A problem of ord (mod $q$)

Question

Let $p\in \mathbb{Z}$ a prime, then exist $q\not\equiv 1 $ (mod $9$), $q$ prime ,such that 3 $\mid ord_{q}(p)$??? I don't know how to prove it, thank you for any help!!!.

Answer 1

A reference for my answer is this.

We look for prime $q\equiv 1 \ \mathrm{mod} \ 3$, $q\not\equiv 1 \ \mathrm{mod} \ 9$ such that $p$ is not a cubic residue modulo $q$. Then this forces $3|\mathrm{ord}_q(p)$.

Consider $K=\mathbb{Q}(\sqrt[3]{p}, \zeta_3)=\mathbb{Q}(\sqrt[3]{p}+\zeta_3)$. Then $K$ over $\mathbb{Q}$ is a Galois extension. We need to have $q$ splitting completely in $\mathbb{Q}(\zeta_3)$ and inert in $\mathbb{Q}(\sqrt[3]{p})$. Thus, the prime $q$ must be decomposed as $\mathcal{P_1}\mathcal{P_2}$ in $K$, each having residue field degree $3$.

The Galois group is $S_3=\langle \sigma, \tau| \sigma^3=\tau^2=\sigma\tau\sigma\tau=1 \rangle $, with $\sigma(\sqrt[3]{p})=\sqrt[3]{p} \zeta_3$, $\sigma(\zeta_3)=\zeta_3$, $\tau(\sqrt[3]{p})=\sqrt[3]{p}$, $\tau(\zeta_3)=\zeta_3^2$. Then the Frobenious for rational prime $q$ has cycle structure $3,3$ acting on the conjugates of $\sqrt[3]p + \zeta_3$. The Frobenious must be in the conjugacy class of $\sigma$ in $S_3$. In other words, it must be of order $3$ in $S_3$.

By Chebotarev density theorem, the natural density of those primes $q$ with $q\equiv 1 \ \mathrm{mod} \ 3$, $p$ is not cubic residue mod $q$, is $\frac{2}{6} = \frac13$.

By Dirichlet's thoerem on arithmetic progressions, the natural density of the primes $q\equiv 1 \ \mathrm{mod} \ 9$ is $\frac16$.

Therefore the proportion of primes with $q\equiv 1 \ \mathrm{mod} \ 3$, $q\not\equiv 1 \ \mathrm{mod} \ 9$, $p$ is not cubic residue mod $q$, is at least $\frac13-\frac16 = \frac16$.

Hence, we conclude that there are infinitely many such prime $q$.