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This article gives the statement and possibly, proof, of an implication relation between two group properties. That is, it states that every group satisfying the first group property (i.e., group whose order has at most two prime factors) must also satisfy the second group property (i.e., solvable group)
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Contents 1 Statement 2 Related facts 3 Facts used 4 Proof 4.1 First step: a group whose order has only two prime factors is not simple unless it is cyclic of prime order 4.2 Second step: using induction

Statement

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

Related facts

• Hall's theorem: This states that if π-Hall subgroups exist for all prime sets π, 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.
• Order has only two prime factors implies prime divisor with larger class two subgroups is core-nontrivial

Facts used

1. Sylow subgroups exist
2. Prime power order implies not centerless
3. Conjugacy class of prime power order implies not simple: This is the meat of the proof. The shortest known proof of this is using linear representation theory. No easy purely group-theoretic proof is known.
4. Lagrange's theorem
5. Order of quotient group divides order of group

Proof

First step: a group whose order has only two prime factors is not simple unless it is cyclic of prime order

Given: A 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 simple and there is nothing to prove).

1. Either the center of G is nontrivial, or G has a conjugacy class of prime power order: By Sylow's theorem (fact (1)), G has a p-Sylow subgroup P. Since P is a nontrivial p-group, fact (2) yields that the center Z(P) is a nontrivial subgroup. Let g be a non-identity element of Z(P). Then C_G(g) contains P, so the index [G:C_G(g)], which is also equal to the size of the conjugacy class of g in G, is a divisor of q^b. In particular, either or the size of the conjugacy class of g is a nontrivial power of q. Thus, either the center is nontrivial or there is a conjugacy class of prime power order.
2. G is either cyclic of prime order or it is not simple:
   • If the center is nontrivial, the only way the group can be simple is if it is simple Abelian -- hence cyclic of prime order. Otherwise, it is not simple.
   • If there is a conjugacy class of prime power order, fact (3) yields that the group is not simple.

Second step: using induction

Given: A group G whose order has at most two prime factors.

To prove: G is solvable.

Proof: We prove this by induction on the order of G. In other words, we assume that the result is true for smaller orders (the base case of the trivial group is easily handled). We now prove it for G:

1. Case that G is simple: In this case, the previous result yields that G is cyclic of prime order, hence solvable.
2. Case that G is not simple: In this case, there is a proper nontrivial normal subgroup N of G. By facts (4) and (5), 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, G is solvable.