$t = n/gcd(a,n)$, prove that if $am \equiv 0 \, (\operatorname{mod} \, n)$ then $t \vert m$

Question

Let $n \in \Bbb{Z_+}$, and $a \in \Bbb{Z}_n$.

for $t = \frac{n}{\operatorname{gcd}(a,n)}$ and $m \in \Bbb{Z}_+$, prove that if $am \equiv 0 \,(\operatorname{mod} \,n)$

then: $t\,\vert\,m$ (there exists $k$ such that $m = tk$)

Here is what I tried:

if $am \equiv 0 \,(\operatorname{mod} \,n)$ then there exists a $k \in \Bbb{Z}$ such that $am = nk$.

From $t = \frac{n}{\operatorname{gcd}(a,n)}$ we get that $n = t \cdot \operatorname{gcd}(a,n)$.

Put that in what we got: $\quad am = t \cdot \operatorname{gcd}(a,n) \cdot k$.

Divide both sides by $a$ ( that is the questionable choice im not sure is correct).

We get $m = t \cdot ( \operatorname{gcd}(a,n) \cdot k \cdot (1/a))$.

now we choose a $k$ such that $( \operatorname{gcd}(a,n) \cdot k \cdot (1/a)) \in \Bbb{Z}$ and we get $t \,\vert\, m$.

is this correct?