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Let $\gcd(a,m)=1$ for some $m\ge1$. Then we know that $a$ has a multiplicative inverse modulo $m$. Let $b$ be such an inverse. Argue that $\mathrm{ord}_m(a)=\mathrm{ord}_m(b)$.

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closed as off-topic by barto, MagicMan, Davide Giraudo, quid, Erick Wong Apr 12 at 14:24

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What have you tried? –  wj32 Oct 14 '12 at 1:30

2 Answers 2

Hint: Since for any positive integer $d$, $$1\equiv (ab)^d \equiv a^db^d \pmod{m},$$ we have $b^d\equiv 1\pmod{m}$ if and only if $a^d\equiv 1\pmod{m}$.

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A hint. Let $x$ be the order of $a$. Then you have $$ab \equiv aa^{x-1} \equiv 1 \pmod m$$ From this you can conclude that $b\equiv a^{x-1} \pmod m$. Note that $\gcd(x,\ x-1)=1$. Can you conclude anything from this?

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