In a finite cyclic group, each subgroup has size dividing the size of the group. Conversely, given a positive divisor of the size of the group, there is a subgroup of that size

Prove that in a finite cyclic group, each subgroup has size dividing the size of the group. Conversely, given a positive divisor of the size of the group, there is a subgroup of that size.

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Expert Solution

Proof. Let G = hgi, with size m. Any subgroup has the form hgki for some k. The size ofthis subgroup is the order of g

k, which is m/(k, m) by the handout on orders of elementsin a group. Therefore the size of each subgroup divides m.

Given a positive divisor d of m, we can write down an element of order d in terms of thechosen generator of G: gm/d has order m/(m/d) = d. Therefore hgm/di has size d..

Let G = hgi, with size m. Any subgroup has the form hgki for some k. The size of this subgroup is the order of gk, which is m/(k, m) by the handout on orders of elementsin a group. Therefore the size of each subgroup divides m.

Given a positive divisor d of m, we can write down an element of order d in terms of thechosen generator of G: gm/d has order m/(m/d) = d. Therefore hgm/di has size d.

Nipun Maity answered 2 years ago

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