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For a finite abelian group g of order n, is it true that g has a subgroup of order m for each divisor of n? please prove with examples
For a finite abelian group G of order n, is it true that G has a subgroup of order m for each divisor m of n (where m is not necessarily prime)? Please prove with examples
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Alka gupta![best-answer]()
this statement is not ture for finite abelian group it is ture for finite cyclic group. " Lagrange's theorem" states that " The order of every subgroup of a finite group G is a divisor of the order of the group ." but convers of Lagrange theorem is not holds Note- convers holds for only finite cyclic group.
But isn't every cyclic group abelian as well?
yes every cyclic group is abelian....but every abelian group is not cyclic
here abelian group is given....not cyclic
Please provide an example of an abelian group where the converse of Langrange's theorem does not hold true
see this