Section 11 Direct products and finitely generated abelian groups

Section 11 Direct products and finitely generated abelian groups Instructor: Yifan Yang Fall 2006

Outline Direct products Finitely generated abelian groups

Cartesian product Definition The Cartesian product of sets S 1, S 2,..., S n is the set of ordered n-tuples (a 1, a 2,..., a n ), where a i S i for i = 1, 2,..., n. It is denoted by either S 1 S 2... S n or n S i. i=1

Direct products Theorem Let G 1, G 2,..., G n be groups. For (a 1, a 2,..., a n ) and (b 1, b 2,..., b n ) in n i=1 G i define (a 1, a 2,..., a n )(b 1, b 2,..., b n ) to be (a 1 b 1, a 2 b 2,..., a n b n ). Then n i=1 G i is a group under this binary operation. Proof. Straightforward. See the textbook. Definition The group n i=1 G i defined above is called the direct product of the groups G i.

Direct sums When the groups G i are all abelian with operation +, we sometimes use the notation G 1 G 2 G n = n i=1 G i instead of G 1 G 2 G n = n i=1 G i. Also, we call the group n i=1 G i the direct sum of the groups G i. Remark The changing of the order of the factors in a direct product yields a group isomorphic to the original one. For example, define φ : G 1 G 2 G 2 G 1 by φ((g 1, g 2 )) = (g 2, g 1 ). It is easy to verify that φ is an isomorphism.

Example Consider Z 2 Z 3. It has 6 elements, namely (0, 0), (0, 1), (0, 2), (1, 0), (1, 1), and (1, 2). Since it is abelian, it must be isomorphic to Z 6. (A group of order 6 is isomorphic to either Z 6 or S 3.) In fact, if we set g = (1, 1), then we have 2g = (1, 1) + (1, 1) = (0, 2) 3g = 2g + (1, 1) = (0, 2) + (1, 1) = (1, 0) 4g = 3g + (1, 1) = (1, 0) + (1, 1) = (0, 1) 5g = 4g + (1, 1) = (0, 1) + (1, 1) = (1, 2) 6g = 5g + (1, 1) = (1, 2) + (1, 1) = (0, 0). In other words, Z 2 Z 3 is cyclic of order 6, and hence isomorphic to Z 6.

Example Consider Z 2 Z 2. It has 4 elements. Thus, it is isomorphic to either Z 4 or Z 8 = {1, 3, 5, 7}. Now for all (a, b) Z 2 Z 2, we have 2(a, b) = (0, 0). Thus, there is no element having order 4. The group must be isomorphic to Z 8.

When is a direct sum of abelian groups cyclic? Theorem (11.5) The group Z m Z n is cyclic and isomorphic to Z mn if and only if m and n are relatively prime. Proof. Let (a, b) be an element in Z m Z n. The order of a in Z m divides m, and the order of b in Z n divides n. Thus, if k is a multiple of lcm(m, n), then we have k(a, b) = (0, 0). If lcm(m, n) mn, i.e., if gcd(m, n) 1, then Z m Z n cannot be cyclic. Conversely, assume that gcd(m, n) = 1. Consider the element (1, 1). If k(1, 1) = (0, 0), then k must be both a multiple of m and a multiple of n, i.e., lcm(m, n) k. Since gcd(m, n) = 1, we have lcm(m, n) = mn. Thus (1, 1) = mn = Z m Z n. Therefore Z m Z n is cyclic and isomorphic to Z mn.

When is a direct sum of abelian groups cyclic? Corollary (11.6) The group n i=1 Z m i is cyclic and isomorphic to Z m1 m 2...m n if and only if the numbers m i are pairwise coprime. Example 1. The group Z 30 is isomorphic to Z 2 Z 3 Z 5, also to Z 6 Z 5, Z 2 Z 15, and so on. 2. The group Z 12 Z 21 is isomorphic to Z 84 Z 3 because Z 12 Z 21 Z 12 Z 7 Z 3 Z 84 Z 3.

Order of an element in a direct product Theorem (11.9) Let (a 1, a 2,..., a n ) n i=1 G i. If a i is of finite order r i in G i, then the order of (a 1, a 2,..., a n ) in n i=1 G i is equal to the least common multiple of all r i. Proof. Since (a 1, a 2,..., a n ) k = (a k 1, ak 2,..., ak n), if (a 1, a 2,..., a n ) k = (e 1, e 2,..., e n ), then r i k for all i. That is, k must be a multiple of lcm(r 1, r 2,..., r n ). This is a necessary and sufficient condition. Therefore, the order of (a 1, a 2,..., a n ) is lcm(r 1, r 2,..., r n ).

