Let $r$ be a primitive root of an odd prime $p$. Prove that
$$ \text{ind}_r (p-1) = \frac{p-1}{2} $$
Since $r$ is a primitive root of $p$, and noting that $\phi(p)=p-1$, we have by definition
$$ \begin{align} r^{p-1} & \equiv 1 \pmod p \\ \\ (r^{\frac{p-1}{2}} )^2 & \equiv 1 \pmod p \end{align} $$
Lemma (4.3) tells us
$$ r^{\frac{p-1}{2}} \equiv \pm 1 \pmod p $$
However $r^{\frac{p-1}{2}} \equiv 1 \pmod p$ does not hold because $\phi(p)=p-1$ is the smallest index $k$ such that $r^k \equiv 1 \pmod p$, and $\frac{p-1}{2}$ is smaler thabn $p-1$.
This leaves $ r^{\frac{p-1}{2}} \equiv -1 \equiv p-1 \pmod p $ as the only possibility, and is equivalent to
$$ \text{ind}_r (p-1) = \frac{p-1}{2} $$