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Proving converse of fundamental theorem of cyclic groups

 
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May7-09, 08:59 AM   #1
 

Proving converse of fundamental theorem of cyclic groups


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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May7-09, 02:56 PM   #2
 
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Post what you've done on this problem please.
 
May7-09, 03:31 PM   #3
 
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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.
 
May7-09, 03:58 PM   #4

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Proving converse of fundamental theorem of cyclic groups


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.
 
May8-09, 01:12 AM   #5
 
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And that's why we have the word 'exactly'.
 
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