Examples 1. Consider (9, 5, 7) in the group Z 12 Z 15 Z 20. The element 9 in Z 12 is of order 12/ gcd(9, 12) = 4, the element 5 in Z 15 has order 15/ gcd(5, 15) = 3, and the element 7 in Z 20 has order 20/ gcd(7, 20) = 20. The least common multiple of 4, 3, and 20 is 60. Thus, the order of (9, 5, 7) is 60. 2. Consider (8, 4, 10) in the group Z 12 Z 60 Z 24. The element 8 in Z 12 has order 3 since gcd(8, 12) = 4, the element 4 in Z 60 has order 15, and element 10 in Z 24 has order 12. The least common multiple of 3, 15, and 12 is 60. Thus, the order of (8, 4, 10) is 60.

In-class exercises 1. Determine whether Z 18 Z 15 Z 12 is isomorphic to Z 9 Z 45 Z 8. 2. Determine whether Z 18 Z 15 Z 12 is isomorphic to Z 9 Z 10 Z 36. 3. Find the order of (2, 6, 3) in Z 12 Z 9 Z 15. 4. Find the order of (3, 4, 8) in Z 12 Z 18 Z 20.

Finitely generated abelian groups Definition An abelian group is finitely generated if it can be generated by a finite number of elements. Theorem (Fundamental theorem of finitely generated abelian groups) Suppose that G is a finitely generated abelian group. Then G is isomorphic to a direct product of cyclic groups in the form Z p e 1 1 Z p e 2 2... Z p en n Z Z... Z, where p i are primes, not necessarily distinct. The direct product is unqiue except for possible rearrangement of the factors.

Finitely generated abelian groups Definition Let G be a finitely generated abelian group. The number of copies of Z in the above theorem is the Betti number of G. In some occasions, it is also called the rank of G.

Example Example Using the above theorem, we find that the only abelian groups of order 360 = 2 3 3 2 5, up to isomorphism, are Z 2 3 Z 3 2 Z 5 ( Z 360 ), Z 2 2 Z 2 Z 3 2 Z 5, Z 2 Z 2 Z 2 Z 3 2 Z 5, Z 2 3 Z 3 Z 3 Z 5, Z 2 2 Z 2 Z 3 Z 3 Z 5, Z 2 Z 2 Z 2 Z 3 Z 3 Z 5.

Finitely generated abelian groups In general, if the prime factorization of a positive integer n is n = p e 1 1 pe 2 2... pe k k, where p i are distinct primes, then the number of all abelian groups of order n, up to isomorphism, is equal to p(e 1 )p(e 2 )... p(e n ), where p(e) is the number of ways to write an integer e as a sum of positive integers. (For example, Thus p(4) = 5.) 4 = 4 = 3 + 1 = 2 + 2 = 2 + 1 + 1 = 1 + 1 + 1 + 1.

Example Consider n = 200 = 2 3 5 2. We have 3 = 3 = 2 + 1 = 1 + 1 + 1, and 2 = 2 = 1 + 1. Thus, there are 3 2 non-isomorphic abelian groups of order 200. They are Z 2 3 Z 5 2 ( Z 200 ) Z 2 3 Z 5 Z 5 Z 2 2 Z 2 Z 5 2 Z 2 2 Z 2 Z 5 Z 5 Z 2 Z 2 Z 2 Z 5 2 Z 2 Z 2 Z 2 Z 5 Z 5.

Applications Theorem (11.16) If m divides the order of a finite abelian group G, then G has a subgroup of order m. Proof. By the fundamental theorem of finitely generated abelian groups, we have G Z e p 1 Z e 1 p 2 Z p en. If m G, then 2 n m = p f 1 1... pn fn for some non-negative integers f i. (There may be more than one way to write m since p i are not necessarily distinct.). Then the subgroup has order p f 1 1... p fn n = m. p e 1 f 1 1 p e 2 f 2 2 p en fn n

Remark The statement in Theorem 11.16 in general is false if G is not abelian. For example, A 4 has 12 elements, but it does not have a subgroup of order 6. To see this, suppose that H < A 4 has order 6. Then it has an element τ of order 2 and an element σ of order 3 with τστ 1 = σ i, where i = 1 or 2. Without loss of generality, we may assume that σ = (1, 2, 3). Now there are 3 elements of order 2 in A 4, namely, τ 1 = (1, 2)(3, 4), τ 2 = (1, 3)(2, 4), and τ 3 = (1, 4)(2, 3). However, τ 1 στ 1 1 = (1, 2)(3, 4)(1, 2, 3)(3, 4)(1, 2) = (1, 4, 2) τ 2 στ 1 2 = (1, 3)(2, 4)(1, 2, 3)(2, 4)(1, 3) = (1, 3, 4) τ 3 στ 1 3 = (1, 4)(2, 3)(1, 2, 3)(2, 3)(1, 4) = (2, 4, 3). Thus, there is no subgroup of order 6 in A 4.

Exercises In-class exercises 1. Let p(e) be the number of ways to write a positive integer e as a sum of positive integers. Find p(8). 2. Find all abelian groups of order 144, up to isomorphism. 3. Find all abelian groups of order 216, up to isomorphism. Homework Problems 6, 8, 10, 12, 16, 18, 24, 26, 29, 38, 39, 47 of Section 11.