We define $\mathbb{Z}[1/p]=\left\{\frac{a}{p^k}|a \in \mathbb{Z},k \in \mathbb{Z}^+\cup\{0\} \right\}$ where $p$ is a prime. In other words, $\mathbb{Z}[1/p]$ consists of all rational numbers whose denominator is a power of the prime number $p$. $\mathbb{Z}[1/p]$ is a subgroup of rational numbers (with addition operation). I am trying to prove that $\mathbb{Z}[1/p]$ has cyclic $p'$ factor groups of all orders, that is, there exists a proper normal subgroup $N$ of $\mathbb{Z}[1/p]$ such that $\dfrac{\mathbb{Z}[1/p]}{N}$ is a cyclic group with order (it can be any order) relatively prime to $p$.
Any help will be appreciated. Thanks.
Let be $n$ relatively prime with $p$. As $\Bbb Z[1/p]$ is abelian, $N = n\Bbb Z[1/p]$ is obviously a proper normal subgroup of $\Bbb Z[1/p]$. The class $[1] = 1 + N$ is obviously a generator of the quotient and has order exactly $n$ because $$m[1] = [0]\implies m\in n\Bbb Z[1/p]\implies m = na/p^k\implies n|mp^k\implies n|m.$$