number theory - How to show the congruence involving the order

If $\alpha$ is the lower positive integer such as $x^{\alpha}\equiv 1\mod m$ and the order of $x$ in modulus $m$ is define by $\alpha = \textrm{ord}_{m} x$

Prove that :

$$a \cdot b \equiv 1\mod m \Longrightarrow \textrm{ord}_{m}a = \textrm{ord}_{m}b$$

I have no idea how to start it!!

Thanks in advance!!

Answer

Hint: Suppose $\textrm{ord}_m a = r$ and $\textrm{ord}_m b = s$; then $a\cdot b\equiv 1\mod m$ implies that $a^rb^r\equiv 1\mod m$, so that $b^r\equiv 1\mod m$. What does that tell you about $r$ as it relates to $s=\textrm{ord}_m b$? Now repeat the process, raising both sides to $s$.