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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.2012-04-21

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

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    @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](http://math.stackexchange.com/q/134387/742), which can be used, *mutatis mutandis* ("changing what needs to be changed") to show that $\mathbb{Z}_5$ cannot be a free abelian group.2012-04-21