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The order of a unit $a \pmod m $ is the least $n \geq 1$ such that $a^n \equiv 1 \pmod m$.
my question is : Is true that number and its inverse have the same order?

Martin Sleziak
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Salma Ali
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2 Answers2

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Notice that we have $\ a^n \equiv 1 \iff (a^{-1})^n \equiv 1\ $ since $\ a^n (a^{-1})^n \equiv 1$

Thus $\ \{ n\ge 1 :\ a^n \equiv 1\} = \{ n\ge 1\ :\ (a^{-1})^n\equiv 1\}\ $ so they have equal least element = order.

Bill Dubuque
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  • Thanks for your immediate help , but should we prove that there is no smaller power that sends a^-1 to 1 ? – Salma Ali Apr 20 '15 at 04:01
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    @Salma Both sets are equal so their least elements are equal, i.e. the order of $a$ equals the order of $,a^{-1}.,$ That completes the proof. – Bill Dubuque Apr 20 '15 at 04:06
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Suppose not. Then what can you say about $a^n (a^{-1})^n \pmod m$?

davidlowryduda
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