Are there infinitely many primes such that $4\notin\langle 6,8\rangle $?

Question

Suppose $p\equiv 1\pmod{5}$, $\langle 6,8\rangle$ denotes the subgroup of $\mathbb{Z}_p^{*}$ generated by $6,8$. Is there any way to prove or disprove that:

There are infinitely many primes $p\equiv 1\pmod{5}$ such that $4\notin\langle 6,8\rangle$?

Answer 1

Let $F=\mathbf{Q}(\sqrt[3]{6},\sqrt[3]{2},\zeta_{15})$. This is Galois with Galois group the semi direct product of $(\mathbf{Z}/15)^*$ by $(\mathbf{Z}/3)^2$. By Cebotarev, there are infinitely many primes whose a Frobenius is the element which is trivial in the quotient and is the order three element which fixes the first cube root and not the second. The fact that the Frobenius is trivial in the quotient implies that $p \equiv 1 \mod 15$, and the other property implies that $6$ is a cube module $p$ but $2$ (and thus $4$) is not. But now $4$ can’t be in the group generated by $6$ and $8$ since the latter are cubes. Hence there exist a positive density of such primes p.

The inverse is also true since if $p \not\equiv 1 \mod 3$ then $2$ and thus $4$ is already generated by $8$.