# 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.

## Definitions used

### Sufficiently large field

Further information: 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 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).