Show that if $(a,b)=1$, $a\mid c$ and $b\mid c$, then $a\cdot b\mid c$ 
Show that if $(a,b)=1$, $a\mid c$ and $b\mid c$, then $a\cdot b\mid c$.

Tried
$c=a\cdot k$ and $c=b\cdot j$ with $k,j\in\mathbb{N}$ then $a\cdot b\mid c^2=c\cdot c$.
 A: By Bezout theorem $(a,b)=1\iff \exists u,z\in\mathbb Z:\ ua+bv=1(*)$ and since $a|c$ and $b|c$ so $c=ka$ and $c=k'b$ so multiplying $(*)$ by $c$ we find
$$uac+bvc=uk'ab+vkab=ab(uk'+vk)=c\Rightarrow ab|c$$
A: $\rm\ a\mid c=b(c/b)\overset{(a,b)=1}{\color{#c00}\Rightarrow}  a\mid c/b \,\Rightarrow\, ab\mid c,\ $ the $\rm\color{#c00}{red}$ inference by Euclid's Lemma. $\ \ $ QED
A: Get the prime factors of c. Then some of them form a, some of them are the factors of b, and because $(a,b)=1$ no factors of a and b occur twice. Then the remaining factors of c are what you have to multiply to $a\cdot b$ to get c, so $a\cdot b|c$.
A: $c=a.k=b.j$ 
But $(a,b)=1$, and $a$ divides $b.j$, so $a$ divides $j$. Hence $a.b$ divides $c$.
A: You can generalize this theorem further:
If $a \mid c$ and $b \mid c$ then $\operatorname{lcm}(a,b) \mid c$ where $\operatorname{lcm}(a,b)$ is the least common multiple of $a$ and $b$. This is obvious by definition of lcm but you also need to prove that $\displaystyle \operatorname{lcm}(a,b) = \frac{|a \cdot b|}{\gcd(a,b)}$.
