Some comments quote a theorem that any finite subgroup of the multiplicative group of any field is finite cyclic and hence has a generator.
This theorem applies to $n$-th roots of unity and so they form a cyclic subgroup and have a generator. But is that generator a primitive $n$-th root of unity?
The problem of existence of $n$-th roots of unity depends on the field. For example, if the field has $4$ elements, then all $3$ non-zero elements are $3$-th roots of unity and, aside from $1$ itself, all roots of unity are $3n$-th roots of unity. So if $k=3n>3$, then there exist $k$-th roots of unity, but none of them are primitive, because if they were, then number of $k$-th roots of unity would be $k$.