Let $D$ be a gcd domain. For any $a\in D$ denote $[a]$ the set of all elements in $D$ that are associates of $a$ $\GCD(a,b)$ the set of all gcd's of elements $a$ and $b$ in $D$ and any $S\subseteq D$ $mS:=\lbrace ms\mid s\in S\rbrace$ Then

1. $\GCD(a,b)=[a]$ iff $a\mid b$
2. $m\GCD(a,b)= \GCD(ma,mb)$
3. If $\GCD(ab,c)=[1]$ then $\GCD(a,c)=[1]$
4. If $\GCD(a,b)=[1]$ and $\GCD(a,c)=[1]$ then $\GCD(a,bc)=[1]$
5. If $\GCD(a,b)=[1]$ and $a\mid bc$ then $a\mid c$

Proof. To aid in the proof of these properties, let us denote, for $a\in D$ and $S\subseteq D$ $a|S$ to mean that every element of $S$ is divisible by $a$ and $S|a$ to mean that every element in $S$ divides $a$ We take the following four steps:

1. One direction is obvious from the definition. So now suppose $a\mid b$ Then $a\mid\GCD(a,b)$ But by definition, $\GCD(a,b)\mid a$ so $[a]=\GCD(a,b)$
2. Pick $d \in \GCD(a,b)$ and $x\in \GCD(ma,mb)$ We want to show that $md$ and $x$ are associates. By assumption, $d\mid a$ and $d\mid b$ so $md\mid ma$ and $md\mid mb$ which implies that $md\mid x$ Write $x=mn$ for some $n\in D$ Then $mn\mid ma$ and $mn\mid mb$ imply that $n\mid a$ and $n\mid b$ and therefore $n\mid d$ since $d$ is a gcd of $a$ and $b$ As a result, $mn\mid md$ or $x\mid md$ showing that $x$ and $md$ are associates. As a result, the map $f: m\GCD(a,b)\to \GCD(ma,mb)$ given by $f(d)=md$ is a bijection.
3. If $d\mid a$ and $d\mid c$ then $d\mid ab$ and $d\mid c$ So $d\mid\GCD(ab,c)=[1]$ hence $d$ is a unit and the result follows.
4. Suppose $d\mid a$ and $d\mid bc$ Then $d\mid ab$ and $d\mid bc$ and hence $d\mid\GCD(ab,bc)=b\GCD(a,c)=[b]$ But $d\mid a$ also, so $d\mid\GCD(a,b)=[1]$ and $d$ is a unit.
5. $\GCD(a,b)=[1]$ implies $[c]=\GCD(ac,bc)$ Now, $a\mid ac$ and by assumption, $a\mid bc$ Therefore, $a\mid\GCD(ac,bc)=[c]$

The second property above can be generalized to arbitrary integral domain: let $D$ be an integral domain, $a,b\in D$ with $\GCD(a,b)\ne \varnothing \ne \GCD(ma,mb)$ then $d\in \GCD(a,b)$ iff $md \in \GCD(ma,mb)$