# Sufficiently large implies splitting

## Statement

Let $G$ be a finite group, and let $d$ be the exponent of $G$: in other words, $d$ is the least common multiple of the orders of all elements of $G$. Suppose $k$ is a sufficiently large field for $G$: $k$ is a field whose characteristic does not divide the order of $G$, and such that the polynomial $x^d - 1$ splits completely over $k$.

Then, $k$ is a splitting field for $G$: Every linear representation of $G$ that can be realized over an algebraic extension of $k$ can in fact be realized over $k$.

## Particular cases

Note that if the exponent is $2s$where $s$ is odd, the existence of $s^{th}$ roots guarantees the existence of $(2s)^{th}$ roots. Hence, the smallest $d$ to guarantee sufficiently large in such circumstances is taken as $s$.

The symmetric group of degree three is the first example where the $d$-values for sufficiently large and splitting diverge.

Group Order Smallest $d$ such that existence of $d^{th}$ roots guarantees sufficiently large Smallest $d$ such that existence of $d^{th}$ roots guarantees splitting
trivial group 1 1 1
cyclic group:Z2 2 1 1
cyclic group:Z3 3 3 3
cyclic group:Z4 4 4 4
Klein four-group 4 1 1
cyclic group:Z5 5 5 5
symmetric group:S3 6 3 1
cyclic group:Z6 6 3 3
dihedral group:D8 8 4 1
quaternion group 8 4 4
symmetric group:S4 24 12 1

## Facts used

1. Brauer's induction theorem (this is also called the characterization of linear characters lemma)

## Proof

Given: A finite group $G$, a field $k$ that is sufficiently large for $G$.

To prove: $k$ is a splitting field for $G$.

Proof: By fact (1), every character of $G$ over $K$ is a $\mathbb{Z}$-linear combination of characters induced from characters of elementary subgroups of $G$. Since elementary groups are supersolvable, every character of an elementary subgroup is induced from a linear character on some subgroup of it; hence, every character of $G$ is a $\mathbb{Z}$-linear combination of linear characters on subgroups.

Now, every linear character can be realized over $k$ because $k$ is sufficiently large, and the induced representation from a linear character can be realized over the same field, so there is a collection of representations realized over $k$ whose characters have all the irreducible characters in their $\mathbb{Z}$-span. This forces that all the irreducible representations over any extension of $k$ can be realized over the field $k$.