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curiousmuch
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Homework Statement
If G is a finite abelian group that has one subgroup of order d for every divisor d of the order of G. Prove that G is cyclic.
matt grime said:Can you check the question. C_2 x C_2 has subgroups of orders 1,2 and 4, but is not cyclic.
The converse of the fundamental theorem of cyclic groups states that if a group G is cyclic, then G is isomorphic to Z/nZ, where n is the order of the group.
Proving the converse allows us to show that every cyclic group has the same structure, which in turn helps us understand and classify all cyclic groups.
The proof involves showing that any cyclic group G can be generated by a single element, and that this element has a specific order. Then, using this element, we can construct an isomorphism between G and Z/nZ.
The converse can be used to show that certain groups are cyclic, and also to prove properties of cyclic groups, such as the existence of unique subgroups of a given order. It also has applications in number theory and cryptography.
The converse only applies to finite cyclic groups, and not to all finite groups. It also does not hold for infinite cyclic groups. Additionally, the converse does not provide a complete characterization of all cyclic groups, as there are some non-cyclic groups that can also satisfy its conditions.