + pn- 1b = 0, g is of order less than pn-1. But gp(g) = G and is of order pn. Thus we have a contradiction. (2) Since G is cyclic of order pn, the type of G is (pn; 0). Hence by Theorem 6.21, only one of the direct summands is nonzero, i.e. G cannot be expressed as a direct sum of more than one non trivial group. 6.52. Prove, by considering the direct sum of cyclic groups of order 2, that if G is the |
direct sum of k infinite cyclic groups and also the direct sum of l infinite cyclic groups, then k = l. Solution: Let G = gp(X1) EB... EB gp(Xk)' Let H = gp(2X1'..., 2Xk)' Then by Corollary 6.17, GIH IGIHI = 2k. Clearly He 2G. Also if g E G, is the direct sum of k cyclic groups of order 2. Thus g = r1x1 +... + Tkxk' Then 2g = r1(2x1) +... + rk(2xk) E H, from which 2G C H. Thus H = 2G. Now by a similar argument we conclude that if G is the direct sum of l infinite cyclic groups, IG/2GI = 21. Thus l = k. 6.53. Prove that nG is a subgroup of G where n is a given integer. Solution: If h, k EnG, h = nl, k = ng where I, g E G. Hence h - k = n(f - g) EnG, and so nG is a subgroup. 202 6.54. ABELIAN GROUPS [CHAP. 6 Let G be an abelian group, G = X EB Y. Let x E X, Y E Y. Prove that (1) if x and yare of finite order, then the order of x + y is the least common multiple (lcm) of the orders of x and y; (2) if x is of infinite order, x + y is of infinite order. Solution: (1) Let 1 = lcm of the orders of x and y. Then l(x + y) = lx + ly = O. Now if m = order of x + y, implies mx = 0 and my = O. This in turn implies that the then m(x + y) = mx + my = 0 order of x divides m and the order of y divides m. Thus we have the result. (2) If x is of infinite order and m(x + y) = 0, then mx + my = O. But by the uniqueness of such expressions in direct sums, mx = my = O. Since x is of infinite order, m = O. 6.55. Let G = II EB... EB 1 |
m EB FI EB... EB Fn where each Ii is torsion-free and each Fi finite. Let T(G) be the set of all elements of finite order. Prove that Solution: T(G) = FI EB... EB Fn Clearly T(G);JFIEB···EBFn • If gET(G), g=iI+···+i,.,+/I+···+ln whereiv...,im are elements of II'...,Im respectively, and 11'....In are elements of Fv...,Fn respectively. As g is of finite order r, say, rg = ril + ri2 +... + rim + r/l +... + rl n = 0 By definition of the direct sum, it follows that ril = ri2 =... = rim = r/l =... = rln = 0 Since II,...,Im are torsion-free, we have il = i2 =... = i,., = O. Thus g E FI EB... EB Fn and the result follows. 6.56. If F = Al EB... EB Ak and ¢: F --> G is an isomorphism, then G = A I ¢ EB... EB A k¢. Solution: We must show that every element of G is uniquely of the form aI¢+... ~ +ak¢ where aI,...,ak such that I¢ = g. But belong to AI'...,Ak respectively. Now if g E G, 1= al +... + ak and so g = aI¢ +... + ak¢. If aI¢ +... + ak¢ = a:¢ +... + a~¢, then there exists IE F (al - a~)¢ +... + (ak - a~)¢ = 0 Let k = al - a: +... + ak - a~. k belongs to Ker ¢. Since ¢ is an isomorphism, k = O. By the uniqueness of expression of direct sums, al - a~ = a2 - a~ =... = ak - a~ = 0 so that al = a~, a2 = a~,..., ak = a~. Therefore each element of G is expressible in the form aI¢ +... + ak¢ in one and only one way. d. Subgroups of finitely generated abelian |
groups The purpose of this section is to decide which groups (up to isomorphism) can appear as subgroups of finitely generated abelian groups. We begin with Theorem 6.26: Let G be free abelian of rank n. Then any subgroup H of G is free abelian of rank less than or equal to n. Proof: By Lemma 6.15, page 196, there exist a basis CI, •••, Cn of G and integers UI,.. •,Un such that H = gp(UICI,...,Uncn). If UI,...,Ui are nonzero, and Ui+I = Ui+2 =... = Un = 0, then gp(UICI,..., Uncn ) = gp(UICl) EB... EB gp(UiCi) (See Problem 6.57.) Hence the result. Corollary 6.27: Let A be a finitely generated abelian group. Then every subgroup of A is finitely generated. Proof: As A is a finitely generated abelian group, it is isomorphic to some factor group of a finitely generated free abelian group G, say A ~ GIN. The subgroups of GIN are of the form HIN where H is a subgroup of G. By Theorem 6.26, H is finitely generated and therefore so is HIN. Consequently every subgroup of A is finitely generated. Sec. 6.3] FINITELY GENERATED ABELIAN GROUPS 203 From this corollary we see that only finitely generated abelian groups can occur as subgroups of finitely generated abelian groups. Theorem 6.28: Let G be a finitely generated group and H a subgroup of G. Let G and H be expressed as direct sums of infinite cyclic groups and cyclic groups of prime power order. If the number of infinite cyclic groups in these decom positions for G and Hare m and k respectively, then k ~ m. Proof: Let G = 11 EB... EB 1m EB Fl EB... EB F n, H = 11 EB A A EB Fz A A where the l;, Ii are infinite cyclic groups and the Fi, Fi are cyclic groups of prime power order. Let T(G) and T(H) be the torsion subgroups of G and H respectively, i. |
e. the re spective sets of elements of finite order (Theorem 6.9, page 189). Then T(H) = H n T(G). Now G/T(G) """ /1 EB... EB 1m by Problem 6.11, page 181. Thus the rank of G/T(G) is m. (H + T(G))/T(G) ~ G/T(G), (See the remarks following Theorem 6.11, page 192.) Since (H + T(G))/T(G) is free abelian of rank less than m, by Theorem 6.26. But (H + T(G))/T(G) """ H/Hn T(G) """ H/T(H) by the subgroup isomorphism theorem (Theorem 4.23, page 125). that H/T(H) """ /1 EB... EB h. Thus k ~ m. It follows as before Again let G be finitely generated and H a subgroup of G. (Recall that if F is any abelian group and p a prime, Fp = {f [ f E F and of order a power of p}.) Gp, as a subgroup of a finitely generated abelian group, is finitely generated (Corollary 6.27) and so is Hp. Clearly Gp :2 Hp. Thus we are led to inquire what groups can occur as subgroups of finitely gen erated p-groups. Of course a finitely generated p-group is finite (Problem 6.44). We first require a lemma. Lemma 6.29: Let G have type (pTI,..., pTn; 0). Then the number of elements of order p in G is pn-1. Proof: Let G = Cl EB... EB Cn where each Ci = gp(Ci) and the order of Ci is pTi. If x EGis of order p, and x = tlCl + '" + tncn where ti E Z, then pTi- l divides t;,. Hence the elements of order p are a subset of H, where H = pTelCl EB... EB pTn-1Cn On the other hand every element (# 0) of H is of order p, so H - {O} is the set of all elements of order p. Accordingly, as |
