Proposition. Let $p$ be a prime number congruent to 1 mod 6. Suppose $g_1$ and $g_2$ are primitive roots mod p such that $g_1 \ne g_2$. By $H_1$ we denote the subgroup of $ \mathbb F_{p} ^ {\times} \cong C_{p-1} $ generated by $g_1^3$; that is, $H_1 =<g_1^3>$. Similary, let $H_2 =<g_2^3>$.

  • $g_2 \notin H_1$
  • If $g_2 \in g_1H_1$ then $$H_1 = H_2 , \text{ } g_1 H_1 = g_2 H_2 , \text{ }g^2_1 H_1 = g^2_2 H_2. $$

  • If $g_2 \in g^2_1H_1$ then $$H_1 = H_2 , \text{ } g_1 H_1 = g^2_2 H_2 ,\text{ }g^2_1 H_1 = g_2 H_2. $$

I know that it is a straightforward exercise from elementary group theory. So I don't need a proof I need only some reference.


If $p=6r+1$, $\mathbb{F}_p^*$ is a cyclic group with order $6k$, hence both $H_1$ and $H_2$ are subgroups of order $2k$ and for some $k$ such that $\gcd(k,p-1)=1$ we have $g_2=g_1^k$, since both $g_1$ and $g_2$ are generators. In particular $g_2\not\in H_1$, since otherwise the order of $g_2$ would be at most $2k$. If $g_2\in g_1 H_1$ we have that $k$ is a number of the form $3m+1$, if $g_2\in g_1^2 H_1$ we have that $k$ is a number of the form $3m+2$. It is easy to finish from here.

  • 2
    $\begingroup$ And by the way, congratulations on reaching 200k! I think you have made a tremendous contribution to this site. $\endgroup$ – Kenny Wong Jun 3 '17 at 14:39

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