Section 11 Direct Products and Finitely Generated Abelian Groups One purpose of this section is to show a way to use known groups as building blocks to form more groups. Definition: The Cartesian product of sets S 1, S 2, …,S n is the set of all ordered n- tuples (a 1, a 2, …,a n ), where a i  S i for i=1, 2, …, n. The Cartesian product is denoted by either S 1  S 2  …  S n or by

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, Define (a 1, a 2, …,a n )(b 1, b 2, …,b n ) to be the element (a 1 b 1, a 2 b 2, …,a n b n ). Then is a group, the direct product of the groups G i, under this binary operation. Proof: exercise. Note: In the event that the operation of each G i is commutative, we sometimes use additive notation in and refer to as the direct sum of the groups G i. If the S i has r i elements for i=1, …,n, then has r 1 r 2,…,r n elements.

Example Example: Determine if Z 2  Z 3 is cyclic. Solution: | Z 2  Z 3 |=6 and Z 2  Z 3 = {(0, 0),(0, 1),(0, 2),(1, 0),(1, 1),(1, 2)}. Here the operations in Z 2,, Z 3 are written additively. We can check that (1, 1) is the generator, so Z 2  Z 3 is cyclic. Hence Z 2  Z 3 is isomorphic to Z 6. (there is, up to isomorphism, only one cyclic group structure of a given order.) Example: Determine if Z 3  Z 3 is cyclic. Solution: We claim Z 3  Z 3 is not cyclic. |Z 3  Z 3 |=9, but every element in Z 3  Z 3 can only generate three elements. So there is no generator for Z 3  Z 3. Hence Z 3  Z 3 is not isomorphic to Z 9. Similarly, Z 2  Z 2 is not cyclic, Thus Z 2  Z 2 must be isomorphic to Z 6.

Theorem The group Z m  Z n is cyclic and is isomorphic to Z mn if and only if m and n are relatively prime, that is, the gcd of m and n is 1. Corollary The group is cyclic and isomorphic to Z m1m2..mn if and only if the numbers for i =1, …, n are such that the gcd of any two of them is 1.

Example The previous corollary shows that if n is written as a product of powers of distinct prime numbers, as in Then Z n is isomorphic to Example: Z 72 is isomorphic to Z 8  Z 9.

Least Common Multiple Definition Let r 1 r 2,…,r n be positive integers. Their least common multiple (lcm) is the positive integer of the cyclic group of all common multiples of the r i, that is, the cyclic group of all integers divisible by each r i for i=1, 2, …, n. Note: from the definition and the work on cyclic groups, we see that the lcm of r 1 r 2,…,r n is the smallest positive integer that is a multiple of each r i for i=1, 2, …, n, hence the name least common multiple.

Theorem Let (a 1, a 2, …,a n ) . If a i is of finite order r i in G i, then the order of (a 1, a 2, …,a n ) in is equal to the least common multiple of all the r i.

Example Example: Find the order of (8, 4, 10) in the group Z 12  Z 60  Z 24. Solution: The order of 8 in Z 12 is 12/gcd(8, 12)=3, the order of 4 in Z 60 is 60/gcd(4, 60)=15, and the order of 10 in Z 24 is 24/gcd(10, 24)=12. The lcm(3, 5, 12)=60, so (8, 4, 10) is or order 60 in the group Z 12  Z 60  Z 24.

The structure of Finitely Generated Abelian Groups Theorem (Fundamental Theorem of Finitely Generated Abelian Groups) Every finitely generated abelian group G is isomorphic to a direct product of cyclic groups in the form Where the p i are primes, not necessarily distinct, and the r i are positive integers. The direct product is unique except for possible rearrangement of the factors; that is, the number (Betti number of G) of factors Z is unique and the prime powers are unique.

Example Example: Find all abelian groups, up to isomorphism, of order 360. Solution: Since the groups are to be of the finite order 360, no factors Z will appear in the direct product in the theorem. Since 360= Then by theorem, we get the following: 1.Z 2  Z 2  Z 2  Z 3  Z 3  Z 5 2.Z 2  Z 4  Z 3  Z 3  Z 5 3.Z 2  Z 2  Z 2  Z 9  Z 5 4.Z 2  Z 4  Z 9  Z 5 5.Z 8  Z 3  Z 3  Z 5 6.Z 8  Z 9  Z 5 There are six different abelian groups (up to isomorphism) of order 360.

Application Definition A group G is decomposable if it is isomorphic to a direct product of two proper nontrivial subgroups. Otherwise G is indecomposable. Theorem The finite indecomposable abelian groups are exactly the cyclic groups with order a power of a prime. Theorem If m divides the order of a finite abelian group G, then G has a subgroup of order m. Theorem If m is a square free integer, that is, m is not divisible of the square of any prime, then every abelian group of order m is cyclic.