[H[ = pn, the number of elements of order p in G is pn - 1. Theorem 6.30: Let G be a group with type (pTl,p T 2, •••,pTm;O). Let H be any subgroup. If the type of H is (pSI,..., pSn; 0), then n ~ m and 0 < Si ~ ri, i = 1,..., n. Proof: We proceed by induction on [G[. If [G[ = 1 or p, the result is trivial. Hence assume [G[ > p, and the result holds for groups of order less than [G[. Now H has type (pSl, pS 2, •••, pSn; 0), and so the number of elements of order p in H is, by Lemma 6.29, pn - 1. Similarly, the number of elements of order p in G is, by Lemma 6.29, pm - 1. Clearly pn - 1 ~ pm - 1 and consequently n ~ m. Now [pG[ < [G[. We can therefore assume the result holds by induction for pG and its subgroup pH. At this point we experience a minor notational inconvenience. If for example rm > 1, pG is of type (pTl-t,...,pTm-l;O). However, if rm = 1 and rm-l> 1, pG is of type (pTl-t,...,pTm-l-l;O). Therefore we need additional notation. Define m* = m if rm> 1; otherwise define m* to be an integer such that rm* > 1, but rm*+l = 1. As rl::=" r2::="... ::=" T m* > 1 and, if m* # m, rm*+l=.,. = rm = 1, then pG is of type (pTl-l,...,pTm*-l; 0). 204 ABELIAN GROUPS [CHAP. 6 Similarly, let us define n * = n if Sn > 1; otherwise define n * to be an integer such that S n*-1 ; 0). Sn* > 1 but Sn* +1 = 1. Arguing as in the paragraph above, pH is of type (pSI-I, ••• |
, p Then by the inductive hypothesis we have n * - 1 =0: m * - 1 and Si - 1 =0: ri - 1 for i = 1,..., n * If n* = n, the result follows immediately. If n* =1= n, then Sn*+1 =... = Sn = 1. Hence rn'+I~Sn*+I, •••, rn~Sn. Thus Si=O:ri for i= 1,...,n. With Theorems 6.28 and 6.30 it is easy to determine, knowing the type of a given finitely generated abelian group G, the possible types of subgroups of G. (See Problem 6.60.) It can be shown (Problems 6.62-65) that every factor group of a finite abelian group G is isomorphic to a subgroup of G. Therefore we know the types of homomorphic images of finite abelian groups. Problems 6.57. Let G = A EEl B and let C, D be subgroups of A, B respectively. Show that C + D = C EEl D. (This can obviously be generalized to the direct sum of any number of groups.) Solution: As {O}=AnB;;:)CnD, we have CnD={O}. Thus C+D=CEElD. 6.58. Let G have type (pTl,..., pTn; 8). Suppose that for some i"" j 1 n Pi-l =1= Pi Put P = Pi. Show that the type of Gp is (pTi,..., pTi; 0). Solution: Pi = Pi + 1 =... = Pi' Decompose G into the direct sum of infinite cyclic groups and groups of prime power order. Clearly G = Ai EEl ••• EEl Ai EEl R where Ak is of order pTk for i "" k "" j, and R is the direct sum of the cyclic groups which are not of order a power of this prime p in the given decomposition of G. Then Gp ;;:) Ai EEl ••• EEl Ai. On the other hand, as any nonzero element of finite order from R is of order coprime to that of p, Gp C; Ai EEl ••• EEl Ai. Thus Gp = Ai EEl..• E |
El Ai' and the result follows. 6.59. Let G be of type (pTl,.•., pTm; u). Show that G has subgroups of type (pSI,..., pSn; v) where n "" m and 1 "" 8i "" ri for 1 "" i "" n and v "" u. Solution: Let G = Al EEl ••• EEl Am EEl II EEl ••• EEl Iu where Ai is cyclic of order pTi, and II'..., Iu are infinite cyclic groups. Now each Ai has a subgroup Bi of order p\ 1 "" i "" n. By repeated applica tion of Problem 6.57, Bl +... + Bn + II +... + Iv = Bl EEl ••• EEl Bn EEl II EEl..• EEl Iv This is then a subgroup of G of type (psI,..., pSn; V), as required. 6.60. Let G be of type (33,32,52,73 ; 1). Determine whether G has a subgroup H of type (a) (3,3,72,7; 1), (c) (33,33,52,73 ; 0), (b) (3,3,5,7; 2), (d) (3,3,7; 1). Solution: (a) No, as then G7 is of type (73 ; 0) whereas H7 is of type (72,7; 0) by Problem 6.58. This is a con- tradiction to Theorem 6.30. (b) No. A direct contradiction to Theorem 6.28. (c) No. Compare G3 and H3 as in (a). (d) Yes. 6.61. Give an infinite number of examples of an abelian p-group that contains exactly p + 1 subgroups of order p. Solution: If G is a group with p + 1 subgroups of order p, each of them contributes p - 1 distinct elements (the identity is common to all) of order p. Hence in all there are (p -l)(p + 1) = p2 -1 elements of order p. Sec. 6.4] DIVISIBLE GROUPS 205 By Lemma 6.29 a p-group with p2 - 1 distinct elements of order p contains p2 summands which |
i = 1,..., p2, be integers. Then any are cyclic groups of order a power of p. Let 1 ==: ri' where group of type (prl,..., pTp; u) where u is a nonnegative integer, has exactly p2 - 1 elements of order p and thus exactly p + 1 subgroups of order p. 6.62. Let G be a p-group. Suppose G = gp(a) EEl B. Prove that G = gp(a, + b) EEl B where b E B of order less than or equal to the order of a. is Solution: If x E gp(a + b) n B, x = r(a + b) = b1 where b 1 E Band r is an integer. Thus ra = b1 rb. Since gp(a) nB = {O}, ra = O. Then r is divisible by the power of p which is the order of a. Consequently rb = O. Hence b 1 = 0 and x = O. Clearly gp(a + b) + B = G and the result follows. then 6.63. Let G be a finite p-group. Prove that if g E G and g is of order p, g appears as an element of a cyclic direct summand of G. Solution: G is the direct sum of cyclic groups, say G = gp(Cl) EEl... EEl gp(cn ). If g = 0, the result follows immediately. Otherwise, without loss of generality, suppose that g = rlpw1cI +... + rmpwmcm (ri' p) = 1, r;pw; Ci # 0 and WI ==: W 2 ==: ••, ==: Wn- Put rici = c;. Clearly gp(cD = gp(CI)' where Then g = pWl(C~ + d) where dE gp(C2'..., cn)' As g is of order p and pWIC; # 0, the order of d is less than or equal to the order of c{. By Problem 6.62, on putting c = c; + d, we obtain Since g E gp (c), the result follows. G = gp(c) EEl gp(C2,..., cn) 6.64. Let G be a finite p-group. Let |
