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Suppose that for some $m\ge1$ and $a$ and $b$ with $\gcd(a,m)=\gcd(b,m)=1$ we have $\operatorname{ord}_ma=k$ and $\operatorname{ord}_mb=l$ where $\gcd(k,l)=1$. Prove that $\operatorname{ord}_m(ab)=kl$.

What I have to go on:

If $(ab)^s \equiv 1 \pmod m$ for some $s \ge 1$, raise both sides of this congruence to the power $k$ and see what this tells you about $s$.

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    $ord_m(a)$ means the order of $[a]$ in $(\mathbb{Z}/m)^*$, I guess?2012-10-25
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    $ord_m(a)=k$ means $a^k \equiv 1 \pmod {m}$.2012-10-25
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    @student.llama No. $ord_m(a) = k$ means $k$ is the _smallest_ positive integer such that $a^k \equiv 1 \pmod{m}$. Else, $ord_m(a)$ is not well-defined.2012-10-25
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    @student.llama $ord_m(a)=k$ means that $k$ is the least positive integer satisfying the congruence $a^k \equiv 1 \pmod {m}$.2012-10-25

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