If $\gcd(a, b) = 1$, then $\gcd(ab, c) = \gcd(a, c) \cdot\gcd(b, c)$ 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!
 A: Without using primes. 
We show that $(ab,c) \mid (a,c)(b,c)$ and that $(a,c)(b,c)\mid (ab,c) $.
We have $ax+by=1$ multiplying by $c$ we have 
$acx+bcy=c$
Now $$(a,c)(b,c)\left[\frac{a}{(a,c)}\frac{c}{(b,c)}x+\frac{b}{(b,c)}\frac{c}{(a,c)}y\right]=c$$
where of course $\frac{a}{(a,c)}$ etc are integers. So $(a,c)(b,c)\mid c$.
It is clear that $(a,c)(b,c)\mid ab$ since $(a,c)\mid a$ and $(b,c)\mid b$. And therefore we have 
$(a,c)(b,c)\mid (ab,c) $. 
To show the other direction note that there are $p,q,r,s$ such that 
$$ap+qc=(a,c)$$
and $$br+cs=(b,c)$$
thus 
$$(a,c)(b,c)=abpr +(aps+brq+qsc)c$$ and this latter is divisible by $(ab,c)$
A: It is very easy: $ $by  basic gcd laws (associative, commutative, distributive)
$$  (a,c)(b,c) = (ab,ac,bc,cc) = (ab,(\!\underbrace{{\color{#c00}{a,b}}}_{\color{#c00}{(a,\:\!b)\:\!=1}\!\!\!\!}\!,c)\:\!c) = (ab,c)\qquad\qquad$$
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 $\,x=0=y).$