N = gp(g) be of order p. Prove that GIN is isomorphic to a sub group of G. Solution: By Problem 6.63, G = gp(c) EEl B where g E gp(c). Then GIN == (gp(c)/N) EEl B (Lemma 6.16, page 197). But clearly gp(pc) == gp(c)/N. Thus gp(pc, B) == GIN. 6.65. Let G be a finite group. Prove by induction on IGI that if N is a subgroup of G, GIN is isomorphic to a subgroup of G. Solution: Assume the result is true for all groups of order less than r. Let IGI = r and let N be a sub group of G. If N = {O}, GIN == G and there is nothing to prove. If N is of order a prime p, G = Gp EEl E where Gp is the p-component of G, and N C; Gpo Then GIN == (GpIN) EEl E by Lemma 6.16. Now Gp/N == H, a subgroup of G p' by the preceding problem. Hence GIN == HEEl E and H EEl E is a subgroup of G. If N is not of order a prime, there is an element no of N of order a prime p by Proposition 5.9, page 137. Let No = gp (no). Then (GINo)/(NINo) == GIN. As IGINol < IGI, GINo has a subgroup HINo == GIN. H # G, since otherwise N = No which is not IHI < IGI and by induction it has a subgroup K such that K == HINo == GIN. The true. Thus result follows. 6.4 DIVISIBLE GROUPS a. p-Prufer groups. Divisible subgroups A group G is said to be divisible if for each integer n # 0 and each element g E G there exists hE G such that nh = g. Both the additive group of rationals and p-Priifer groups are divisible in this sense (Problem 6.66 below). If the groups Gi, i E I, are divisible, then ~ Gi is divisible. For if n # 0 is any integer and g |
E ~ Gi, then g = gl +... + gk, say. So there exist hI,..., hk such that nhl = gl,..., nhk = gk. Then i E I i E I n(hl +... + hk) gl +... + gk g 206 ABELIAN GROUPS [CHAP. 6 It follows that direct sums of p-Priifer groups and copies of the additive group of ra tionals are also divisible. We show that in fact this exhausts all divisible groups. To prove this we need several facts. First we remark that a homomorphic image of a divisible group is divisible. For sup pose G is divisible and H is a subgroup of G. Let g + HE GIH (g E G) and let n be a positive integer. Then there exists g' E G such that ng' = g. Accordingly n(g' + H) = g + H, and so GIH is divisible. Secondly we remark that if G = H EB K and G is divisible, so also are Hand K, since H e= GIK (Problem 6.11, page 181) is a homomorphic image of a divisible group. Similarly K is divisible. Next we need the following theorem which classifies p-Priifer groups. Theorem 6.31 (Main Theorem on p-Priifer Groups): Let p be a prime. Let G be a group which is the union of an ascending sequence of subgroups C1 <: C2 <:... where Cr is cyclic of order pr for r = 1,2,.... Then G is isomorphic to the p-Priifer group. Proof: We may suppose that Cr = gp(c T ) and that PCr+l = Cr for r = 1,2,... (see Problem 6.67 for the details). Define B: G ~ (QIZ)p by (mcr)B = mlpr + Z for all integers m. We must prove that B is an isomorphism. We are however not even certain that B is a mapping. The snag is this: If g = mCr and if also g = ncs, is gB = mlpr + Z or is gB = nips + Z? We will show that mlpr + Z = nips + Z, thus proving that B is uniquely defined. Assume without loss of gener |
ality that r:=" s. It follows that pr-scr = cs. Then from which (m - npr-s)c T = 0 mCr = npr-scr As Cr is of order p', m - npT-S = kp' for some integer k. Thus m = npr-s + kp' from which mlpr = np-s + k. We therefore conclude that mlpT + Z = nips + Z. Thus B is a mapping. Next we show that B is a homomorphism. If g, hE G, then g, hE Cr for some integer r, so that g = SCr, h = tCr, with s, t E Z. Then (g + h)B = ((s + t)cr)B = (s + t)lpT + Z = (slpr + Z) + (tlpr + Z) = gB + hB Finally B is one-to-one as Ker B {g I gB = Z} = {SC, I s an integer, r a positive integer and (scr)B = slpr + Z = Z} = {SCr I slpT E Z} = {SCr I s divisible by pr} = {O} Thus the result follows. In the future we will call any group isomorphic to (QIZ)p a p-Priifer group. The following result is not only the main tool in Section 6Ab, but is also of interest in itself. Theorem 6.32: Let G contain a divisible subgroup D. Then there exists a subgroup K of G such that G = DEB K, i.e. a divisible subgroup is a direct summand. Proof: We accomplish this proof by Zorn's lemma. Let cP be the collection of all sub groups L of G such that LnD = {O}. (Our idea is to pick a maximal subgroup K which meets D in {O}. Then D + K = D EB K and we need only show that D + K = G which will turn out to be true because of the maximality of K.) Can we apply Zorn's lemma to CP? Suppose {Li liE I} is a chain in CPo Is i ld [Li in CP? We require Sec. 6.4) DIVISIBLE GROUPS 207 (i) D n U |
