Prove $\gcd(a,m) \mid \gcd(ab,m)$ $\forall a,b,m \in \Bbb Z$

Question

I named $\gcd(a,m) = d$ and $\gcd(ab,m) = d' $

So I know that $d\mid a$, $d\mid m $ and $d'\mid ab $ , $d' \mid m$

But I can't use the transitive property of divisibility here.

How can I prove that $d \mid d'$?

Answer 1

For each $x\mid \gcd(a,m)$, $x\mid a,m$ so $x\mid ab,m$ and $x\mid\gcd(ab,m)$.