Cyclic subgroup

Fundamental theorem of cyclic groups

Every subgroup of a cyclic group is cyclic. Moreover, if then the order of any subgroup of is a divisor of ; and, for each positive divisor of , the group has exactly one subgroup of order , namely .¹ group

Proof every subgroup is cyclic

Suppose is a subgroup. Clearly the trivial group is cyclic, so assume that contains at least one non-identity element. Since all elements of are some power of , and cannot contain only negative powers by closure.; Let be the smallest positive integer such that . By closure . We will prove . Let some . Then there exist and such that and thus , whence . Since by closure But and is the least positive integer such that . Therefore and . Hence implying .

Proof the order of a subgroup divides that of the group

Let and . From above, where is the smallest positive integer such that . Letting as above, it follows .

Proof each divisor has a unique commensurate subgroup

Let be some positive divisor of . From the theorem on Order of powers of a group element, it follows that . This proves existence, now we must show uniqueness. Let be an arbitrary subgroup of order . From above where divides . Then , whence . Therefore and .

The first part of this theorem is clearly the only that may be applied to infinite cyclic groups.

Corollary for modular arithmetic

For each positive divisor of the unique subgroup of of order is .

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Footnotes

1. 2017, Contemporary Abstract Algebra, p. 81 (thm. 4.3) ↩