I want to show that $x=\operatorname{ord}(a) \bmod m$ and $y=\operatorname{ord}(b)\bmod m$ then $\operatorname{ord}(ab)\equiv xy \pmod{\phi(m)}$

Question

Given:$\DeclareMathOperator{\ord}{ord}$

$\phi(m)$ is Euler's totient function

$\ord(a)$ is the least solution of $a^t\equiv 1\pmod{m}$

$\operatorname{gcd}(x,y)=1$

$x=\ord(a)$

$y=\ord(b)$

Show $\ord(ab)\equiv xy \pmod{\phi(m)}$

Answer 1

Lemma: If $a^r\equiv b^s\pmod{m}$

and $r,s<\phi$.

Then since $(a^r)^{ord(a)}\equiv 1 \pmod{m}$,

$ord(a^r)|x$

Similiary $ord(b^s)|y$.

Since $gcd(x,y)=1$, $ord(a^r)=1$ and

Then $a^r \equiv 1$

$a^{ord(ab)}\equiv b^{-ord(ab)}$

By the lemma, $a^{ord(ab)}\equiv 1\equiv b^{-ord(ab)}$

$x|ord(ab)$ and $y|ord(ab)$

Since $gcd(x,y)=1$, $xy=lcm(x,y)|ord(ab)$

But since $(ab)^{xy}\equiv 1$, $ord(ab)|xy$,

So we have $xy=ord(ab)$

This begs a question: can we have $xy>\phi$, requiring us to say that $xy\equiv ord(ab) \pmod{\phi}$ (as in the given problem) rather than saying that they are exactly equal? But $xy<=ord(ab)<=\phi$