if $a|bc$ and $gcd(a,b) = 1$, then $a|c$
We know that $b|bc$
and $(a,b)= 1 \rightarrow \big(ab=lcm(a,b\big)\big| bc$
so $a | c$.
Is this proof correct?
Edit: I think assuming $lcm(a,b)=ab$ is too much so here is another elementary proof:
$(a,b)=1\rightarrow \exists p,q \in Z \ni pa +qb = 1 \rightarrow pac + qbc = c $
now $a|ac$ and $a|bc$ so $a|c$