If c|ab And gcd(c, a) = 1, Then c|b

Since gcd(c, a) = 1:

$$1=cx+ay$$$$b=b(cx)+b(ay)$$

Since c|bc and c|ab, then c|(bcm + abn), where m and n can be any integers, including x and y:

$$\begin{align} c|(bcm + abn) & \implies c|(bcx + bay)\\ & \implies c|(b) \end{align}$$