Sufficiently large implies splitting

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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 2swhere 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

Related facts

Facts about minimal splitting fields

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.

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