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.

A quick way of going at it is using Gauss Lemma:
If $a$ and $b$ are coprime ($\gcd(a,b)=1$) then  $a|bc\implies a|c$.
Reminder of the proof: By Bézout you have $u,v\in\mathbb{Z}$ such that
$ua+vb=1$ and you just multiply by $c$ to see it through.
Now your exercice: $b|c$ so $c=bk$; $a|c$ implies by Gauss Lemma that $a|k$, so $ab|c$.
