The theorem was proved by 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 for groups of odd order, and by for groups of even order. simplified the proofs.
Proof
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 subgroupH, then, 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 nilpotent. As G is simple, its center must therefore be trivial. By the first statement of Sylow's theorem, G has a subgroupS of order pa. Because S is a nontrivial p-group, its center Z is nontrivial. Fix a nontrivial element. The number of conjugates of g is equal to the index of its stabilizer subgroupGg, which divides the index qb of S. Hence this number is of the form qd. Moreover, the integerd is strictly positive, since g is nontrivial and therefore not central in G. Let 1 ≤ i ≤ h be the family of irreducible characters of G over ℂ. Because g is not in the same conjugacy class as 1, the orthogonality relation for the columns of the group's character table gives: Now the χi 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, which is absurd. This proves the statement. The set of integer-valued class functions on G, Z, is a commutative ring, finitely generated over ℤ. All of its elements are thus integral over ℤ, in particular the mapping u which takes the value 1 on the conjugacy class of g and 0 elsewhere. The mapping which sends a class function f to is a ring homomorphism. Because ρ−1Aρ = A for all s, Schur's lemma implies that A is a homothety λIn. Its tracenλ is equal to Because the homothety λIn is the homomorphic image of an integral element, this proves that the complex numberλ = qdχ/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 ζ be the complex number χ/n. It is an algebraic integer, so its norm N is a nonzero integer. Now ζ is the average of roots of unity, hence so are its conjugates, so they all have an absolute value less than or equal to 1. Because the absolute value of their product N is greater than or equal to 1, their absolute value must all be 1, in particular ζ, which means that the eigenvalues of ρ are all equal, so ρ is a homothety. Let N be the kernel of ρ. The homothety ρ is central in Im, whereas g is not central in G. Consequently, the normal subgroup N of the simple groupG is nontrivial, hence it is equal to G, which contradicts the fact that ρ is a nontrivial representation. This contradiction proves the theorem.
