# Thread: Finitely generated subgroup implies cyclic

1. ## 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!

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: $\frac{K}{\mathbb{Z}}=<\frac{1}{m} + \mathbb{Z} >=\{\frac{j}{m}+\mathbb{Z}: \ 0 \leq j < m \}. \ \ \Box$