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In M. Ram Murty's paper Artin's Conjecture on Primitive Roots I am not able to understand a statement.

Let $k$ be the order of $a \bmod p$ where $p$ is a prime and also we have $a^{p-1} \equiv 1 \mod p$ then $k \mid (p-1)$ if $k \neq p-1$ then $k\mid(p-1)/q$ for some prime divisor $q$ of $p-1$.

I am not able to understand how the author was able to make the second claim i.e if $k \neq p-1$ then $k\mid(p-1)/q$ for some prime divisor $q$ of $p-1$.

Bill Dubuque
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cabmetric
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1 Answers1

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This is a special case of a more general statement: if $k\mid n$ and $k\ne n$ then $k\mid \frac nq$ for some prime divisor $q$ of $n$.

Indeed, this statement holds for any prime $q$ that divides $\frac nk$, which I invite you to verify.

Greg Martin
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  • We have $k \mid n$ and $k \neq n$. let $j$ be an integer such that $kj =n$ now if $q \mid n$ then $q \mid kj$ so, $q\mid j$ or $q \mid k$ now if $q \mid k$ such that $qm = k$ then we have $mj \mid \frac{n}{q}$. How can I conclude from here that $k \mid \frac{n}{q}$ ? – cabmetric Dec 27 '20 at 03:25