I know they are $\{1, \gamma, \gamma^2, ..., \gamma^{n-1}\}$, where $\gamma$ is the primitive n-th root of unity. To show that this is a group (the four axioms), I did this:
Since $\gamma^i$ is a root, $(\gamma^i)^n = 1$ and similarly since $\gamma^j$ is a root, $(\gamma^j)^n = 1$, thus ($\gamma^i\gamma^j)^n = (\gamma^i)^n(\gamma^j)^n = 1*1 = 1$ The identity element is $1$, the associativity is inherited from the underlying field and for inverses, suppose $\gamma^i$ is a root, then we can divide by it (since we're in a field) and get $\frac{1}{\gamma^i}$ and that raised to the n-th power yields $\frac{1}{1}$ which is just $1$.
Does that seem legit?