Li = {OJ. (ii) U Li is a subgroup of G. i E I i E I iEI Part (i) is true because Dn/dILi ¥= {OJ implies DnL j ¥= {OJ for some L j • To prove (ii) we must show that if g, hE U L i, then g - hE U L i. Now g, hE U Li iEI implies g E L j and h ELk. Either Lk d L j or L j d L k, so without loss of generality as sume that LkdLj • Then since Lk is a subgroup of G, g-h ELk. Therefore g-h E ildILi and (ii) holds. So cP has a maximal element, say K, which satisfies DnK = {OJ. Thus D+K = D (f)K. Suppose now that D + K ¥= G. Then G/(D (f) K) is nonzero. We prove first that G/(D (f) K) is a torsion group. Suppose the contrary. Then we can find x E G such that is of infinite order in G/(D (f) K). Now x f1. K. If we put Kl = gp(K, x), gp(x + (D (f) K» it consists of all elements of the form nx + k, where n is an integer and k is an element of K. If nx+kED, then nXE(K+D). So n=O is the only possibility. But KnD={O}. So k = 0 follows also. Therefore DnKl = {OJ. This contradicts the maximality of K. Thus we have proved that G/(K (f) D) is a torsion group. iEI Let x E G, x f1. K (f) D. Then gp(x + (K (f) D» is a subgroup of G/(K (f) D) of finite order. Suppose that x + (K (f) D) is of order w. It follows then that wx E K (f) D, but rx f1. K (f) D for 0 < r < w. Suppose wx = k + d. Since D is divisible, we can find d l |
ED so that wdl = d. Put Xl = X - d l. Then WXl = WX - wdl == k + d - d = k. Notice that Xl f1. K but WXl E K. Put Kl = gp(Xl, K). Then Kl = {rxl+kl r=O,l,...,w~l, and all kEK} We claim that KlnD = {OJ. For if rXl + kED, r E {O, 1,..., w-1}, and k E K, then' D (f) K = rXl + (D (f) K) = r(xl + (D (f) K» Since Xl + (D (f) K) = x + (D (f) K), we must have r = O. So kED and thus k = O. Therefore KlnD = {OJ. But K is maximal. This contradiction shows that our original assumption, i.e. G ¥= D (f) K, is false, thus proving the theorem. Problems 6.66. Use a proof different from that of Problem 6.36, page 192, to prove that a p-Priifer group is divisible. Solution: As Q is divisible, so is QIZ. But as QIZ = ~ (QIZ)p, each (QIZ)p is itself divisible, since pEII every direct summand of a divisible group is divisible. 6.67. Let G be as defined in Theorem 6.31. Prove that we can choose elements cr such that Cr = gp(cr) and PCr+l = Cr'r = 1,2,.... Solution: Assume by induction that Cl' •.., cn have been chosen with pcr + 1 = cr r = 1,..., n - 1 and Ci = gp(Ci), i = 1,..., n. Let Cn + 1 = gp(c). Then gp(pc) is a cyclic group of order pn and, since cyclic groups have only one subgroup of any given order, gp (pc) = Cn" Hence r(pc) = cn for some integer r. As Cn is of order pn, rand p are coprime. Thus gp(rc) = Cn + l' Now put |
Cn + 1 = rc. Then pcn + 1 = Cn and it is possible to choose the elements ClJ C2,..., as required. for 6.68. A p-group G is a p-Priifer group if and only if it has the following two properties: (1) every proper subgroup of G is cyclic, (2) there is a cyclic subgroup of G of order pi for every i = 1,2,.... (Hard.) Solution: First we shall prove that if G satisfies (1) and (2), it is a p-Priifer group. Let Cl ~ C2 ~ • " be a sequence of subgroups of G where each Ci is a cyclic group of order pi. If the sequence is infinite, D = U Ci is a p-Priifer group (Theorem 6.31). Hence it is divisible and G = D (f) K by Theorem 6.32. However, if K ¥= {O}, G has a subgroup which is the direct sum of two cyclic p-groups, and hence is not cyclic, contrary to the hypothesis. Accordingly, G = D is a p-Priifer group. i=l 00 208 ABELIAN GROUPS [CHAP. 6 Suppose now that C1 C C2 C... C Cn is a sequence of subgroups, each Ci cyclic of order pi, and there exists no subgroup C n + 1 ;;) C n where Cn + 1 is of order pn+1. Let C n = gp(a). We know that there exists a subgroup B = gp(b) of order pn+1. Consider gp(a, b). As a finitely generated abelian group, it is the direct sum of cyclic groups. If it is the direct sum of two or more cyclic groups, it is not cyclic (Theorem 6.21). Thus gp(a, b) is cyclic, and as G is a p-group, gp(a, b) is cyclic of order pm where m "" n + 1. But then gp(a, b) contains a cyclic subgroup of order pn+ 1 containing Cn' contrary to the hypothesis. The result follows. Next we note that the p-Priifer group satisfies condition (2). As for condition (1), let H be a subgroup of (Q/Z) |
p' If H # {Z}, then let x E H, x # 0 + Z. Let x = m/pT + Z where m is an integer between 1 and pT - 1. Clearly gp(x) = gp(l/pT + Z). If there is no integer n for which for r "" n, then H = (Q/Z)p' If there exists such an integer n, He gp(l/pn + Z). l/pT + Z El H Thus H is cyclic as required. 6.69. Show that if U is the subgroup of the mUltiplicative group of the complex numbers consisting of all the nth roots of unity, then U == Q/Z. Solution: pEII U = ~ Up by Theorem 6.10, page 190. Now Up is the union of cyclic groups of order pi, namely Up is the union of Ci = gp ({x j x is a pith root of unity}). But Ci is cyclic of order pi. Then by Theorem 6.31, Up == (Q/Z)p' Thus, as Q/Z = ~ (Q/Z)p (by Theorem 6.10), Q/Z == U by Theorem 6.5, page 185. p E II 6.70. Show that the additive group of rationals is the union of an ascending sequence of infinite cyclic groups. Solution: Let Qn = gp(l/n!). 00 Q the additive group of rationals. 6.71. Show that if G is a p-Priifer group and H, K are subgroups of G, then either H;;) K or K;;) H. Solution: If one of H or K is G, the result is true. Assume both Hand K are proper subgroups. Then by Problem 6.68, Hand K are both finite cyclic groups. Suppose pT = jHj "" jKj = pS. Then H con tains a subgroup of order jKj, say H 1• Now both H1 and K are contained in some cyclic subgroup of G, as G is the union of cyclic subgroups. But then it follows that H1 = K, as there is one and only one subgroup of order pS in a cyclic p-group of order exceeding pS. Therefore H;;) K. 6.72. Let G be an ascending union |
