Let $a$ and $n$ be integers with $n>0$.
Show that the additive order of $a$ modulo $n$ is $\frac{n}{\gcd(a,n)}$. Let $a$ and $n$ be integers with $n>0$.
Show that the additive order of $a$ modulo $n$ is $\frac{n}{\gcd(a,n)}$.
attempt:
Let $a$ and $n$ be integers with $n>0$.
Let $m=\frac{n}{\gcd(a,n)}$.
Then,
\begin{equation*}
ma = \left(\frac{n}{\gcd(a,n)}\right)a =
n\left(\frac{a}{\gcd(a,n)}\right)
\equiv 0 \pmod{n}.
\end{equation*}
Since $d=\gcd(a,n)$ is the greatest positive integer such that $d\mid a$ and $d \mid n$, then $m$ is the smallest positive integer such that $n \mid ma$.
Hence, $m$ is the additive order of $a$ modulo $n$.
Am I true?

The additive order of $a$ modulo $n$ is the smallest positive integer $m$ such that $ma \equiv 0 \pmod{n}$.
Now, $ma = kn$ for some integer $k$.
Then, $m = \frac{kn}{a}$.
Let $\gcd(a,n) = d$. Then,
$m = k\frac{\frac{n}{d}}{\frac{a}{d}}$.
Since $\frac{a}{d}$ and $\frac{n}{d}$ are relative prime, then $\frac{a}{d} \mid k$. Hence, the order of $a$ is a multiple of
$\frac{n}{d}$. Since the order must be minimum, then the order of $a$ is
$\frac{n}{d}$.
Is the approach above true?
