Question about the solution to the exercise: prove that $\mathbb{Q}/\mathbb{Z}$ contains elements of every possible finite order

Question

For the solution to the following question:

Prove that $\mathbb{Q}/\mathbb{Z} $ contains elements of every possible finite order.

Proof: For any positive integer $n$ consider the element $\frac{1}{n}+\mathbb{Z}\in \mathbb{Q}/\mathbb{Z}$. Then $n(\frac{1}{n}+\mathbb{Z})=1+n\mathbb{Z}=\mathbb{Z}$. Now, let $k$ be the order of $\frac{1}{n}+\mathbb{Z}$. Then $\frac{k}{n}\in \mathbb{Z}$, so that $n\mid k$. Thus $n$ is the order of $\frac{1}{n}+\mathbb{Z}$.

I am confused about: "let $k$ can be the order of $\frac{1}{n}+\mathbb{Z}$. Then $\frac{k}{n}\in \mathbb{Z}$, so that $n\mid k$. Thus $n$ is the order of $\frac{1}{n}+\mathbb{Z}$."

I thought if I have say $n=5$, then $\frac{1}{5}+\mathbb{Z}$ would have order $5$. If I translate the portions in quotes using numerical examples, say with $k=75$ and $n=5$, we have "let $75$ be the order of $\frac{1}{5}+\mathbb{Z}$. Then $\frac{75}{5}\in \mathbb{Z}$, so that $5\mid 75$. Thus $5$ is the order of $\frac{1}{5}+\mathbb{Z}$.".

Then $\frac{1}{5}+\mathbb{Z}$ are of both orders $5$ and $75$. Either I am missing something in the text of the solution or I am misreading something.

Thank you in advance.

Answer 1

The order of $r+\Bbb Z$ in the quotient group $\Bbb Q/\Bbb Z$ is the least positive integer $n$ such that $nr\in\Bbb Z$. But $r\in\Bbb Q$, and hence $r=\frac{a}{b}$, where $a\in\Bbb Z$, $b\in\Bbb Z_{>0}$ and $\gcd(a,b)=1$. Therefore, while $r$ runs in $\Bbb Q$, $n=b$ runs over the whole $\Bbb Z_{>0}$.