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 $k$.

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 $K$ can be realized over the field $k$.

References

Textbook references

• Linear representations of finite groups by Jean-Pierre Serre, 10-digit ISBN 0287901906 (English), ISBN 3540901906 (French), Page 94, Corollary to Theorem 24, Section 12.3, ^{More info}