Suppose $d,e,f \in \mathbb Z, d $ | $f, e $ | $f$ and $g = \gcd(d,e)$. Prove that $de$ | $fg$

Question

Here is the statement:

Suppose $d,e,f \in \mathbb Z, d $ | $f, e $ | $f$ and $g = \gcd(d,e)$. Prove that $de$ | $fg$

Here is my attempt at it:

$g=\gcd(d,e) \implies g | d$ and $g|e$

$g|d, d|f \implies g|f$ (by TD)

$d|f \implies f=d(k), k\in \mathbb Z$

$f(e) = d(e)(k)$

$fe = de(k) \implies de|fe$

I'm stuck after this. My guess is that I need to show $g=e$ so I can substitute and get $de|fg$, but I'm not sure.

Answer 1

Since

$$|de|=(\gcd(d,e))(\operatorname{lcm}(d,e))=g(\operatorname{lcm}(d,e)),$$

it suffices to show that $\operatorname{lcm}(d,e)$ divides $f$.

Notice that $f$ is a common multiple of $d$ and $e$, since every common multiple of $d$ and $e$ is a multiple of $\operatorname{lcm}(d,e)$, we have $\operatorname{lcm}(d,e)|f.$

Hence $$de|fd$$

Answer 2

Let $l=\textrm{lcm}(d, e)$. Since $d|f$ and $e|f$, so $l|f$ and therefore $lg|fg$. But $lg=de$.

Without lcm:

Since $d|f$ and $e|f$ so let $f=\alpha_1d=\alpha_2e$. Now $g=\gcd(d, e)$ so $g$ can be expressed as $$g=c_1d+c_2e$$ for some integers $c_1$ and $c_2$. So $$fg=fc_1d+fc_2e=(c_1\alpha_2+c_2\alpha_1)de.$$