## Burnside’s Theorem

**Point of post: **In this post we put together a lot of our rep theory to prove one of the fundamental (pure) group theoretic results amenable to the subject.

*Motivation*

In this post we finally use representation theory to prove something in pure group theory that is near impossible to do without representation theory. We have seen on our thread about solvable groups that every -group is solvable. In this thread we prove *Burnside’s Theorem *an amazing generalization which says that every group of order where and are primes. As a corollary we will be able to conclude that every non-abelian simple group is divisible by three distinct primes which, of course, will eliminate a respectable amount of group orders for analyzing simplicity. This is really one of the most beautiful applications of representation theory.

*Propositions*

Before we prove the actual theorem we shall need to prove a series of propositions which make the proof of Burnside’s theorem possible.

**Proposition:**

Before proving the full theorem we shall need to prove some lemmas which are interesting in and of themselves.

**Proposition #1: ***Let be a finite group and . Suppose that there existed a conjugacy class of such that then for every either or where is the center of the character .*

**Proof: **Since we have by Bezout’s identity there exists such that and thus multiplying by for gives dividing both sides by gives

we know them from a previous theorem that is an algebraic integer and since is evidently an algebraic integer, and the algebraic integers are a ring that .

Suppose then that , then by definition we have that . Let and let be the splitting field of , and note then that necessarily . Now, note that is a sum of roots of unity, and since any necessarily takes roots of unity to other such roots of unity we may conclude that is a sum of such roots of unity for each . From this it follows that and so for each . Now from this we may conclude that

(since by assumption) but note that since elements of necessarily take into itself we must have that each is in and so their product, as above, is in . That said, since any just permutes the relevant roots of unity we clearly have that is in the fix field . That said, since is the splitting field of the separable polynomial we know that is Galois and so and thus is rational. But, since the only rational algebraic integers are integers we know that this product is an integer, and since it’s modulus is less than one it must be zero. Thus, this tells us that for some and so , so as desired.

**Proposition #2:*** Let be a non-abelian simple group. Then, is the only conjugacy class of which is a prime power.*

**Proof: **Suppose that is such that where is the conjugacy class containing and is prime. Note that for any distinct from the trivial character we have that . The first of these is true since by simplicity and so or and it must be otherwise a quick argument (considering that is really the kernel of any character inducing ) shows that is the trivial character. The second is true since once again by simplicity or , and if then using the fact that (since ) and the fact that so we may conclude that contradicting that is not abelian. Lastly, the fact that is clear since otherwise is abelian, appealing to simplicity again of course.

So, if then by proposition #1 (since and so ) we have that . So, using the second orthogonality relation we see that

So that

and since each we may conclude from the above that . But, since this is a non-integer rational this is a contradiction.

*Burnside’s Theorem*

And finally the result:

**Theorem (Burnside):*** Let be a finite group with where and are primes. Then, is solvable.*

**Proof: **We note that we can assume without loss of generality that otherwise is a -group and the conclusion follows from a previous theorem.

We first show that if is a simple group of order where and are primes then it is abelian. Indeed, by Sylow’s first theorem we know that must have a subgroup of order . If has only one of these such subgroups, then by Sylow’s second theorem we have that this subgroup is normal, and so has a non-trivial (via our opening remark) proper normal subgroup contradicting simplicity, so assume not. So, let be one of these Sylow -subgroups. By first principles we know that is non-trivial. So, we may choose a non-trivial . From the orbit stabilizer theorem we know that where is the conjugacy class containing and ‘s centralizer. It follows from Lagrange’s theorem that and thus is a prime power. But, by proposition # 2 this implies that can’t be non-abelian and simple and since it is simple it follows that is abelian.

We now use this to prove the full result. We induct on . If then is a cyclic group and thus trivially solvable. So, assume that every group of order at most where is solvable and let be a group of order where . If is simple we’re done since by the first part of the proof is abelian and so trivially solvable. So, assume that is not simple. Then, has . But, this implies that and are groups of order where and thus by the induction hypothesis they themselves are solvable. But, by a prior theorem this implies that itself is solvable. Thus, the induction is complete and the conclusion follows.

**References:**

1. Isaacs, I. Martin. *Character Theory of Finite Groups*. New York: Academic, 1976. Print.

[…] very difficult (if not impossible) to prove without using representation theory– of course Burnside’s Theorem is the quintessential example. In this post we prove another such theorem, namely that for a finite […]

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