Handout#5 Math 545 Fall, 2010 1. List the Elements of the Subgroups

Handout#5 Math 545 Fall, 2010

1. List the elements of the subgroups < 6 > and < 18 > in Z24. < 6 >= {0, 6, 12, 18} and note that 18 is also a generator of < 6 >, from this or otherwise, < 18 >= {0, 18, 12, 6}.

2. Let G =< a > and let |a| = 45. (a) List all generators of G. By Theorem 4.2, the powers of a that are relatively prime to 45, a, a2, a4, a7, a8, a11, a13, a14, a16, a17, a19, a22, a23, a26, a28, a29, a31, a32, a34, a37, a38, a41, a43, and a44, are the generators. (b) List all generators for the subgroup, of G, of order 9. 45 3 5 1 5 2 5 3 5 4 The subgroup, of G, of order 9 is: < a 9 >=< a >= {e, (a ) , (a ) , (a ) , (a ) , (a5)5, (a5)6, (a5)7, (a5)8}. By Theorem 4.2, the elements of order 9 are the powers of a5 that are relatively prime to 9: (a5)1, (a5)2, (a5)4, (a5)5, (a5)7, (a5)8. Thus the generators for the subgroup are a5, a10, a20, a25, a35, and a40. (c) Compute the orders of the elements a8, a10, and a15. Since gcd(8, 45) = 1, the element a8 is a generators of G (Theorem 4.2). Thus |a8| = |a| = 15 45 k |a| |G| = 45. Also, since 15|45, |a | = 15 = 3. Apply the formula |a | = gcd(k,|a|) to compute 10 10 45 45 the orders of a . Thus |a | = gcd(10,45) = 5 = 9. 3. Suppose that a has infinite order. Find all generators of the subgroup < a11 >. If G = < b > is infinite, then the only generators of G are b and b−1. Now |a| = ∞ =⇒ |a11| = ∞ (since (a11)m = e ⇐⇒ a11m = e ⇐⇒ 11m = 0 ⇐⇒ m = 0 ). Hence < a11 > is infinite. Thus a11 and (a11)−1 = a−11 are the only generators of the subgroup < a11 >. 4. Suppose that a cyclic group has exactly three subgroups: G itself, {e}, and a subgroup of order 5. What is |G|? Since |{e}| = 1, and the order of a sugroup of G divides the order of G, we need to find a positive integer |G| such that 1||G|, 5||G|, and |G|||G|, and no other positive integer divides |G|. 5||G| =⇒ |G| = 5k, for some k ∈ Z+. We must have k=5 because otherwise |G| will have positive divisor d other than 1, 5 and |G| and G does not have a subgroup of order d. Hence |G| = 25. 5. Let G be a group and a be an element of G of order n. For each integer k between 1 and n, show that |ak| = |an−k|. First, note that an−k = ana−k = a−k (since |a| = n) = (ak)−1. So an−k = (ak)−1 and we know (by Problem 4/page 65) that: In any group, an element and its inverse have the same order. Thus |ak| = |(ak)−1| = |an−k|.

6. Suppose that G is a cyclic group and that 10 divides |G|. How many elements of order 10 does G have? If 7 divides |G|, how many elements of order 7 does G have? If a is one element of order 10, list the other elements of order 10. 7 divide |G| =⇒ G has φ(7) = 6 elements of order 7 and 10 divide |G| =⇒ G has φ(10) = 4 elements of order 10 (since If d is a positive divisor of |G|, then G has φ(d) elements of order d). Since 10||G|, the elements of order 10 are the generators of the subgroup, of G order 10. This subgroup of order 10 can be written as < a > (given). The elements of order 10 of < a > are ak, 1 ≤ k < 10 and gcd(10, k) = 1. Thus a, a3, a7 and a9 are all elements of order 10 in G. 7. Suppose that a cyclic group has exactly four subgroups: G itself, {e}, a subgroup of order 5, and a subgroup of order 25. What is |G|? Since |{e}| = 1, and the order of a sugroup of G divides the order of G, we need to ﬁnd a positive integer |G| such that 1||G|, 5||G|, 25||G|, and |G|||G|, and no other positive integer divides |G|.

1 25||G| =⇒ |G| = 25k, for some k ∈ Z+. We must have k=5 because otherwise |G| will have positive divisor d other than 1, 5, 25 and |G| and G does not have a subgroup of order d. Hence |G| = 125.

