If g is a primitive root modulo m and k | ( m-1), show that g^((m-1)/k) has order k? What theorem do we need to use to prove this proposition.
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2If ord$_ma=d,$ ord$_m(a^k)=\dfrac d {(d,k)}$ (Proof @Page$#95$ of http://archive.org/details/NumberTheory_862 – lab bhattacharjee May 31 '17 at 07:40
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1Is $m$ a prime? Otherwise the group order is $\phi(m)$ instead of $m-1$. There are good reasons for automatically assuming $m$ to be prime, when discussing primitive roots, but should $m$ be a power of an odd prime, the group $\Bbb{Z}_m^*$ is still cyclic, and we can choose to call any generator of that group a primitive root modulo $m$. And in that case $g^k$ has order $\phi(m)/\gcd(\phi(m),k)$. If $m=p^a$, then $\phi(m)=(p-1)p^{a-1}$. All in accordance with the general result lab bhattacharjee cited (valid in any group actually). – Jyrki Lahtonen May 31 '17 at 07:44
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@lab No need to refer offsite when there are plenty of proofs here, e.g. this answer. – Bill Dubuque May 31 '17 at 13:26