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Note: $\text{ord}_ma = k$ here is the smallest $k$ such that $a^k \equiv 1 \pmod m$, not the highest power of $m$ that divides $a$.

Is it always that case that if $\text{ord}_ma=k$ and if $\text{ord}_mb=l$ then $\text{ord}_m(ab)=kl$ should be the $\text{lcm}(k,l)$ of which is true if $\gcd(k,l)=1$?

How would I prove this?

Perhaps I could consider the case where $b$ is a multiplicative inverse of $a$?

So, $ab \equiv 1\pmod{m}$

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    This seems false in general. $\mathrm{ord}_5(3) = \mathrm{ord}_5(2) = 4$ but $\mathrm{ord}_5(2\cdot 3) = 1$.2012-10-25
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    @EuYu This isn't what I said. The example you used has the order being the same, $4$, but I need the order to be different. Perhaps, I read it wrong and it should be $\gcd(a,b)=1$. Maybe that is what's giving me so much trouble.2012-10-25
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    Could you clarify your question? Especially as your idea of $b=a^{-1}$ leads to order 0.2012-10-25

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