8. List the elements of the subgroup < 8 > in the group Z20, of integers modulo 20 (under addition modulo 20). < 8 >= {0, 1.8, 2.8, 3.8, 4.8} = {0, 8, 16, 4, 12}. 9. Let G =< a > and let |a| = 16. (a) List all generators of G. Since the integers 1, 3, 5, 7, 9, 11, 13 and 15 are relatively prime to 16, the set of generators of G is {a, a3, a5, a7, a9, a11, a13, a15}. (b) List all elements of the subgroup, of G, of order 4. 16 4 4 1 4 2 4 3 4 4 The subgroup, of G, of order 4 is : < a 4 >= < a > = {(a ) , (a ) , (a ) , (a ) = e} = {a4, a8, a12, e} =. (c) List all generators of the subgroup, of G, of order 8. 16 2 2 1 2 2 2 3 2 4 2 5 2 6 2 7 2 8 The subgroup, of G, of order 8 is : < a 8 >= < a > = {(a ) , (a ) , (a ) , (a ) , (a ) , (a ) , (a ) , (a ) = e}. Since 1, 3, 5, and 7 are relatively prime to 8, the set of generators for this group is {(a2)1, (a2)3, (a2)5, (a2)7} = {a2, a6, a10, a14}.

10. List the elements of the subgroup < 4 > in the group Z16, of integers modulo 16 (under addition modulo 16). Note 4.4 = 16 ≡16 0 and |4| = 4. Thus < 4 >= {0, 4, 8, 12}. 11. List the elements of the subgroup < 16 > in U(21), the group of positive integers less than 21 and relatively prime to 21, under multiplication modulo 21. 1 2 3 Note 16 = 16, 16 = 256 ≡21 4, 16 ≡21 64 ≡21 1. Thus |16| = 3 and < 16 >= {1, 4, 16}. 12. Let G =< a > and let |a| = 20. (a) List all generators of G. Since the integers 1, 3, 7, 9, 11, 13, 17 and 19 are relatively prime to 20, the set of generators of G is {a, a3, a7, a9, a11, a13, a17, a19}. (b) List all elements of the subgroup, of G, of order 4. 20 5 5 1 5 2 5 3 5 4 The subgroup, of G, of order 4 is : < a 4 >= < a > = {(a ) , (a ) , (a ) , (a ) = e} = {a5, a10, a15, e}. (c) List all generators of the subgroup, of G, of order 5. 20 4 4 1 4 2 4 3 4 4 4 5 The subgroup, of G, of order 5 is : < a 5 >= < a > = {(a ) , (a ) , (a ) , (a ) , (a ) = e} = {a4, a8, a12, a16, e}. Since 1, 2, 3 and 4 are relatively prime to 5, the set of generators for this group is {(a4)1, (a4)2, (a4)3, (a4)4} = {a4, a8, a12, a16}. 13. List all subgroups of U(16) = {1, 3, 5, 7, 9, 11, 13, 15, 16}. 3 3 Note |1| = 1, |3| = 4 (and since 3 ≡16 11, |11| = 4), |5| = 4 (and since 5 ≡16 13, |13| = 4), |7| = |9| = |15| = 2. Also, note that U(16) is not cyclic (since it does not has an element of order |U(16)| = 8). U(16) has 8 subgroups, 6 cyclic and 2 noncyclic. These 8 subgroups are listed below. Subgroups, of U(16), of order 8: U(16) Subgroups, of U(16), of order 4: (1) < 3 >=< 11 >= {1, 3, 9, 11} (2) < 5 >=< 13 >= {1, 5, 9, 13} (3) < 7, 9 >= {1, 7, 9, 15}. Subgroups, of U(16), of order 2: (1) < 7 >= {1, 7} (2) < 9 >= {1, 9}

2 (3) < 15 >= {1, 15}. Subgroups, of U(16), of order 1: < 1 >= {1}. 14. Let G =< a >, |a| = 18. (a) List all generators of G. (b) How many subgroups does G have? List generators for each of these subgroups. (c) Find elements of order 6 in G. (d) List all elements of order 9 in G. (e) Does G have elements of order 8?

15. Let G = Z18. (a) List all generators of G. (b) How many subgroups does G have? List generators for each of these subgroups. (c) Find elements of order 6 in G. (d) List all elements of order 9 in G. (e) Does G have elements of order 8? 16. Let G be a cyclic group. If 10||G|, and a is an element of order 10 then list all elements of G of order 10. 17. If |a| = 40, then ﬁnd all elements of order 8 in < a >. 18. If |a4| = 15, then ﬁnd all possibiliies for |a|.

19. List all elements of order 4 in Z8000000. 20. List all cyclic subgroups of U(30). 21. Prove that every cyclic group G =< a > is abelian.

22. Find all generators of Z.