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Math Help - Finitely generated subgroup implies cyclic

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    Finitely generated subgroup implies cyclic

    Let G = \frac { \mathbb {Q} } { \mathbb {Z} } under +, so the elements are the equivalence classes  \hat {r} = \{ s \in \mathbb {Q} : s-r \in \mathbb {Z} \} . Write  r \equiv s \ (mod \ 1 ) if  r - s \in \mathbb {Z} . If H is a finitely generated subgroup of G = \frac { \mathbb {Q} } { \mathbb {Z} } , then H is a finite cyclic subgroup. Also determine a generator for H.

    proof so far.

    Suppose that H is a finitely generated subgroup, so I need to find an element h such that <h> = H. How should I process with this? Thanks!
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    Quote Originally Posted by tttcomrader View Post
    Let G = \frac { \mathbb {Q} } { \mathbb {Z} } under +, so the elements are the equivalence classes  \hat {r} = \{ s \in \mathbb {Q} : s-r \in \mathbb {Z} \} . Write  r \equiv s \ (mod \ 1 ) if  r - s \in \mathbb {Z} . If H is a finitely generated subgroup of G = \frac { \mathbb {Q} } { \mathbb {Z} } , then H is a finite cyclic subgroup. Also determine a generator for H.
    a finitely generated subgroup of G is in the form \frac{K}{\mathbb{Z}}, where \mathbb{Z} \subseteq K and K is a finitely generated subgroup of \mathbb{Q}. suppose K=\sum_{j=1}^n r_j\mathbb{Z}, where r_j=\frac{a_j}{b_j} \in \mathbb{Q}. let \prod_{j=1}^n b_j=c.

    then K=\frac{1}{c}\sum_{j=1}^nr_jc \mathbb{Z}. since \sum_{j=1}^nr_jc \mathbb{Z} is a subgroup of \mathbb{Z}, it's cyclic. thus K=\frac{b}{c}\mathbb{Z}. since \mathbb{Z} \subseteq K, we must have c=bm, for some integer m. therefore: K=\frac{1}{m}\mathbb{Z}. to finish

    the proof show that: \frac{K}{\mathbb{Z}}=<\frac{1}{m} + \mathbb{Z} >=\{\frac{j}{m}+\mathbb{Z}: \ 0 \leq j < m \}. \ \ \Box
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