# 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 &lt; 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&quot;/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.