Every finite group admits infinitely many sufficiently large finite prime fields

Definition
For any finite group, there exist infinitely many prime fields (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 $$q - 1$$.

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 infinitely many primes $$p$$ such that $$p$$ is congruent to $$1$$ modulo the exponent of the group. The existence of such primes is guaranteed by fact (1).