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

 

Post what you've done on this problem please.

 

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.

 

And that's why we have the word 'exactly'.