of infinite cyclic groups Ci such that Ci = gp(Ci) and (i+ 1)Ci+1 = ci, for i = 1,2,.... Prove that G is isomorphic to the additive group of rationals. (Hard.) Solution: Let Q denote the additive group of rationals and let Qi = gp(l/i!), i = 1,2,.... Clearly, Qi+1 ~ Q i and U Qi = Q. We shall prove that fJ below defines a mapping of G to Q. Define (zCi)fJ = z/i! where Z E Z. We must prove that fJ is uniquely defined. i=l Suppose Zlci = Z2Cj' zvzz and i,j integers. If i === j, zl(j!/i!)cj = ZZCj. Since C j is infinite cyclic, zl(j!/i!)cj = ZZCj ci = (j!/i!)cj. Hence ZlCi = then implies that zl(j!/i!) = z2' To prove that fJ is well defined, we must show that zl/i! = zz/j!, i.e. zl(j!/i!) = zz. But this is what we have just shown. Since CifJ = Qi = gp (l/i!), it follows that GfJ;;) U Qi = Q. Hence fJ is an onto m ap p ing. · ;=1 00 Is fJ a homomorphism? Let f, g E G. We may as well suppose that f, g E Ci for some integer i. Hence f = ZlCi' g = ZZci' say. f + g = (Zl + Z2)Ci'.Then 1 (f + g)fJ = (Zl + Z2) 7f and 1 Z2 ffJ + gfJ = -:-;- + -:-;- = (Zl + Z2)"7f Zl t. to t. to Thus fJ is a homomorphism. Finally, to show that fJ is an isomorphism, it is sufficient to show that Ker fJ = {O}. Suppose f is such that ffJ = O. We have that f = zCi for some integers Z and i. Then ffJ = |
z/i! = 0 only if Z = O. Hence f = 0 and the result follows. Sec. 6.4] DIVISIBLE GROUPS 209 b. Decomposition theorem for divisible groups The results of Section 6.4a enable us to deduce the following decomposition theorem for divisible groups. Theorem 6.33: A divisible group is the direct sum of p-Priifer groups and copies of the additive group of rationals. LhL\r·<--,.'",~,~L~<.> £.~j. Proof: Let G be divisible and let T be the torsion subgroup of G. Now for any integer n and element t E T, there exists an element g E G such that ng = t. Since t is of finite order, so is g, and hence gET. Thus T is itself divisible. A divisible subgroup is a direct summand (Theorem 6.32); so G = T EEl F. Since Tn {F} = 0, F ~ (T EEl F)/T; hence F is torsion-free by Theorem 6.9, page 189. Moreover, as F is a direct summand, F is itself divisible. We now consider F and T separately. r: (a) F. - ~ Q,' I.;,) Q \r-. J Q. 1."\. \1..b S ;" 'l-t. lrc. \J. We show that F is a direct sum of copies of the additive group of rationals. To this end let S be a maximal independent set (Theorem 6.13, page 194). For each S E S we shall define a subgroup Cs of F. Let r!,s = s. For a given S E S and positive integer i, there exists by the divisibility of F an element ri+Ls E F such that (i + l)ri+Ls = ri,s. We put Cs = gp(ri,s I i = 1,2,... ). It follows then from Problem 6.72 that Cs is iso morphic to the additive group of rational numbers. Note that if x E Cs, x =F 0, then there is a nonzero multiple of x which is also a multiple of s, as S =F 0. (This is true for any two |
nonzero rational numbers.) As we remarked above, there exists a nonzero multiple k j of each Cj which is then a multiple of Sj. Hence there exists a nonzero integer k, namely ki... kn, such that We claim that F is actually the direct sum of these subgroups Cs as S ranges over S. To prove this, suppose that SI, S2,..., Sn are distinct elements of S and that CI + C2 +... + Cn = 0, where Cj E CSj, Cj =F ° (j = 1,..., n). kCj = ljsj, where lj is a nonzero integer. As kCI +... + kCn = ° is a consequence of C1 + C2 +... + Cn = 0, we find therefore, on substituting for the elements kCj the elements ljsj, that llSl +... + lnsn = 0. But the elements Sl,..., Sn are independent. From this contradiction it follows that the Cs generate the direct sum C = gp(Cs I S E S) = 1: Cs s E S But C is divisible since each summand Cs is divisible. Hence C is a direct summand (Theorem 6.32), i.e. F = C EEl D, say. then let dE D (d =F 0). Clearly the set SU {d} is definitely larger than S since dEl S (d does not even lie in C) and S U {d} is independent. This is a contradiction as S is a maximal independent set. So F = C is a direct sum of copies of the rationals. If D =F {O}, (b) T. First of all T = 1: Tp by Theorem 6.10. Since T is divisible, so also are the summands Tp. It is sufficient then to assume that T is a divisible p-group and to prove that T is a direct sum of p-Priifer groups. pElT Let P be the set of elements of T of order at most p. P is clearly a subgroup. Let S be a maximal independent subset of P (Theorem 6.13). For each S E S define (b) PCi+l,s = ci,s for i = 1,2,.... (a) pc!,s = s, c1,s, |
c2,s,... inductively as follows: This is possible since T is divisible. Clearly gp(cl,s) C gp(c 2,s) c.. '. Since gp(ci,s) is cyclic of order pi+I, Cs = gp(c!,s, c 2,s,... ) is an ascending union of cyclic groups of order.pi, one for each i = 1,2,.... Accordingly by Theorem 6.31, Cs is a p-Priifer group. Then C = gp(Cs I S E S) is divisible. By Theorem 6.32, C is a direct summand of T, i.e. T = C EEl D. If D =F {O}, there is an element of dE D of order p. Since SeC, SU {d} is an independent subset of P larger then S, and so we must have D = {O} by the maximality of S. Thus T = C. 210 ABELIAN GROUPS [CHAP. 6 It remains only to prove that G = ~ Gs• If 81,82,..., 8 n are distinct elements of S, and Cl + C2 +... + Cn = 0 where Ci E GSi i = 1,..., n, then, similarly to (a) above, we arrive at a dependent relation between 81,..., 8 n, unless Cl = C2 =... = Cn = 0 (Problem 6.78). for s E S The proof of the theorem is complete. Problems 6.73. Prove that (a) every free abelian group is isomorphic to a subgroup of a divisible abelian group, and (b) every abelian group is isomorphic to a subgroup of a divisible group. Solution: (a) A free abelian group F is a direct sum of infinite cyclic groups Ci we choose one copy of the rationals Qi for each (i E I): F = ~ Ci• Now i E I. Let K = ~ Qi and let di ;6 0 be iEI iEI chosen in each Qi' F is clearly isomorphic to gp({di liE I}). But ~ Q i is divisible since each Qi is divisible. The result follows. i E I (b) If G is any group, G == FIN for |
