Burnside's theorem

The theorem was proved by William Burnside (1904) using the representation theory of finite groups. Several special cases of it had previously been proved by Burnside, Jordan, and Frobenius. John Thompson pointed out that a proof avoiding the use of representation theory could be extracted from his work on the N-group theorem, and this was done explicitly by Goldschmidt (1970) for groups of odd order, and by Bender (1972) for groups of even order. Matsuyama (1973) simplified the proofs.

This proof is by contradiction. Let paqb be the smallest product of two prime powers, such that there is a non-solvable group G whose order is equal to this number.

If G had a nontrivial proper normal subgroup H, then (because of the minimality of G), H and G/H would be solvable, so G as well, which would contradict our assumption. So G is simple.

If a were zero, G would be a finite q-group, hence nilpotent, and therefore solvable.

Similarly, G cannot be abelian, otherwise it would be solvable. As G is simple, its center must therefore be trivial.

Now the χi(g) are algebraic integers, because they are sums of roots of unity. If all the nontrivial irreducible characters which don't vanish at g take a value divisible by q at 1, we deduce that

is an algebraic integer (since it is a sum of integer multiples of algebraic integers), which is absurd. This proves the statement.

is a ring homomorphism. Because ρ(s)−1A(u)ρ(s) = A(u) for all s, Schur's lemma implies that A(u) is a homothety λIn. Its trace is equal to

Because the homothety λIn is the homomorphic image of an integral element, this proves that the complex number λqdχ(g)/n is an algebraic integer.

Since q is relatively prime to n, by Bézout's identity there are two integers x and y such that:

Because a linear combination with integer coefficients of algebraic integers is again an algebraic integer, this proves the statement.

Let N be the kernel of ρ. The homothety ρ(g) is central in Im(ρ) (which is canonically isomorphic to G/N), whereas g is not central in G. Consequently, the normal subgroup N of the simple group G is nontrivial, hence it is equal to G, which contradicts the fact that ρ is a nontrivial representation.