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

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 $$4\notin\langle 6,8\rangle$$?

• What did you try? Commented Jul 4, 2019 at 16:52
• Do you have a list of the first few primes satisfying your condition? Commented Jul 4, 2019 at 18:19
• You have $\langle 6,16\rangle$ in the title and $\langle 6,8\rangle$ in the text.
– YCor
Commented Jul 4, 2019 at 18:22
• @YCor Thanks. I have corrected this typo. Commented Jul 5, 2019 at 0:17
• $\mathbb{Z}_p^{*}$ is a field. What subgroup do yo want to have, additive or multiplicative? Commented Jul 5, 2019 at 0:49

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 $$8$$.
• If you don’t know how to compute the Galois group of $F$ as a semi-direct product then I won’t be able to explain what a Frobenius element is let alone the Cebotarev density theorem. Plus, I am on an iPhone. The details are all there. Commented Jul 6, 2019 at 15:24