some free abelian group F and some subgroup N. Now in (a) we have proved that there exists a divisible group D containing F. Hence D contains N, and so DIN contains as a subgroup FIN, i.e. G. Now DIN is divisible since a homomorphic image of a divisible group is divisible. Thus the result follows. 6.74. Suppose G has the property that if H is any group such that H d G, of H. Prove that G is divisible. then G is a direct summand Solution: G is a subgroup of some divisible group D by Problem 6.73. Thus D = G EB T. But every direct summand of a divisible group is divisible. Therefore G is divisible. 6.75. Let G be an infinite group whose proper subgroups are all finite. Prove that G is a p-Priifer group by using the theorem which states: if G is a group such that for some integer n;6 0, nG = {O}, then G is a direct sum of cyclic groups (of finite order). (This theorem is not proved in this text.) Solution: Consider the subgroups nG for all positive integers n. If nG = G for all such n, then G is divisible, and so G is the direct sum of p-Priifer groups and copies of the additive group of rationals. As all the proper subgroups of G are finite and the additive group of rationals has an infinite cyclic group as a proper subgroup, only p-Priifer groups are involved. Since each p-Priifer group is infinite, G must in fact be a p-Priifer group. If on the other hand nG is a proper subgroup of G for some n, then nG is a finite group of order m, say, and so mnG = {O}. Using the theorem quoted in the statement of the problem, G is the direct sum of finite cyclic groups. As G is infinite, it must be the direct sum of an infinite number of cyclic groups Ci• But then the subgroup generated by all but one of the Ci must be infinite. This contradiction proves the result. 6.76. (a) Let G be any group and let S be the subgroup generated by all the divisible subgroups of G. Prove that S is divisible. (b) Prove that G is |
the direct sum of a divisible group and a group which has no divisible subgroups other than the identity subgroup {O}. Solution: (a) Let 8 E S. Then 8 = hi + h2 +... + hn where each hi belongs to a divisible subgroup Hi of G. is any integer, there exist k i E Hi such that zki = hi' As S d Hi for each Thus if z;6 0 i E I, Hence S is divisible. Sec.6.4J DIVISIBLE GROUPS 211 (b) Since S is divisible, we can apply our direct summand theorem for divisible subgroups (Theorem 6.32) to find that G = S EEl T. As T contains no divisible subgroup other than {O}, the result follows. 6.77. (a) Prove that the additive group of rationals Q has a proper subgroup which is not free abelian. (b) Let G be a torsion-free group, every proper subgroup of which is free abelian. Prove that G is free abelian. Solution: (a) Consider the subgroup H generated by 1/2,1/4,1/8,..., 1/2i,... in the additive group of ra tionals. H is of rank 1 as Q is of rank 1 (Problem 6.38, page 194). So if H is free abelian, it must be infinite cyclic. But H is not infinite cyclic; for if it were, H = gp(z/2i) for some inte gers z and i. But then 1/2i + 1 E Hand 1/2i + 1 ~ gp(z/2i). We have only to prove that H is not G. This is obvious since 1/3 ~ H. (b) Suppose that nG = G for all positive integers n. Then G is divisible and is the direct sum of copies of the additive group of rationals. (As G is torsion-free, no p-Priifer group is involved.) But by (a) above, G will have a non-free subgroup. Thus for some n oF 0, nG oF G, and nG is free abelian. Now (J: g ~ ng is a homomorphism of G onto nG. Ker (J = {g I |
ng = O} = {O} as G is torsion-free. Hence (J is an isomorphism, and so G is free abelian. 6.78. Prove the result stated at the end of the proof of Theorem 6.33, i.e. prove that if C1 +... + cn = 0, then c1 = C2 =... = cn = O. Solution: Assume the contrary. As the order of C1'..., Cn is immaterial, we may assume that C1 oF 0 and C1 is of highest order. Say C1 is of order p!. Then p!-l ci = mi8i where mi is an integer, i = 1,2,...,no Thus we obtain p!-l (C1 +... + cn) = m181 + m282 +... + m n8n Since m181 oF 0, we have a contradiction to the fact that S is an independent set. A look back at Chapter 6 This chapter was mainly concerned with the structure of divisible and finitely gen erated abelian groups. Direct sums of groups were discussed. Given a family Ai (i E J) of groups, there is always a group which is the direct sum of groups isomorphic to each of the groups Ai. Any homomorphism of the direct summands of a group extends to a homomorphism of the whole group. From this it follows that if two groups are direct sums of isomorphic sub groups, they are isomorphic. Direct sums of infinite cyclic groups are all the free abelian groups. An important fact is that every abelian group is a homomorphic image of a free abelian group. The torsion group T(G) of a group G was defined, and it was shown that G/T(G) is tor sion-free. It was proved that if G is a torsion group, it is the direct sum of its p-components. This led to the definition of the p-Priifer group as the p-component of Q/Z. An application of Zorn's lemma proves every group has a maximal independent subset. The rank of a group was defined and proved an invariant of the group by the Steinitz exchange theorem. In the fundamental theorem of abelian groups, finitely generated abelian groups were shown to be expressible as the direct sum of |
