Order has only two prime factors implies solvable

Statement
The Burnside's p^aq^b theorem states that a group whose order has at most two prime factors (viz., a group whose order is of the form $$p^aq^b$$ where $$p,q$$ are primes and $$a,b$$ are nonnegative integers), must be solvable.

This includes the case of the trivial group as well as of a group of prime power order, though much stronger statements can be made in these cases.

Order has only one prime factor

 * Prime power order implies not centerless
 * Prime power order implies nilpotent

Similar conditions for solvability

 * Hall's theorem: This states that if $$\pi$$-Hall subgroups exist for all prime sets $$\pi$$, then the group is a solvable group.
 * Odd-order implies solvable: Also known as the odd-order theorem and as the Feit-Thompson theorem, this states that any group of odd order is solvable.
 * Order has only two prime factors implies prime divisor with larger prime power is core-nontrivial except in finitely many cases: This result is also called Burnside's other $$p^aq^b$$-theorem.
 * Every Sylow subgroup is cyclic implies metacyclic
 * Square-free implies solvability-forcing

More on order with few prime factors

 * Classification of finite simple groups whose order has at most five prime factors counting multiplicities: These are precisely the projective special linear groups for primes $$p$$ where $$p \in \{ 5,7,11,13\}$$.
 * Neumann's open problem on number of projective special linear groups whose order has exactly six prime factors counting multiplicities

More on groups whose order has only two prime factors

 * Order has only two prime factors implies prime divisor with larger class two subgroups is core-nontrivial

First part of proof: a nontrivial group whose order has only two prime factors is not simple unless it is cyclic of prime order
Given: A nontrivial group $$G$$ whose order is $$p^aq^b$$ for primes $$p,q$$ and nonnegative integers $$a,b$$.

To prove: $$G$$ is either cyclic of prime order or it is not simple.

Proof: Without loss of generality, assume that $$a > 0$$ (if both $$a = b = 0$$, $$G$$ is trivial and there is nothing to prove).

Second part of proof: using induction
We prove the statement by a strong form of induction on the order. Specifically, the statement we are trying to prove is:

Statement to be proved by induction on $$n$$: If $$n$$ is of the form $$p^aq^b$$ where $$p,q$$ are primes and $$a,b$$ are nonnegative integers, then any group of order $$n$$ is solvable. Note that this includes the values of $$n$$ that are powers of only a single prime and it also includes $$n = 1$$, because $$a,b$$ are allowed to be zero.

Our strong form of induction uses the truth of the statement for all smaller orders. In other words:

Inductive hypothesis: The statement to be proved by induction is true for all $$m < n$$. In particular, it is true for all proper divisors of $$n$$.

Inductive goal: If $$n = p^aq^b$$ with $$p,q$$ primes and $$a,b$$ nonnegative integers, and $$G$$ is a group of order $$n$$, then $$G$$ is solvable.

Proof: We make two cases:


 * 1) Case that $$G$$ is simple: In this case, the first part of proof yields that $$G$$ is cyclic of prime order, hence abelian, and hence solvable.
 * 2) Case that $$G$$ is not simple: In this case, there is a proper nontrivial normal subgroup $$N$$ of $$G$$. By facts (5) and (6), the orders of $$N$$ and $$G/N$$ both divide the order of $$G$$, and hence both have at most two prime factors. Hence, the induction hypothesis applies to both (since $$N$$ is proper and nontrivial), and we obtain that $$N$$ and $$G/N$$ are both solvable. Thus, by fact (7), $$G$$ is solvable.