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How can I find the subgroups of abelian group $\mathbb{Z}_5$?

From Lagrange's theorem, the size of the subgroup should divide 5 in this case. So the size of the subgroup should be 1 or 5 ($\mathbb{Z}_5$ itself). Can there be any size 1 subgroups of $\mathbb{Z}_5$?

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Every group has one and only one subgroup of size $1$: the subgroup that consists only of the identity element of the group. – Arturo Magidin Apr 21 '12 at 4:20

The trivial group -- consisting of the identity only -- is the only one-element subgroup of $\mathbb{Z}_5$.

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Any subgroup of a free abelian group should also be free abelian. So, if I want to prove that Z5 is not free, will it be valid if I use the fact that the trivial group is not free? – Gunbuddy Apr 21 '12 at 4:13
@Gunbuddy: The trivial group is free: it is free on $0$ generators. So the argument you are trying to use is based on a false premise. Read the comments and answers to your previous question, which can be used, mutatis mutandis ("changing what needs to be changed") to show that $\mathbb{Z}_5$ cannot be a free abelian group. – Arturo Magidin Apr 21 '12 at 4:19

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