cyclic groups. Two finitely generated abelian groups were shown to be isomorphic if and only if they have the same type. Finally, the type of a subgroup of a group was shown to be, roughly speaking, "less than" the type of the group. 212 ABELIAN GROUPS [CHAP. 6 Divisible groups were discussed. Any group which is the union of cyclic groups of order a power of p turns out to be a p-Prufer group. Any divisible subgroup of a group is also a direct summand. This led to the proof that every divisible group is the direct sum of isomorphic copies of the additive group of rationals and p-Prufer groups. Supplementary Problems DIRECT SUMS AND FREE ABELIAN GROUPS 6.79. If the mapping a --> a-l, a E G, is an automorphism of the group G, prove G is abelian. 6.80. Suppose that G is a finite group, a E aut (G), a is of order 2, and ga -#- g for all g (-#- 1) E G. Show that G is an abelian group. (Hint: First prove G = {g-l(ga) I g E G} and then use Problem 6.79.) 6.81. Denote the set of all homomorphism of an abelian group G into an abelian group H by Hom (G, H). If ¢,,y E Hom (G, H), we define ¢ +'1' by g(¢ +'1') = g¢ + g,y (1) for all g E G. Show that Hom (G, H) with the operation defined by (1) is an abelian group. 6.82. If A is an abelian group and Z is the group of integers under addition, prove that Hom (Z, A) "'" A. 6.83. Prove that the group of rationals under addition is not the direct sum of cyclic groups. 6.84. If G is the direct sum of cyclic subgroups, show that a factor group of G is not necessarily a direct sum of cyclic subgroups. (Hint: Use the preceding problem and free abelian groups.) f.85. Let N be a normal subgroup of G. Prove that if GIN is free abelian |
, N is a direct summand. TORSION GROUP AND RANK 6.86. If G is a finite group, show that aut (G) "'" II aut (Gp ), where 'iT is the set of all primes. pE1T 6.87. Prove the first Sylow theorem, page 130, for abelian groups using the p-components. 6.88. Let G denote the group of rotations of the plane (see Section 3Ac, page 68). As G is an abelian group is Pe + Pe = Pe +6 • we may use additive notation. Thus the rotation Pe 1 Let CJ{ = {Pe I 0 = 21Tmln radians where m, n are integers, n > O}. Prove that (a) CJ{ is a subgroup of G; (b) every element of CJ{ has finite order; (c) if CJ{p denotes the p component of CJ{ for any prime p, then CJ{p = {Pe I 0 = 21TmlpT radians where m and r are integers} and CJ{p "'" the p-Priifer group. followed by rotation Pe 2 2 2 1 1 6.89. Let a, b be elements of an abelian group. Let the order of a plus the order of b be n. Prove by induction on n.that a + b is of order the least common multiple of the orders of a and b. 6.90. Let G be an abelian group. Suppose every element of G is of order less than some fixed integer n and there are elements of order n in G. Prove that the elements of order n generate G. CHAP. 6] SUPPLEMENTARY PROBLEMS 213 FUNDAMENTAL THEOREM FOR FINITELY GENERATED ABELIAN GROUPS 6.91. Prove that if G is a finitely generated abelian group, then G = 11 EB... EB 1m EB C1 EB •.• EB Cn where (i = 1,..., n), I j is an infinite cyclic group (j = 1,..., m), Ci is a finite cyclic group of order Vi i = 1,..., n - 1. (Hint: Use the fundamental theorem and then first look and vi divides Vi + 1 for at the highest power of each different prime in the decomposition.) 6. |
92. Prove that the automorphism group of a finitely generated abelian group is finite if and only if there is at most one infinite cyclic summand in a cyclic decomposition of G. 6.93. Find the type of the additive group of integers modulo m for any integer m > O. 6.94. Let G be a non-cyclic finite abelian group. Show that G has a subgroup of type (p, p; 0) for some prime p. 6.95. Prove that the automorphism group of a finite non-cyclic abelian group G is non-abelian. (Hint: Use Problem 6.91 to find suitable elements a, bEG such that the order of a divides the order of b. Then look at the mappings a1: asb t -> as+tbt; a2: asbt -> awb- t and a3: asbt -> atbs where 8 and t are integers.) 6.96. Let G be a finite abelian group. Suppose that for each divisor d of G there are at most d ele ments in G of order d. Prove that G is cyclic. 6.97. Let G be a finitely generated abelian group. Prove by induction on the number of generators of G that every subgroup of G is finitely generated. DIVISIBLE GROUPS 6.98. Show that a divisible abelian group has no subgroup of finite index. 6.99. If G is a nondivisible abelian group, then G has a subgroup of prime index. (Hint: Use the fol lowing theorem (not proved in this book): An abelian group G for which nG = {O}, n of= 0, is the direct sum of cyclic groups.) 6.100. Prove that the additive group of the real numbers is the direct sum of isomorphic copies of the additive group of rationals. 6.101. (a) Let G = Hom (A, B) (see Problem 6.81) where B is a torsion-free divisible group. Prove that G is the direct sum of copies of the additive group of rationals. (b) Let G be as in (a) but with B a divisible p-group. Prove that if A is finite, G is the direct sum |
of p-Priifer groups. 6.102. A subgroup H of an abelian group A is a pure subgroup of A if whenever na = h E H for some a E A, then there is an h' E H such that nh' = h. Prove that (a) a direct summand of an abelian group is a pure subgroup, and (b) the torsion subgroup of an abelian group is a pure subgroup. 