# Corollary 3.3. Let G = (a), where lal = n < co. Then the order of every subgroup of G is a divisor of n. Furthermore, if m is a positive divisor of n, then G has exactly one subgroup of order m, namely (a"/my. Theorem 3.18. Let G = (a) be a cyclic group of order n.

Corollary 3.3. Let G = (a), where lal = n < co. Then the order of every subgroup

of G is a divisor of n. Furthermore, if m is a positive divisor of n, then G has exactly

one subgroup of order m, namely (a"/my.

Theorem 3.18. Let G = (a) be a cyclic group of order n. Let m be a positive divisor

of n. Then the number of elements of order m in G is p(m).

Example 3.33. Let G = (a) be cyclic of order 50. Then we know that there are

(10) = 4 elements of order 10 in G. They lie in the subgroup of order 10, namely

(a50/10) = (as). Indeed, the precise elements will be (a*), where (k, 10) = 1. This

means that k c {1, 3, 7, 9}, so the elements of order 10 are a, al, a35 and a45. It is

worth noting that the number of elements of order 10 in a cyclic group of order one

million is also 4 (10) = 4.

3.44. 1. Let G = (a) be a cyclic group of order 120. List all of the elements of

order 12 in G.

2. How many elements of order 12 are there in a cyclic group of order 1200?

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The post Corollary 3.3. Let G = (a), where lal = n < co. Then the order of every subgroup of G is a divisor of n. Furthermore, if m is a positive divisor of n, then G has exactly one subgroup of order m, namely (a”/my. Theorem 3.18. Let G = (a) be a cyclic group of order n. appeared first on Superb Professors.