If $(ord_m(a), ord_m(b)) = 1$ prove that $ord_m(ab) = ord_m(a)*ord_m(b) $

Question

$\DeclareMathOperator\ord{ord}$Let $a, b$, and $m$ be positive integers such that $(a,m) = (b,m) = 1$. Assume that $(\ord_m(a), \ord_m(b)) = 1$. Prove that $\ord_m(ab) = \ord_m(a)*\ord_m(b)$.

So I got $(ab)^{\ord_m(a)*\ord_m(b)} = 1 \bmod m$

so $\ord_m(ab) \mid \ord_m(a) * \ord_m(b)$.

I am stuck on how to proceed from here though.

Answer 1

$\DeclareMathOperator\ord{ord}$Let $x=\ord_m(a)$, $y=\ord_m(b)$ and $z=\ord_m(ab)$. You already found that $z\mid xy$. It remains to prove that $xy\mid z$.

We have that $$\begin{align}1&\equiv((ab)^z)^y\\ &=a^{zy}(b^y)^z\\ &\equiv a^{zy}.\end{align}$$ Therefore, $x\mid zy$. As $\gcd(x,y)=1$, it follows that $x\mid z$.

Analogously we can show that $y\mid z$. Because $\gcd(x,y)=1$, this means that $xy\mid z$, which completes the proof. $\square$