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If $r$ is a primitive root of $n$, then so is $r^{-1}$ modulo $n$.

Note by definition, a number $r$ is a primitive root modulo $n$ if every number that is co-prime to $n$ is congruent to a power of $r$ modulo $n$.

Therefore $r$ being a primitive root of $n$, we see $r$ generates the group $\mathbb{Z}^{\times}_n$, that is for any $x\in\mathbb{Z}^{\times}_n\implies x=r^k\pmod n$ and we see $k\leq \phi (n)$ that is $\mathbb{Z}^{\times}_n={1,r,r^2,\dots,r^{\phi(n)−1}}$. Therefore we see that from $r^{\phi(n)}=1\implies r\cdot r^{\phi(n)-1}=1\implies r^{-1}=r^{\phi(n)-1}\implies (r^{-1})^{\phi(n)}=r^{(\phi(n)-1)\phi(n)}=1$. Therefore $o(r^{-1})|\phi(n)$.

I'm having trouble showing that $\phi(n)|o(r^{-1})$ or that it will generate the whole set.

HighSchool15
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2 Answers2

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If $g$ is a primitive root modulo $n$, then $g^k$ is a primitive root modulo $n$ if and only if $gcd(\phi(n),k)=1$. Hence $g^{-1}$ is a primitive root modulo $n$. For a proof, see for example André Nicolas answer.

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Dietrich Burde
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I think this is easier if you prove and then use a more general fact: $r$ and $r^{-1}$ have the same order in the multiplicative group. Just write down the smallest power of each that gives the identity and see that they must be the same.

Ethan Bolker
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  • Yes, this is true in any group. It is (sort of an) duplicate, see here in general – Dietrich Burde Apr 03 '17 at 19:05
  • @DietrichBurde Of course it's true in any group. Whether that extra generality helps this OP with this question depends on his or her background. Sometimes number theory motivates abstract algebra, sometimes abstract algebra helps with number theory. – Ethan Bolker Apr 03 '17 at 19:14