I was playing around with some polynomials over finite fields (specifically $\mathbb{F}_p^*$ where $p$ is prime), and I was wondering if there is in general a condition for the roots of a polynomial to form a group.
As an example, the polynomial that prompted the question was $x^3 - 1$. Playing around with the roots in different fields, I noticed that the set of solutions is always either $\{1\}$ or $\{1,k,k^{-1}\}$, which are obviously subgroups of the finite field we work in. It's easy to show that this is always true. Next came the generalization to polynomials of the form $x^q-1$ for any $q\in \mathbb{N}$, and it again it's fairly straightforward to show that their roots form a group.
However after investigating some more, it looks like I got lucky with those polynomials, as other ones I've looked don't have that nice structure in their roots. The base condition that already limits our choice quite a bit is that $1$ must be root, but I haven't found any more restrictions. What are some more examples of such polynomials and where could I learn some more about them?
To have a multiplicative group structure on zeros (of a polynomial over some field $\Bbb{F}$) we necessarily need three conditions:
1) $1$ must be a zero.
2) Polynomial must be reciprocally symmetric; in other words $$p(z)=\text{constant}.z^{\text{deg}(p)}p(1/z).$$ (This condition guarantees that the existence of inverses)
3) If $a,b$ are zeros of $p,$ then $p(ab)=0.$
(This condition guarantees the closed-ness under multiplication)
When three conditions are satisfied, one can easily show that zeros of $p$ form a subgroup of the multiplicative group of some field extension of $\Bbb{F}.$
EDIT:
Here the fist condition is redundant as it can be deduced from other two conditions.
Also the third condition implies if $a$ is a zero then so does $a^2, a^3, \cdots.$ Since we have only finitely many zeros for a polynomial, this implies that there are $m, n\in\Bbb{N}$ such that $a^m=a^n.$ Thus $a$ must be a root of unity.