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.

Related facts

 * Sufficiently large implies splitting for every subquotient
 * Splitting not implies sufficiently large
 * Splitting field for a group implies splitting field for every quotient
 * Splitting field for a group not implies splitting field for every subgroup

Facts about minimal splitting fields

 * Minimal splitting field need not be unique
 * Minimal splitting field need not be cyclotomic
 * Field generated by character values is splitting field implies it is the unique minimal splitting field

Facts used

 * 1) uses::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 $$k$$.