Every finite group admits a sufficiently large finite prime field

Definition
For any finite group, there exists a prime field (not of characteristic zero) that is sufficiently large with respect to the finite group.

Sufficiently large field
A field $$k$$ is termed sufficiently large with respect to a finite group $$G$$ if the following are true:


 * The characteristic of $$k$$ does not divide the order of $$G$$.
 * $$k$$ contains $$d$$ distinct $$d^{th}$$ roots of unity, where $$d$$ is the exponent of $$G$$. In other words, the polynomial $$x^d - 1$$ splits completely into linear factors over $$k$$.

Since the multiplicative group of a prime field is cyclic, a prime field with $$p$$ elements is sufficiently large with respect to the finite group $$G$$ iff the exponent of $$G$$ divides $$p - 1$$. Similarly, since the multiplicative group of a finite field is cyclic, a finite field of order $$q = p^r$$ is sufficiently large with respect to the finite group $$G$$ iff the exponent of $$G$$ divides $$q - 1$$.

Related facts

 * Every finite group admits infinitely many sufficiently large finite prime fields

Facts used

 * 1) There are infinitely many primes that are one modulo any modulus: This is the easy case of Dirichlet's theorem on primes in arithmetic progressions, which states that given any positive integer $$m$$, there exist infinitely many primes $$p$$ such that $$m|p-1$$.

Proof
By the definition of sufficiently large, it suffices to find a prime $$p$$ such that $$p$$ is congruent to $$1$$ modulo the exponent of the group. The existence of such a prime is guaranteed by fact (1).