I'm working through a lecture video on abstract algebra where the lecturer points out that an element $\overline{a} \in \mathbb{Z}/n\mathbb{Z}$ is invertible if and only if $\gcd(a,n) = 1$. I'm fine with this fact, but typically every time I've proved it, I take the simplest representative of the equivalence class, i.e., an element of $\{0,1, \ldots, n-1\}$. He commented that this is true for any representative of the class. I'm not fully certain I understand why.
Trying to reformulate the claim: if $a \equiv b \pmod n$ and $\gcd(a,n) = 1$, then $\gcd(b,n) = 1$.
After checking around for some properties of the $\gcd$, I came across: $$ \gcd(a,b) = \gcd(a + mb, b) $$ for any $m \in \mathbb{Z}$. This appears useful. As $a \equiv b \pmod n$, there exists $t \in \mathbb{Z}$ such that $a - b = nt$, hence $a = b + nt$. By the above result, we have $$ \gcd(a,n) = \gcd(b + nt, n) = \gcd(b,n), $$ so $\gcd(b,n) = 1$.
Is this argument sound?
