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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.
Proving converse of fundamental theorem of cyclic groups
- Thread starter curiousmuch
- Start date
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Homework Help Overview
The discussion revolves around a proof concerning finite abelian groups and their subgroups, specifically addressing the conditions under which such a group is cyclic. The original poster presents a statement that requires proving that a finite abelian group with one subgroup of order d for every divisor d of its order must be cyclic.
Discussion Character
- Conceptual clarification, Assumption checking
Approaches and Questions Raised
- Participants discuss the implications of the subgroup structure in finite abelian groups, questioning the original statement and its assumptions. There is a focus on the necessity of having exactly one subgroup of each order, as highlighted by examples that challenge the initial understanding.
Discussion Status
The discussion is ongoing, with participants actively questioning the clarity of the problem statement and exploring the implications of subgroup counts. Some guidance has been offered regarding the interpretation of the subgroup condition, but no consensus has been reached yet.
Contextual Notes
Participants note that the example of the group C_2 x C_2 illustrates a case where multiple subgroups exist for certain orders, which raises questions about the validity of the original claim regarding subgroup uniqueness.
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