(0, 0) : order 1; (0, 1) : order 4; (0, 2) : order 2; (0, 3) : order 4; (1, 0) : order 2; (1, 1) : order 4; (1, 2) : order 2; (1, 3) : order 4.


 Frederick Dalton
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1 11.01 List the elements of Z 2 Z 4. Find the order of each of the elements is this group cyclic? Solution: The elements of Z 2 Z 4 are: (0, 0) : order 1; (0, 1) : order 4; (0, 2) : order 2; (0, 3) : order 4; (1, 0) : order 2; (1, 1) : order 4; (1, 2) : order 2; (1, 3) : order 4. This group is not cyclic since no element can generate the whole group List the elements of Z 3 Z 4. Find the order of each of the elements is this group cyclic? Solution: The elements of Z 3 Z 4 are: (0, 0) : order 1; (0, 1) : order 4; (0, 2) : order 2; (0, 3) : order 4; (1, 0) : order 3; (1, 1) : order 12; (1, 2) : order 6; (1, 3) : order 12; (2, 0) : order 3; (2, 1) : order 12; (2, 2) : order 6; (2, 3) : order 12. This group is cyclic since it can be generated by either of the elements (1, 1), (1, 3), (2, 1), and (2, 3). 18
2 11.13 Disregarding the order of the factors, write direct products of two or more groups of the form Z n so that the resulting product is isomorphic to Z 60 in as many ways as possible. Solution: There are 4 different ways: Z 60 = Z 2 2 Z 3 Z 5 = Z 4 Z 3 Z 5, Z 60 = Z Z 5 = Z 12 Z 5, Z 60 = Z Z 3 = Z 20 Z 3, Z 60 = Z 2 2 Z 3 5 = Z 4 Z a. The cyclic subgroup of Z 24 generated by 18 has order 4. b. Z 3 Z 4 is of order 12. c. The element (4, 2) of Z 12 Z 8 has order 12. d. The Klein 4group is isomorphic to Z 2 Z 2. e. Z 2 Z Z 4 has 8 elements of finite order Find the maximum possible order for some element of Z 4 Z 6. Solution: (1, 1) in Z 4 Z 6 has the maximum order lcm(4, 6) = Are the groups Z 8 Z 10 Z 24 and Z 4 Z 12 Z 40 isomorphic? Why or why not? Solution: We decompose both groups into indecomposible ones: Z 8 Z 10 Z 24 Z 8 (Z 2 Z 5 ) (Z 8 Z 3 ) = Z 2 (Z 8 ) 2 Z 3 Z 5, Z 4 Z 12 Z 40 Z 4 (Z 4 Z 3 ) (Z 8 Z 5 ) = (Z 4 ) 2 Z 8 Z 3 Z 5. So they are not isomorphic How many abelian groups (up to isomorphism) are there of order 24? of order 25? of order (24)(25)? Solution: 24 = = = So there are 3 abelian groups of order 24: Z 2 3 Z 3, Z 2 Z 2 2 Z 3, Z 2 Z 2 Z 2 Z = 5 2 = 5 5. So there are 2 abelian groups of order 25: Z 5 2, Z 5 Z 5. 19
3 Because gcd(24, 25) = 1, there are 3 2 = 6 abelian groups of order (24)(25): Z 2 3 Z 3 Z 5 2, Z 2 3 Z 3 Z 5 Z 5, Z 2 Z 2 2 Z 3 Z 5 2, Z 2 Z 2 2 Z 3 Z 5 Z 5, Z 2 Z 2 Z 2 Z 3 Z 5 2, Z 2 Z 2 Z 2 Z 3 Z 5 Z Mark each of the following true or false: a. (T) If G 1 and G 2 are any groups, then G 1 G 2 is always isomorphic to G 2 G 1. b. (T) Computation in an external direct product of groups is easy if you know how to compute in each component group. c. (F) Groups of finite order must be used to form an external direct product. d. (T) A group of prime order could not be the internal direct product of two proper nontrivial subgroups. e. (F) Z 2 Z 4 is isomorphic to Z 8. f. (F) Z 2 Z 4 is isomorphic to 8. g. (F) Z 3 Z 8 is isomorphic to 4. h. (F) Every element in Z 4 Z 8 has order 8. i. (F) The order of Z 12 Z 15 is 60. j. (T) Z m Z n has mn elements whether m and n are relatively prime or not a. How many subgroups of Z 5 Z 6 are isomorphic to Z 5 Z 6? Solution: No subgroup of Z 5 Z 6 is isomorphic to Z 5 Z 6. b. How many subgroups of Z Z are isomorphic to Z Z? Solution: There are infinite many subgroups of Z Z that are isomorphic to Z Z. They are of the form mz nz for positive integers m and n with m 1 or n Mark each of the following true or false: 20
4 a. (T) Every abelian group of prime order is cyclic. b. (F) Every abeliang roup of prime power order is cyclic. c. (F) Z 8 is generated by {4, 6}. d. (T) Z 8 is generated by {4, 5, 6}. e. (T) All finite abelian groups are classified up to isomorphism by Theorem f. (F) Any two finitely generated abelian gruops witht he same Betti number are isomorphic. g. (T) Every abelian group of order divisible by 5 contains a cyclic subgroup of order 5. h. (F) Every abelian group of order divisible by 4 contains a cyclic subgroup of order 4. i. (T) Every abelian group of order divisible by 6 contains a cyclic subgroup of order 6. j. (T) Every finite abelian group has a Betti number of Prove that a direct product of abelian groups is abelian. Solution: Suppose G i are abelian groups. We prove that n i=1 G i is an abelian group. Let (a 1,, a n ) and (b 1,, b n ) be elements of n i=1 G i. Then (a 1,, a n )(b 1,, b n ) = (a 1 b 1,, a n b n ) = (b 1 a 1,, b n a n ) = (b 1,, b n )(a 1,, a n ). This shows that the binary operation on n i=1 G i is commutative. So n i=1 G i is an abelian group Let G be an abelian group. Let H be the subset of G consisiting of the identity e together with all elements of G of order 2. Show that H is a subgroup of G. Solution: We show that H meets the criteria of a subgroup of G: 1. (Closed) If a, b H, then a 2 = b 2 = e. So (ab) 2 = a 2 b 2 = e. This implies that ab H. 2. (Identity) The identity is in H by definition. 21
5 3. (Inverse) If a H, then a 2 = e and so a 1 = a H. Therefore, H is a subgroup of G Prove that if a finite abelian group has order a power of a prime p, then the order of every element in the group is a power of p. Can the hypothesis of commutativity be dropped? Why, or why not? Solution: Suppose a group G has the order p k for some prime p and some positive integer k. Then the order m of every element a of G divides the order p k of G. So m = p r for some integer 0 r k. That is, the order of a is a power of p. The hypothesis of commutativity can be dropped, since we do not use the commutativity in the above argument. 22
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