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

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. 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}