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Sufficiently large implies splitting

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Contents

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 xd − 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.

Related facts

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

Facts about Sufficiently large implies splittingRDF feed
Page classFact  +
Proved inBook:Serre (94, Corollary to Theorem 24, Section 12.3, ?)  +
Referenced inBook:Serre (94, Corollary to Theorem 24, Section 12.3, ?)  +
Stated inBook:Serre (94, Corollary to Theorem 24, Section 12.3, ?)  +
UsesBrauer's induction theorem  +
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