(a) Let $r$ be a primitive root modulo $n$. Show that if $a$ and $n$ are relatively prime then $1 \le \text{ind}_r (a) \le \phi (n)$.
(b) Prove Proposition (6.16) (c).
(a) Definition (6.12) tells us that if $r$ is a primitive root of $n$, and $\gcd(a,n)=1$, then the smallest positive index $k$ such that
$$ r^{k} \equiv 1 \pmod n $$
is called the index of $a$ relative to $r$, or $\text{ind}_r(a)$.
Since $k$ is the smallest positive index (non-zero), this means $1 \le k$.
By Euler's Theorem $r^{\phi(n)} \equiv 1 \pmod n$, and so $k \le \phi(n)$, because it is the smallest positive index satisfying the above congruence and so can't be larger than $\phi(n)$.
Combining these two facts gives us $1 \le k \le \phi(n)$, or
$$ 1 \le \text{ind}_r(a) \le \phi(n) $$
(b) Let's remind ourselves of Proposition (6.16)(c).
Let $r$ be a primitive root of $n$ and $\text{ind}_r(a)$ be the index of $a$ relative to $r$.
(c) $\text{ind}_r (1) \equiv 0 \pmod {\phi (n)}$ and $\text{ind}_r (r) \equiv 1 \pmod {\phi (n)}$.
Let's first consider $\text{ind}_r(1)$,
$$ r^{\text{ind}_r(1)} \equiv 1 \equiv r^0 \pmod n $$
By Proposition (6.6)
$$ \text{ind}_r(1) \equiv 0 \pmod {\phi(n)} $$
Now let's consider $\text{ind}_r(r)$
$$ r^{\text{ind}_r(1)} \equiv r^1 \equiv r^0 \pmod n $$
By Proposition (6.6)
$$ \text{ind}_r(1) \equiv 1 \pmod {\phi(n)} $$