6.103. Prove that all the subgroups of a group in which every element has square-free order is pure. 6.104. Let H be a pure subgroup of an abelian group G. Prove that if g + A E G/A, there is an element ii E G such that ii + A = g + A and the order of ii is equal to the order of g + A. Chapter 7 Permutational Representations Preview of Chapter 7 There are three main divisions of this chapter. In the first we generalize Cayley's theorem, that every group is isomorphic to a permutation group. As consequences of this generalization we prove the following theorems for G, a group generated by a finite number of elements: (2) The number of subgroups of fixed finite index in G is finite. (3) If the subgroups of finite index of G intersect in the identity, then every homomorphism of G onto G is an automorphism. (1) A subgroup of finite index in G is itself finitely generated. The second main division of this chapter appears in Section 7.7. We call a group G an extension of a group H by a group K if there is a normal subgroup il of G such that GIN"" K and il "" H. We examine G to see how it is built up from Hand K. The most general case is complicated and we restrict ourselves to a special extension called "the splitting extension." Reversing our analysis, we are able to build a group G that is the splitting extension of a given group H by a given group K. A particular example of a splitting extension is the direct product, used in Chapter 5. Our third division, which begins in Section 7.8, defines a homomorphism of a group into one of its abelian subgroups. This homomorphism is called the transfer. We use it to show that a group G with center of finite index has finite |
derived group. 7.1 CAYLEY'S THEOREM a. Another proof of Cayley's theorem We saw in Chapter 2 that every groupoid is isomorphic to a groupoid of mappings. In particular, every group is isomorphic to a group of permutations. The consequences of this theorem are important. We repeat the proof here for the case of groups alone. Theorem 7.1 (Cayley): Every group is isomorphic to a group of permutations. Proof: Let G be a group. Let p be the mapping which assigns to each element g in G ·the following mapping of G into G: Thus the image of gin Gunder p is, by definition, gp where gp: x ~ xg (x E G) x ~ xg for each x E G The definition of p is unambiguous. To prove that p defines an isomorphism of G onto a subgroup of Sa, we have to check that: (i) gp is a permutation of G for every g E G; (ii) p is a homomorphism, i.e. if g, hE G, then (gh)p = gp. hp; (iii) p is also an isomorphism, i.e. p is one-to-one. 214 Sec. 7.1] CAYLEY'S THEOREM 215 We deal first with (i). Thus we must prove gp is a one-to-one mapping of G onto G. If x(gp) = y(gp), then xg = yg. So, multiplying by g-l on the right, we find x = y. Next we prove gp is a mapping of G onto G. Suppose x E G; then (xg-l)(gp) = (xg-l)g = x and so gp is onto. Secondly we prove (ii). For x E G, x((gh)p) = x(gh) = (xg)h = (x(gp))(hp) = x(gphp) Since (gh)p and gphp have precisely the same effect on every element of G, (gh)p = gphp (by the definition of equality of mappings). It remains to prove (iii), i.e. p is one-to-one. Suppose gp = hp; identity element of G, g = l(gp) = l(hp) = h, i.e. g |
= h. Therefore p is one-to-one. then if 1 is the (Note: In the proof of Cayley's theorem p is a mapping of G into Sc, so that gp is itself a mapping of G to G. Caution and patience are required to avoid confusion in some of our subsequent equations.) h. Cayley's theorem and examples of groups Cayley's theorem tells us that there is an isomorphic image of every group among the per mutation groups of suitably chosen sets. If one demands that a permutation group satisfy further conditions, one frequently comes across interesting groups (see Chapter 3). His torically many important groups arose in precisely this way. Problem 7.1. Describe in detail the isomorphisms given by Cayley's theorem for (i) a cyclic group of order 2, (ii) a cyclic group of order n (n "'" 3), (iii) the symmetric group on three letters. Solution: (i) Let G be cyclic of order 2. Then G consists of two elements, 1 and a, where a2 = 1 and l'a = a = a'!. Let P be as in Theorem 7.1. 1p is the mapping 1p: 1 ~ 1, a -> a and ap is the mapping given by ap: 1 ~ a, a -> 1 Clearly p is one-to-one, as ap oF 1p. (ii) Let G be cyclic of order n. Then G consists of n elements 1, a, a2,..., an-I, say (see Lemma 4.5, page 102). Then 1p: 1 -> 1, 1 -> a, 1 ~ a2, a -> a,.. 0' an-l -> a n - l •• 0, an- l -> 1 •. 0' a n - 2 -> 1, a n - 1 -> a an-Ip: 1-> an-I, a -> 1,..., a n - 1 -> a n - 2 (iii) The symmetric group on the set {1, 2, 3} consists of the permutations PI : 1 -> 1, 2 -> 2, 3 -> 3 P2: Pa: 1 -> 1, 2 -> 3, 3->2 1 -> 2, 2 -> 3, 3->1 P4 : Ps: Pa: 1 -> 2, 2 -> 1, 3 -> 3 1 ~ 3, 2 -> 2, 3->1 |
1 -> 3, 2 -> 1, 3->2 Then PaP: PIP: P2P: PI -> PI, P2 -> P2' Pa -> Pa, P4 -> P4, Ps -> Ps, P6 -> P6 PI -> P2, P2 -> PI' Pa -> Ps, P4 -> Pa, Ps -> Pa, P6 -> P4 PI -> Pa, P2 -> P4' Pa -> P6' P4 -> Ps, Ps -> P2, P6 -> PI PI -> P4' P2 -> Pa, Pa -> P2' P4 -> PI, Ps -> P6, P6 -> Ps PI -> Ps, P2 -> Pa, Pa -> P4, P4 -> Pa, Ps -> PI, P6 -> P2 PI -> Pa, P2 -> Ps' Pa -> PI> P4 -> P2' Ps -> P4' P6 -> Pa It is worth checking (PiP)(PjP) = (PiPj)P for some i and j between 1 and 6. P4P: PsP: P6P: 216 PERMUTATIONAL REPRESENTATIONS [CHAP. 7 7.2 PERMUTATIONAL REPRESENTATIONS Definition of a permutational representation A homomorphism of a group G into the symmetric group on the set X is called a per mutational representation of G on X. So if p is the isomorphism provided by Cayley's theorem for the group G, then p is a permutational representation of G on G. Repeating the definition of a permutational representation of a group G in detail, we say that a mapping p. of G into the symmetric group on some set X is a permutational representation of G if (gh)p. = gp.hp. for all g and h in G. The permutational representation provided by Cayley's theorem is called the right-regular representation (the adjective right is used because the representa tion is obtained by multiplication on the right), Example 1: (il Let G be the symmetric group on {I, 2, 3}, and let p be as in the solution to Problem 7.1(iii). Then p itself is a representation of G as a permutation group on six elements. (ii) There is another representation of the symmetric group G on {I, 2, 3}, the most natural one. This is the identity isomorphism |
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