Order has only two prime factors implies solvable
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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- 1 Statement
- 2 Related facts
- 3 Facts used
- 4 Proof
- 5 References
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 where are primes and are nonnegative integers), must be solvable.
Order has only one prime factor
Similar conditions for solvability
- Hall's theorem: This states that if -Hall subgroups exist for all prime sets , 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 -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 where .
- 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
The table below lists key facts used directly and explicitly in the proof. Fact numbers as used in the table may be referenced in the proof. This table need not list facts used indirectly, i.e., facts that are used to prove these facts, and it need not list facts used implicitly through assumptions embedded in the choice of terminology and language.
|Fact no.||Statement||Steps in the proof where it is used||Qualitative description of how it is used in the proof||What does it rely on?||Other applications|
|1||Sylow subgroups exist||Step (1) of first part of proof||Constructs a -Sylow subgroup, which is then shown to be contained in the centralizer of a non-identity element.||Basic group theory, group actions||click here|
|2||Prime power order implies not centerless||Step (1) of first part of proof||Find a central element in a -Sylow subgroup, which must therefore have a fairly big centralizer in the whole group.||Basic group theory, class equation of a group||click here|
|3||Simple abelian implies cyclic of prime order||Step (2) of first part of proof||Takes care of the case of nontrivial center||Basic group theory||click here|
|4||Conjugacy class of prime power size implies not simple||Step (3) of first part of proof||This fact itself is the hardest part of the proof, though its application to the proof is straightforward.||Character theory; hardest part of proof||click here|
|5||Lagrange's theorem||Step (1) of first part of proof, inductive step of second part of proof||Reducing from a group to a normal subgroup||Basic group theory||click here|
|6||Order of quotient group divides order of group||Inductive step of second part of proof||Reducing from a group to a quotient group||Basic group theory||click here|
|7||Solvability is extension-closed||Inductive step of second part of proof||Putting together solvability of normal subgroup and quotient group to get solvability of whole group||Basic definition of solvability||click here|
This proof uses a tabular format for presentation. Provide feedback on tabular proof formats in a survey (opens in new window/tab) | Learn more about tabular proof formats|View all pages on facts with proofs in tabular format
This proof uses the principle of mathematical induction in a nontrivial way (i.e., it would be hard to write the proof clearly without explicitly using induction).
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 whose order is for primes and nonnegative integers .
To prove: is either cyclic of prime order or it is not simple.
Proof: Without loss of generality, assume that (if both , is trivial and there is nothing to prove).
|Step no.||Assertion/construction||Facts used||Given data used||Previous steps used||Explanation|
|1||Either the center of is nontrivial, or has a conjugacy class of prime power order||Facts (1), (2), (5)||[SHOW MORE]|
|2||If the center of is nontrivial, then is either cyclic of prime order or not simple.||Fact (3)||[SHOW MORE]|
|3||If has a conjugacy class of prime power order, then is not simple.||Fact (4)||Fact-direct.|
|4||is either cyclic of prime order or not simple.||Steps (1),(2),(3)||Step-combination-direct.|
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 : If is of the form where are primes and are nonnegative integers, then any group of order is solvable. Note that this includes the values of that are powers of only a single prime and it also includes , because 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 . In particular, it is true for all proper divisors of .
Inductive goal: If with primes and nonnegative integers, and is a group of order , then is solvable.
Proof: We make two cases:
- Case that is simple: In this case, the first part of proof yields that is cyclic of prime order, hence abelian, and hence solvable.
- Case that is not simple: In this case, there is a proper nontrivial normal subgroup of . By facts (5) and (6), the orders of and both divide the order of , and hence both have at most two prime factors. Hence, the induction hypothesis applies to both (since is proper and nontrivial), and we obtain that and are both solvable. Thus, by fact (7), is solvable.
- 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