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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 2.1 Order has only one prime factor 2.2 Similar conditions for solvability 2.3 More on order with few prime factors 2.4 More on groups whose order has only two prime factors 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 5 References 5.1 Textbook references 5.2 Journal references

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. Note that the case where the order

Related facts

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 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.
• 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: These are precisely the projective special linear groups for primes p where .
• Neumann's open problem on number of projective special linear groups whose order has exactly six prime factors

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

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.

References

Textbook references

• Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347, Page 886, Section 19.2 (Theorems of Burnside and Hall), (The proof of the theorem spans over pages 886-890, covering several definitions and lemmas.)^{More info}

Journal references

• On groups of order p^αq^β by William Burnside, Proceedings of the London Mathematical Society, ISSN 1460244X (online), ISSN 00246115 (print), Volume 1, Page 388 - 392(Year 1904): ^{More info}