TITLE: If $\gcd(a, b) = 1$, then $\gcd(ab, c) = \gcd(a, c) \cdot\gcd(b, c)$
QUESTION [6 upvotes]: How can I prove that if $\gcd(a, b) = 1$, then
$\gcd(ab, c) = \gcd(a, c) \times \gcd(b, c)$?
By eea there exists $ax+by=1$ from $\gcd(a,b)=1$ so a and be are co-primes there also exists $dk=a$ and $dj= b$ where $d=\gcd(a,b)=1$ this is all the information I have gathered from the question but I dont know how to approach and solve it. Can anyone help explain to me how to arrive at the answer? Thanks!

REPLY [4 votes]: Hint $ $ By basic gcd laws (associative, commutative, distributive)
$$  (a,c)(b,c) = (ab,ac,bc,cc) = (ab,\!\!\!\!\!\underbrace{\color{c00}{(a,b,c)}}_{\large =1\ {\rm by}\ (a,b)=1}\!\!\!\!\!c) = (ab,c)\qquad\qquad$$
Remark $ $  This proof is more general than proofs using the Bezout equation for the gcd (e.g. other answer) since there are rings with gcds that don't satisfy a Bezout equation, e.g. the above proof still works in the polynomial rings $\,\Bbb Z[x]\,$ and $\,\Bbb Q[x,y]\,$ where Bezout fails, e.g. $\,x,y\,$ are coprime but $\, ax+by= 1\Rightarrow 0 = 1$ by eval at $\,x=0=y).$