Find all elements of $\mathbb{Z}_{19}^*$ of multiplicative order $18$.
I started by using Euler's Theorem and since gcd(18, 19) = 1 it implies that $a^{\phi (19)} \equiv 1 \pmod n$. Which means that the multiplicative order divides $\phi(19) = 18$, but I don't know where to go from here.
Best Answer
$\mathbb{Z}_{19}^{*}$ is cyclic of order $18$. According to this page, $g = 2$ is a primitive root modulo $19$ (i.e. a generator of $\mathbb{Z}_{19}^{*}$). The set of all primitive roots modulo $19$ is $\{g^k|\gcd(k,18) = 1\} = \{g,g^5,g^7,g^{11},g^{13},g^{17}\} = \{2,-6,-5,-4,3,-9\}$.
(Since every proper divisor of $18$ divides either $6$ or $9$, the fact that $g = 2$ is a primitive root can be verified by checking that $g^6, g^9 \ne 1$.)