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Show that every finite subgroup of the quotient group $\mathbb{Q}/\mathbb{Z}$ (under addition) is cyclic.

Note: there is a related problem which I just proved: "Let $G$ be a finite abelian group, then $G$ is non-cyclic iff $G$ has a subgroup isomorphic to $C_p \times C_p$ for some prime $p$."

Since $\mathbb{Q} /\mathbb{Z}$ is abelian, so based on the related problem it suffices to show it has no elementary abelian subgroup group. I tried to start prove by contradiction: Let $\mathbb{Z} <A<\mathbb{Q}$ such that

$A$/$\mathbb{Z} \simeq C_p \oplus C_p$ for some prime p, but I can't proceed further.

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Let $G < \mathbb Q/\mathbb Z$ be finite, can't you just take a minimal element of $\{q \in (0,1]\mid q + \mathbb Z \in G\}$ and prove that $q + \mathbb Z$ generates $G$? – martini Jun 20 '12 at 9:09
I don't think your idea of proving that there is no subgroup isomorphic to $C_p \oplus C_p$ is the best method for this problem. Follow martini's hint. More concretely, let $n$ be the LCM of all denominators in the elements of $G$ and show that $G$ is a subgroup of the cyclic subgroup $\{ \mathbb{Z} + a/n \mid a \in \mathbb{Z} \}$. – Derek Holt Jun 20 '12 at 9:17
While I agree with above comments, I think your way should be possible too, but harder. (Following their hints it is very straightforward.) @user1729 the homomorphic image is never finite, except when trivial! – tomasz Jun 20 '12 at 9:50
I think I should withdraw somewhat. In the end, I believe you need to do pretty much the same observations as in the comments to show what you intended to show. – tomasz Jun 20 '12 at 10:11
@user1729 how so? $\mathbf Q^2$ is certainly divisible and has noncyclic finite subgroups. – tomasz Jun 20 '12 at 10:31
up vote 8 down vote accepted

Not in the direction you wish but you can identify $\mathbb{Q}/\mathbb{Z}$ as a subgroup of the complex numbers: $\mathbb{Q}/\mathbb{Z} < \mathbb{R}/\mathbb{Z} \cong S^1 \subset \mathbb C^\times$, and then use that every finite multiplicative subgroup of a field is cyclic.

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See also… – lhf Jun 20 '12 at 10:39

Consider the natural homomorphism $\mathbb{Q}\rightarrow\mathbb{Q}/\mathbb{Z}$ and lift a set $\overline{q}_1,\ldots ,\overline{q}_r$ of generators of the finite subgroup $\overline{G}$ under consideration to $\mathbb{Q}$. The group $\overline{G}$ then is the image of $G:=\mathbb{Z}q_1 +\ldots \mathbb{Z}q_r$, where $q_i$ is the preimage of $\overline{q}_i$. The $\mathbb{Z}$-module $G$ is free of rank $1$ - one only needs the existence of gcd's in $\mathbb{Z}$ to see this. Thus $G$ and hence $\overline{G}$ are cyclic.

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This is pretty much the the solution in comments. – tomasz Jun 20 '12 at 10:51

Following your characterisation of cyclic groups it suffices to show that $\mathbb{Q}/\mathbb{Z}$ contains only $p-1$ elements of order $p$ for any given prime $p$. Those are easy to find and it is straightforward to show that there cannot be any more elements of this order.

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