Burnside theorem

For the counting result sometimes called "Burnside's theorem", see Burnside's lemma.

In mathematics, Burnside's theorem in group theory states that if G is a finite group of order

p^a q^b\

where p and q are prime numbers, and a and b are non-negative integers, then G is solvable. Hence each non-Abelian finite simple group has order divisible by at least three distinct primes.

History

The theorem was proved by William Burnside (1904) using the representation theory of finite groups. Several special cases of it had previously been proved by Burnside, Jordan, and Frobenius. John Thompson pointed out that a proof avoiding the use of representation theory could be extracted from his work on the N-group theorem, and this was done explicitly by Goldschmidt (1970) for groups of odd order, and by Bender (1972) for groups of even order. Matsuyama (1973) simplified the proofs.

Outline of Burnside's proof

  1. If Χ is an irreducible complex character of any finite group G then |K|Χ(k)/Χ(1) is an algebraic integer, where k is in a conjugacy class K.
  2. Use step 1 to show that if Χ(1) and |K| are coprime then Χ(k) is either 0 or has absolute value Χ(1).
  3. Use step 2 to show that if a finite group has a nontrivial conjugacy class K of size pn for some prime p then the group is not simple. In fact by the orthogonality relations there must be a nontrivial character Χ of degree coprime to |K| whose value Χ(k) is coprime to p, so by step 2 Χ(k) has absolute value Χ(1), which means that k acts as a scalar in the irreducible representation of Χ so G cannot be simple.
  4. Using the class equation, a group G of order paqb (b>0) has a non-identity conjugacy class of size prime to q. Hence G has a nontrivial conjugacy class of size p^r for some integer r so is not simple by the previous step.
  5. Induction on the order of G then shows that as no group of such order can be simple, any group of such an order must be solvable.

References

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