1 / 11

Section 11 Direct Products and Finitely Generated Abelian Groups

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:

dionysius-hilary
Download Presentation

Section 11 Direct Products and Finitely Generated Abelian Groups

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 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 productof sets S1, S2, …,Sn is the set of all ordered n-tuples (a1, a2, …,an), where aiSi for i=1, 2, …, n. The Cartesian product is denoted by either S1S2 …Sn or by

  2. Theorem Theorem Let G1, G2, …,Gn be groups. For (a1, a2, …,an) and (b1, b2, …,bn) in , Define (a1, a2, …,an)(b1, b2, …,bn) to be the element (a1 b1, a2 b2, …,an bn). Then is a group, the direct productof the groups Gi, under this binary operation. Proof: exercise. Note: • In the event that the operation of each Gi is commutative, we sometimes use additive notation in and refer to as the direct sum of the groups Gi. • If the Si has ri elements for i=1, …,n, then has r1r2,…,rn elements.

  3. Example Example: Determine if Z2  Z3 is cyclic. Solution: | Z2  Z3 |=6 and Z2  Z3 ={(0, 0),(0, 1),(0, 2),(1, 0),(1, 1),(1, 2)}. Here the operations in Z2, , Z3 are written additively. We can check that (1, 1) is the generator, so Z2  Z3 is cyclic. Hence Z2  Z3 is isomorphic to Z6. (there is, up to isomorphism, only one cyclic group structure of a given order.) Example: Determine if Z3  Z3 is cyclic. Solution: We claim Z3  Z3 is not cyclic. |Z3  Z3|=9, but every element in Z3  Z3 can only generate three elements. So there is no generator for Z3  Z3. Hence Z3  Z3 is not isomorphic to Z9. Similarly, Z2  Z2 is not cyclic, Thus Z2  Z2 must be isomorphic to Z6.

  4. Theorem Theorem The group ZmZn is cyclic and is isomorphic to Zmn 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 Zm1m2..mn if and only if the numbers for i =1, …, n are such that the gcd of any two of them is 1.

  5. Example The previous corollary shows that if n is written as a product of powers of distinct prime numbers, as in Then Zn is isomorphic to Example: Z72 is isomorphic to Z8 Z9.

  6. Least Common Multiple Definition Let r1r2,…,rn be positive integers. Their least common multiple (lcm) is the positive integer of the cyclic group of all common multiples of the ri, that is, the cyclic group of all integers divisible by each ri for i=1, 2, …, n. Note: from the definition and the work on cyclic groups, we see that the lcm of r1r2,…,rn is the smallest positive integer that is a multiple of each ri for i=1, 2, …, n, hence the name least common multiple.

  7. Theorem Theorem Let (a1, a2, …,an) . If ai is of finite order ri in Gi, then the order of (a1, a2, …,an) in is equal to the least common multiple of all the ri.

  8. Example Example: Find the order of (8, 4, 10) in the group Z12  Z60  Z24. Solution: The order of 8 in Z12 is 12/gcd(8, 12)=3, the order of 4 in Z60 is 60/gcd(4, 60)=15, and the order of 10 in Z24 is 24/gcd(10, 24)=12. The lcm(3, 5, 12)=60, so (8, 4, 10) is or order 60 in the group Z12  Z60  Z24.

  9. 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 pi are primes, not necessarily distinct, and the ri 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.

  10. 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=23325. Then by theorem, we get the following: • Z2  Z2  Z2  Z3  Z3  Z5 • Z2  Z4  Z3  Z3  Z5 • Z2  Z2  Z2  Z9  Z5 • Z2  Z4  Z9  Z5 • Z8  Z3  Z3  Z5 • Z8  Z9  Z5 There are six different abelian groups (up to isomorphism) of order 360.

  11. 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.

More Related