Do the $n$-th roots of unity of an *arbitrary* field form a cyclic group?

Do the $n$-th roots of unity of an arbitrary field form a cyclic group?

Or stated differently, can we always find a primitive $n$-th root of unity? Because if we have this element we can generate the group $<\zeta_n>$, and we are done.

In particularly I'm interested in the case where $n$ is not a prime.

• Yes, because $x^n = 1$ can have at most $n$ solutions for any $n$. Say if the roots of unity is isomorphic as a group to $\mathbb Z/n\mathbb Z \times \mathbb Z/nn'\mathbb Z$, then $x^n=1$ has $n^2$ solutions. – Hw Chu Apr 20 '18 at 14:58
• @HwChu I understand that it can have at most $n$ solutions, because we have a polynomial ring over a field. But I don't get the point that you are making, can you elaborate? – Jens Wagemaker Apr 20 '18 at 15:02
• I mean, for instance if $F$ is a field whose root of unity is isomorphic to $\mathbb Z/n\mathbb Z \times \mathbb Z / nn' \mathbb Z$, then the equation $x^n - 1=0$ has at least $n^2$ roots (count the $n$-torsions in $\mathbb Z/n\mathbb Z \times \mathbb Z / nn' \mathbb Z$), and this is absurd. – Hw Chu Apr 20 '18 at 15:10
• I didn't cover torsion groups yet. Is it a theorem that the group of $n$-th roots of unity are always isomorphic to a group $\mathbb{Z} / n \mathbb{Z} \times \mathbb{Z} / n n' \mathbb{Z}$? You say nothing about $n'$, what is this, or does some theorem state the existence? – Jens Wagemaker Apr 20 '18 at 15:19
• Every finite subgroup of a field's multiplicative group is cyclic. – Lord Shark the Unknown Apr 20 '18 at 15:42

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 $k$.
• I don't see how this answers the question. As the question is merely about showing the existence of the $n$-th primitive roots. But as someone already pointed out it would follow from the general theorem that every finite subgroup of the multiplicative group of a field is finite. – Jens Wagemaker Apr 20 '18 at 19:36