1
$\begingroup$

Prove that if $g$ and $h$ are primitive roots modulo $m$ so $\text{ind}_g (h)$ is the inverse of $\text{ind}_h (g)$ modulo $\phi(m)$

My attempt:

I need to prove that $\text{ind}_h (g)\cdot \text{ind}_g (h)\equiv 1 \pmod{\phi (m)}$

Because $g$ and $h$ are primitive roots modulo $m$

$$\text{ord}_m (g)=\text{ord}_m (h)=\phi(m)$$

$$\Longrightarrow g^{\phi(m)}\equiv h^{\phi(m)}$$

$$\Longrightarrow h=g$$

$$\Longrightarrow \text{ind}_h (h)\cdot \text{ind}_h (h)=1$$

I am not sure at all about what I did

$\endgroup$
2
$\begingroup$

Let $a=\text{ind}_g(h)$ and $b=\text{ind}_h(g)$. Then $h\equiv g^a\pmod{m}$ and $g\equiv h^b\pmod{m}$. So $h\equiv (h^b)^a\equiv h^{ab}\pmod{m}$. It follows that $ab\equiv 1\pmod{\varphi(m)}$.

Remarks: $1.$ The solution proposed in the OP does not work. It is true that $g^{\varphi(m)}\equiv h^{\varphi(m)}\pmod{m}$, they are both congruent to $1$. But from this one cannot conclude that $g=h$.

$2.$ Note that the relationship we have established is analogous to the relationship between $\log_x y$ and $\log_y x$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.