Burnside's p^a q^b theorem

Statement

Let $p$ and $q$ be prime numbers, let $a, b \geq 0$ be integers, and let $G$ be a finite group of order $p^a q^b$. Then $G$ is solvable.

Remarks

A consequence is that the order of each finite simple group that is not abelian has at least three distinct prime divisors. For example, $A_5$ has order $60 = 2^2 \cdot 3 \cdot 5$. For $b = 0$, the result says that finite $p$-groups are solvable, which follows from their nilpotency.

History

William Burnside proved the theorem in 1904, with the representation theory of finite groups. Burnside, Jordan and Frobenius had proved special cases before, in 1897, 1898 and 1902. John G. Thompson later showed that his work on the N-group theorem gives a proof without representation theory.

Explanations

L1 human

If the order of a group has only two prime factors, then the group is solvable. So a simple group that is not abelian needs at least three primes in its order. The smallest such group, $A_5$, uses the primes $2$, $3$ and $5$.