# Thread: Proper subgroups of fraction group

1. ## Proper subgroups of fraction group

Let p be a prime and let $\displaystyle \mathbb {Z} (p^ \infty ) = \{ \hat { \frac {a}{b} } \in \frac { \mathbb {Q} }{ \mathbb {Z} } : a,b \in \mathbb {Z} \ , \ b=p^i \ , \ i \geq 0 \}$.
Prove that the only proper subgroups of $\displaystyle \mathbb {Z} (p^ \infty )$ are the finite cylic groups $\displaystyle J_k = < \hat { \frac {1}{p^k} } >$

Proof so far.

Suppose to the contrary that H is a subgroup of $\displaystyle \mathbb {Z} (p^ \infty )$ such that $\displaystyle H \neq \mathbb {Z} (p^ \infty )$ and $\displaystyle H \neq < \hat { \frac {1}{p^k} } > \ \ \ \forall k$. Now, suppose that $\displaystyle H = < \frac {a}{b}>$, then $\displaystyle \frac {a}{b} \neq \frac {1}{p^k}$. But how would I get a contradiction? Thanks.

Let p be a prime and let $\displaystyle \mathbb {Z} (p^ \infty ) = \{ \hat { \frac {a}{b} } \in \frac { \mathbb {Q} }{ \mathbb {Z} } : a,b \in \mathbb {Z} \ , \ b=p^i \ , \ i \geq 0 \}$.
Prove that the only proper subgroups of $\displaystyle \mathbb {Z} (p^ \infty )$ are the finite cylic groups $\displaystyle J_k = < \hat { \frac {1}{p^k} } >$
let $\displaystyle J$ be a proper subgroup of $\displaystyle \mathbb{Z}_{p^{\infty}}.$ then there exists an integer $\displaystyle n \geq 1$ such that $\displaystyle \frac{1}{p^n} + \mathbb{Z} \notin J.$ let $\displaystyle m=\min \{n \geq 1: \ \frac{1}{p^n} + \mathbb{Z} \}.$ then $\displaystyle \frac{a}{p^n} + \mathbb{Z} \in J,$ for all $\displaystyle a \in \mathbb{Z}$ and $\displaystyle 0 \leq n \leq m-1.$

in here i showed that $\displaystyle \frac{a}{p^n} + \mathbb{Z} \notin J,$ for all integers $\displaystyle a$ coprime with $\displaystyle p$ and for all $\displaystyle n \geq m.$ therefore: $\displaystyle J=\{\frac{a}{p^n}+\mathbb{Z}: \ \ a \in \mathbb{Z}, \ 0 \leq n \leq m-1 \} = <\frac{1}{p^{m-1}} + \mathbb{Z} >. \ \Box$