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Proving converse of fundamental theorem of cyclic groups

by curiousmuch
Tags: converse, cyclic, fundamental, groups, proving, theorem
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curiousmuch
#1
May7-09, 08:59 AM
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1. The problem statement, all variables and given/known data
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.


2. Relevant equations



3. The attempt at a solution
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Tom Mattson
#2
May7-09, 02:56 PM
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Post what you've done on this problem please.
matt grime
#3
May7-09, 03:31 PM
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Can you check the question. C_2 x C_2 has subgroups of orders 1,2 and 4, but is not cyclic.

Dick
#4
May7-09, 03:58 PM
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Proving converse of fundamental theorem of cyclic groups

Quote Quote by matt grime View Post
Can you check the question. C_2 x C_2 has subgroups of orders 1,2 and 4, but is not cyclic.
I think the point is that it is supposed to have ONE subgroup of each order. Your example has several subgroups of order 2.
matt grime
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May8-09, 01:12 AM
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And that's why we have the word 'exactly'.


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