Consider the quotient group $\frac{\Bbb{Q}}{\Bbb{Z}}$
So $\frac{\Bbb{Q}}{\Bbb{Z}}$ = $\{\frac{p}{q}: p, q \in \Bbb{Z}, q \neq 0\}$
Now consider $\Bbb{Z}$+$\frac{r}{m}$ where $m$ is a natural number and $1 \leq r \leq m$ and $gcd(r,m)$ = $1$.
I want to show that order of $\Bbb{Z}$+$\frac{r}{m}$ is $m$.
Now ($\Bbb{Z}$+$\frac{r}{m}$)$^m$ = $\Bbb{Z}$
This implies that order of $\Bbb{Z}$+$\frac{r}{m}$ divides $m$.
Let $1 \leq p \leq m$
Then how can we show that ($\Bbb{Z}$+$\frac{r}{m}$)$^p$ $\neq$ $\Bbb{Z}$.
The left hand side of the above expression is $\Bbb{Z}$+$\frac{pr}{m}$
How to show that $\frac{pr}{m}$ isn't an integer ?
We have $$m \cdot (\mathbb Z + r/m) = \mathbb Z + m r/m = \mathbb Z + r = \mathbb Z$$
and $$k \cdot (\mathbb Z + r/m) = \mathbb Z + k r/m$$
this only equals $\mathbb Z$ when $kr/m$ is an integer. We saw that this happens when $k=m$. An in fact it wont happen before that because $r$ is coprime to $m$.
Details on the arithmetic:
Let $(r,m)=1$ and $1 \le k < m$, then $kr/m$ is not an integer.
In other words, $m \not | \; kr$.
Since $(r,m)=1$, $m | \; kr$ iff $m | \; k$.