Multiplicative group of integers modulo $n$ is well-defined

Question

In Abstract Algebra by Dummit and Foote, p.10, proposition 4.

The authors indicate that the equivalence classes in the multiplicative group of integers modulo $n$ are:

\begin{equation}
(\mathbb{Z}/n\mathbb{Z})^\times = \{\bar{a}\in\mathbb{Z}/n\mathbb{Z}|(a,n)=1\}
\end{equation}

Then they say that:

It is easy to see that if any representative of $\bar{a}$ is
relatively prime to $n$ then all representatives are relatively prime to
$n$ so that the set on the right in the proposition is well defined.

I can not see why that is the case. Could someone help me ?

Answer 1

Representatives of $\bar a$ are $a+kn$, $k\in \mathbb Z$.

If we take any element $a+k_1n\in \bar a$ such that $\gcd(a+k_1n,n)=1$,

then for any other element $a+k_2n\in \bar a$, by a known property of the greatest common divisor,

$\gcd(a+k_2n,n)=\gcd(a,n)=\gcd(a+k_1n,n)=1$ too.