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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
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SUMMARY
The discussion centers on proving that a finite abelian group G is cyclic if it has exactly one subgroup of order d for every divisor d of the order of G. Participants clarify that the example of C_2 x C_2 is invalid because it contains multiple subgroups of order 2, contradicting the requirement for uniqueness. The emphasis is on the necessity of having precisely one subgroup for each divisor, which is crucial for the proof.
PREREQUISITES- Understanding of finite abelian groups
- Knowledge of group order and subgroups
- Familiarity with the fundamental theorem of cyclic groups
- Basic concepts of group theory
- Study the structure of finite abelian groups
- Learn about subgroup lattice and its implications
- Explore the proof of the fundamental theorem of cyclic groups
- Investigate examples of groups with unique subgroup orders
Mathematics students, particularly those studying abstract algebra, group theorists, and anyone interested in the properties of finite abelian groups.
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