Summary: It is well known that if \(\mathbb{F}\) is a finite field then \(\mathbb{F}^{\ast}\), the set of nonzero elements of \(\mathbb{F}\), is a cyclic group. In this paper we will assume \(\mathbb{F} = \mathbb{F}_p\) (the finite field with \(p\) elements, \(p\) a prime) and \(\mathbb{F}_{p^2}\) is a quadratic extension of \(\mathbb{F}_p\). In this case, the groups \(\mathbb{F}^{\ast}_p\) and \(\mathbb{F}^{\ast}_{p^2}\) have orders \(p-1\) and \(p^2 -1\) respectively. We will provide necessary and sufficient conditions for an element \(u\in \mathbb{F}^{\ast}_{p^2}\) to be a generator. Specifically, we will prove \(u\) is a generator of \(\mathbb{F}^{\ast}_{p^2}\) if and only if \(N(u)\) generates \(\mathbb{F}^{\ast}_p\) and \(\frac{u^2}{N(u)}\) generates \(Ker\, N\), where \(N : \mathbb{F}^{\ast}_{p^2} \rightarrow \mathbb{F}^{\ast}_p\) denotes the norm map.