Burnside's theorem

In mathematics, Burnside's theorem in group theory states that if G is a finite group of order ${\displaystyle p^{a}q^{b}}$ 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.

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

William Burnside.

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.

Proof

This proof is by contradiction. Let p^aq^b be the smallest product of two prime powers, such that there is a non-solvable group G whose order is equal to this number.

• G is a simple group with trivial center and a is not zero.

If G had a nontrivial proper normal subgroup H, then (because of the minimality of G), H and G/H would be solvable, so G as well, which would contradict our assumption. So G is simple.

If a were zero, G would be a finite q-group, hence nilpotent, and therefore solvable.

Similarly, G cannot be abelian, otherwise it would be nilpotent. As G is simple, its center must therefore be trivial.

• There is an element g of G which has q^d conjugates, for some d > 0.

By the first statement of Sylow's theorem, G has a subgroup S of order p^a. Because S is a nontrivial p-group, its center Z(S) is nontrivial. Fix a nontrivial element ${\displaystyle g\in Z(S)}$. The number of conjugates of g is equal to the index of its stabilizer subgroup G_g, which divides the index q^b of S (because S is a subgroup of G_g). Hence this number is of the form q^d. Moreover, the integer d is strictly positive, since g is nontrivial and therefore not central in G.

• There exists a nontrivial irreducible representation ρ with character χ, such that its dimension n is not divisible by q and the complex number χ(g) is not zero.

Let (χ_i)_{1 ≤ i ≤ h} be the family of irreducible characters of G over ℂ (here χ₁ denotes the trivial character). Because g is not in the same conjugacy class as 1, the orthogonality relation for the columns of the group's character table gives:

${\displaystyle 0=\sum _{i=1}^{h}\chi _{i}(1)\chi _{i}(g)=1+\sum _{i=2}^{h}\chi _{i}(1)\chi _{i}(g).}$

Now the χ_i(g) are algebraic integers, because they are sums of roots of unity. If all the nontrivial irreducible characters which don't vanish at g take a value divisible by q at 1, we deduce that

${\displaystyle -{\frac {1}{q}}=\sum _{i\geq 2,~\chi _{i}(g)\neq 0}{\frac {\chi _{i}(1)}{q}}\chi _{i}(g)}$

is an algebraic integer (since it is a sum of integer multiples of algebraic integers), which is absurd. This proves the statement.

• The complex number q^dχ(g)/n is an algebraic integer.

The set of integer-valued class functions on G, Z(ℤ[G]), is a commutative ring, finitely generated over ℤ. All of its elements are thus integral over ℤ, in particular the mapping u which takes the value 1 on the conjugacy class of g and 0 elsewhere.

The mapping ${\displaystyle A:Z(\mathbb {Z} [G])\rightarrow \operatorname {End} (\mathbb {C} ^{n})}$ which sends a class function f to

${\displaystyle \sum _{s\in G}f(s)\rho (s)}$

is a ring homomorphism. Because ρ(s)⁻¹A(u)ρ(s) = A(u) for all s, Schur's lemma implies that A(u) is a homothety λI_n. Its trace nλ is equal to

${\displaystyle \sum _{s\in G}f(s)\chi (s)=q^{d}\chi (g).}$

Because the homothety λI_n is the homomorphic image of an integral element, this proves that the complex number λ = q^dχ(g)/n is an algebraic integer.

• The complex number χ(g)/n is an algebraic integer.

Since q is relatively prime to n, by Bézout's identity there are two integers x and y such that:

${\displaystyle xq^{d}+yn=1\quad {\text{therefore}}\quad {\frac {\chi (g)}{n}}=x{\frac {q^{d}\chi (g)}{n}}+y\chi (g).}$

Because a linear combination with integer coefficients of algebraic integers is again an algebraic integer, this proves the statement.

• The image of g, under the representation ρ, is a homothety.

Let ζ be the complex number χ(g)/n. It is an algebraic integer, so its norm N(ζ) (i.e. the product of its conjugates, that is the roots of its minimal polynomial over ℚ) is a nonzero integer. Now ζ is the average of roots of unity (the eigenvalues of ρ(g)), hence so are its conjugates, so they all have an absolute value less than or equal to 1. Because the absolute value of their product N(ζ) is greater than or equal to 1, their absolute value must all be 1, in particular ζ, which means that the eigenvalues of ρ(g) are all equal, so ρ(g) is a homothety.

• Conclusion

Let N be the kernel of ρ. The homothety ρ(g) is central in Im(ρ) (which is canonically isomorphic to G/N), whereas g is not central in G. Consequently, the normal subgroup N of the simple group G is nontrivial, hence it is equal to G, which contradicts the fact that ρ is a nontrivial representation.

This contradiction proves the theorem.

References

• Bender, Helmut (1972), "A group theoretic proof of Burnside's p^aq^b-theorem.", Math. Z., 126: 327–338, doi:10.1007/bf01110337, MR 0322048
• Burnside, W. (1904), "On Groups of Order p^αq^β" (PDF), Proc. London Math. Soc. (s2-1 (1)): 388–392, doi:10.1112/plms/s2-1.1.388
• Goldschmidt, David M. (1970), "A group theoretic proof of the p^aq^b theorem for odd primes", Math. Z., 113: 373–375, doi:10.1007/bf01110506, MR 0276338
• James, Gordon; and Liebeck, Martin (2001). Representations and Characters of Groups (2nd ed.). Cambridge University Press. ISBN 0-521-00392-X. See chapter 31.
• Matsuyama, Hiroshi (1973), "Solvability of groups of order 2^aq^b.", Osaka J. Math., 10: 375–378, MR 0323890