Multiplicative inverses modulo $m$
The multiplicative group of integers modulo $m$ is defined as: \begin{align} \Z_m^* = \{a\ |\ \gcd(a,m) = 1\} \end{align} But why? This is because Euler’s theorem says that: \begin{align} \gcd(a,m) = 1\Rightarrow a^{\phi(m)} = 1 \end{align} This in turn, implies that every element in $\Z_m^*$ has an inverse, since: \begin{align} a\cdot a^{\phi(m) - 1} &= 1 \end{align} Thus, for a prime $p$, all elements in $\Z_p^* = \{1,2,\dots, p-1\}$ have inverses. Specifically, the inverse of $a \in \Z_p^*$ is $a^{p-2}$.
Finding primitive roots mod $p$
Suppose you have $\Zp^* = \{1,2,3,\ldots,p-2,p-1\}$: i.e., the group of integers mod $p$, where $p$ is a prime.
How do you find a generator $g$ for it? (a.k.a., primitive roots) \begin{align} \langle g\rangle \bydef \{g^0, g^1, g^2,\ldots,g^{p-2}\}= \Zp^* \end{align}
First, you factor its order $p-1$ as \begin{align} p-1=\prod_i q_i^{e_i},\ \text{where}\ q_i\ \text{are primes} \end{align}
Second, you simply brute-force: you pick a potential candidate generator $g$ and ensure: \begin{align} \forall i, g^\frac{p-1}{q_i} \ne 1 \pmod p \end{align} If all these checks pass, then $g$ is a generator.