I need to prove that
$$(ah,bk)=(a,b)(h,k)\left( \frac{a}{(a,b)},\frac{k}{(h,k)}\right)\left( \frac{b}{(a,b)},\frac{h}{(h,k)}\right)$$
I'm most certain I need to use $$\left(\frac{a}{(a,b)},\frac{b}{(a,b)}\right)=1 $$ and $(a ,b)=1 $ and $a \mid bc $ then $a \mid c$ but I'm not sure how to build up the proof.
Cancel $\rm\:(a,b)(h,k),\:$ write $\rm\: A = \dfrac{a}{(a,b)},\ B = \dfrac{b}{(a,b)},\ H = \dfrac{h}{(h,k)},\ K = \dfrac{k}{(h,k)}\:$ it becomes
$$\rm\ (AH,BK) = (A,K)(B,H)$$
$$\rm\iff \left(\frac{A}{(A,K)}\frac{H}{(B,H)},\, \frac{B}{(B,H)}\frac{K}{(A,K)}\right) = 1\phantom{\iff}$$
True since each term on the left is coprime to each term on the right, e.g. the first terms are coprime by $\rm\:(A,B) = (a/(a,b),\,b/(a,b)) = (a,b)/(a,b) = 1\:$ and the first and last terms are coprime by a similar argument (replace $\rm\:a,b\:$ by $\rm\:A,K).$
