Show that if a|c, b|c and gcd(a,b)=1, then ab|c Let $a,b,c \in \mathbb{Z}$ such that $a|c$, $b|c$, gcd$(a,b)=1$. Prove that $ab|c$.
My thoughts so far: By the Unique Factorization Theorem, we can rewrite $a$ as $p_1^{\alpha_1}p_2^{\alpha_2}...p_k^{\alpha_k}$ where $p_i$'s are primes that make up $a$. Similarly, we can write $b=q_1^{\beta_1}q_2^{\beta_2}...q_k^{\beta_k}$, but with the restriction that $p_i\neq q_j$, $\forall i,j$. Then, by definition of "$|$," we can say that $a|c$ implies $\exists k \in \mathbb{Z}$ such that $k(p_1^{\alpha_1}p_2^{\alpha_2}...p_k^{\alpha_k})=c$. (Similarly for $b$).
From here I am unsure where to go, but it feels like I am only a lemma away from my conclusion.

The idea is right. We can first add that it's not restrictive to assume $a,b,c>0$ (verify it).
Since $a\mid c$, we have $c=ak$; every prime power that divides $b$ must so divide $k$.

Alternatively, write $c=abq+r$, with $0\le r<ab$. Since $a\mid c$, we deduce that $a\mid r$ and, similarly, that $b\mid r$. Suppose $r>0$; then $r$ is a multiple of the lowest common multiple of $a$ and $b$, which is $ab$. Hence $r\ge ab$, a contradiction. Therefore $r=0$.
