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**1. Introduction**

Burnside’s theorem, first proved in the early 20th century by William Burnside, shows that a group of order , where and are primes and , is solvable. The original proof of Burnside’s theorem utilized representation theory in an essential way. A proof using purely group theoretical tools was given by D. Goldschmidt around 1970. The aim of this note is to provide yet another proof of Burnside’s theorem, via representation theory. First we establish a fundamental definition: A group , with identity , is solvable, if there exists a chain of subgroups

such that is normal in and is abelian for . Such a chain is called an abelian tower. Thus, for instance, all abelian subgroups are solvable. We also get the easy result: If is a -group, then is solvable and has a non-trivial center. *Proof:* Note that acts on itself by conjugation, so by the class equation:

where the sum is over distinct conjugacy classes, and is the center of . Because each index is divisible and , it follows that . Now, note that a group of order is abelian, so it is solvable. Thus, by induction, is solvable. We can lift any abelian tower in to get an abelian tower in , as desired.

We fix some notation. If is a complex representation of , then we omit the homomorphism and simply write instead. We denote the character of the representation by . If , then , is the centralizer of in and is the conjugacy class of in . Note that , by the orbit-stabilizer formula. We write for the stabilizer of in .

**2. The Proof of Burnside’s Theorem**

The proof will proceed by a series of claims. We assume that is non-abelian and , otherwise is solvable by section one. We wish to derive a contradiction.

Claim 1We can, without loss of generality, assume that is simple.

*Proof:* If is a normal subgroup of , then and are solvable, by induction, therefore, is solvable.

Assuming this, we note that because is normal, and is non-abelian. This also implies that every non-trivial representation of is faithful.

Claim 2Let be an irreducible representation of . Then, for all

is an algebraic integer.

*Proof:* Let be the conjugacy classes of . Then

is in the center of because it is invariant under conjugation. Further, the center of is generated by these elements. Indeed, suppose is in the center of . Write . Suppose that , and choose such that . Then conjugation by fixes , so . It follows that is a class function and so . We have shown that the center of is finitely generated over , and hence it is integral over . Note that is a invariant map. Thus, by Schur’s lemma there exists such that . Because is integral over , and satisfies the same polynomial over that does, it follows that is integral over . Thus, and so is integral over .

Claim 3Let be an irreducible representation of . Then .

*Proof:* We maintain the notation from the previous claim. Note that by the second orthogonality relations:

If we divide both sides by , we get a sum of algebraic integers on the left, and a rational number on the right. Thus, .

Claim 4G has no non-trivial 1-dimensional representations.

*Proof:* If is a non-trivial one 1-dimensional representation, then , so is abelian. This is a contradiction.

Claim 5For every , , there exists an irreducible representation such that and for some .

*Proof:* Let be the irreducible representations of , where is the trivial representation. By claim 2.3, for all , so for some . If is the character of the regular representation, then is if , and otherwise. Now, suppose divides for . Then

because is the only 1-dimensional representation of . This is a contradiction: if we divide both sides by , we get an algebraic integer on the left, and a rational number on the right. In particular, it follows that there is an such that for some and .

The following claim holds for any finite group.

Claim 6If there exists and a representation such that , then .

*Proof:* Because is unitary, it is diagonalizable: . Note that with equality if, and only if, . Thus, because

it follows by induction that for all . Thus, , and so is in the center of the image of , but is simple so it is isomorphic to its image and hence .

By the Sylow theorems, has a -Sylow subgroup . By proposition 1.2, is non-trivial. Thus, we can choose . By taking an appropriate power of , we can assume that . Note that , so is not divisible by , i.e. for some . Let be a representation such that and . Note that the the map

is a 1-dimensional representation of , thus, it is trivial, so , for all . Let be the eigenvalues of listed with multiplicity. Note that is a -th root of unity for all , because but for all . Let and . By the formula

each conjugate of , appear with the constant multiplicity because the product is stable under the action of , in other words, if has multiplicity , then each has multiplicity for . Therefore, because each , has conjugates, we have . In particular, Recall that for any primitive -th root of unity , , thus,

By claim 2.2, it follows that is an algebraic integer, and hence an integer. However, , is relatively prime to , so , and for some , because . Now, because

it follows that and . By claim 2.6, it follows that , which is a contradiction. Thus, is solvable.

**3. Consequences**

There are several key steps in this argument. The most important assumptions are that is simple, and for any , there is a prime power dimensional representation such that . Retracing the steps of the proof yields the following consequence:

Corollary 1Suppose is a simple group and is an irreducible representation of of dimension , for some . Let be a -Sylow subgroup of . If of order , and , then is abelian.