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Is it true that $ord_{n}(a)=ord_{n}(\bar{a})$ $\forall n$?

Here, $\bar{a}$ refers to the multiplicative inverse of $a$ modulo $n$ and $ord_{n}(a)$ refers to the multiplicative order of $a$ modulo $n$.

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  • $\begingroup$ Please define the notation that you're using. What does $\bar a$ mean? Does ord${}_n(a)$ mean the order of $a$ (in the multiplicative group) modulo $n$? $\endgroup$ – Omnomnomnom Apr 30 '15 at 1:16
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Yes, it is. Since $a \bar{a}=1$, it follows that for any positive integer $k$ we have $a^k (\bar{a})^k=1$. It follows that $a^k=1$ if and only if $(\bar{a})^k=1$. In particular, if $k$ is the smallest positive integer such that $a^k=1$, then $k$ is the smallest positive integer such that $(\bar{a})^k=1$.

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The answer is yes.

Hint: note that $a^m (\bar a)^n = a^{m-n}$.

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