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 $\phi(m)$ > 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

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 $\log_y x$.