Finitely generated subgroup implies cyclic

• Nov 28th 2008, 07:23 AM
Finitely generated subgroup implies cyclic
Let $\displaystyle G = \frac { \mathbb {Q} } { \mathbb {Z} }$ under +, so the elements are the equivalence classes $\displaystyle \hat {r} = \{ s \in \mathbb {Q} : s-r \in \mathbb {Z} \}$. Write $\displaystyle r \equiv s \ (mod \ 1 )$ if $\displaystyle r - s \in \mathbb {Z}$. If H is a finitely generated subgroup of $\displaystyle G = \frac { \mathbb {Q} } { \mathbb {Z} }$, 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!
• Nov 29th 2008, 12:52 AM
NonCommAlg
Quote:

Let $\displaystyle G = \frac { \mathbb {Q} } { \mathbb {Z} }$ under +, so the elements are the equivalence classes $\displaystyle \hat {r} = \{ s \in \mathbb {Q} : s-r \in \mathbb {Z} \}$. Write $\displaystyle r \equiv s \ (mod \ 1 )$ if $\displaystyle r - s \in \mathbb {Z}$. If H is a finitely generated subgroup of $\displaystyle G = \frac { \mathbb {Q} } { \mathbb {Z} }$, then H is a finite cyclic subgroup. Also determine a generator for H.
a finitely generated subgroup of $\displaystyle G$ is in the form $\displaystyle \frac{K}{\mathbb{Z}},$ where $\displaystyle \mathbb{Z} \subseteq K$ and $\displaystyle K$ is a finitely generated subgroup of $\displaystyle \mathbb{Q}.$ suppose $\displaystyle K=\sum_{j=1}^n r_j\mathbb{Z},$ where $\displaystyle r_j=\frac{a_j}{b_j} \in \mathbb{Q}.$ let $\displaystyle \prod_{j=1}^n b_j=c.$
then $\displaystyle K=\frac{1}{c}\sum_{j=1}^nr_jc \mathbb{Z}.$ since $\displaystyle \sum_{j=1}^nr_jc \mathbb{Z}$ is a subgroup of $\displaystyle \mathbb{Z},$ it's cyclic. thus $\displaystyle K=\frac{b}{c}\mathbb{Z}.$ since $\displaystyle \mathbb{Z} \subseteq K,$ we must have $\displaystyle c=bm,$ for some integer $\displaystyle m.$ therefore: $\displaystyle K=\frac{1}{m}\mathbb{Z}.$ to finish
the proof show that: $\displaystyle \frac{K}{\mathbb{Z}}=<\frac{1}{m} + \mathbb{Z} >=\{\frac{j}{m}+\mathbb{Z}: \ 0 \leq j < m \}. \ \ \Box$