A weaker version of Burnside’s theorem

Burnside’s theorem in group theory asserts that if p, q are prime numbers and a, b are non-negative integers then a group of order p^aq^b is soluble.

As outlined in the Wikipedia article, the proof of this theorem is not simple and most proofs make use of representation theory. Under a weaker hypothesis, however, we can give the following simpler proof using Sylow’s theorems.

Theorem. Let G be a finite group of order p^aq^b. Suppose that a<\hbox{ord}_qp or b<\hbox{ord}_pq. Then G is soluble.

Note that if p=q then the hypothesis is trivially satisfied.

An immediate corollary is the following.

Corollary. Every group of order pq is soluble.

To prove the Theorem we shall make use of the following two lemmas.

Lemma 1. If H is a normal subgroup of G then G is soluble iff both H and G/H are soluble.

Proof. See here\square

Lemma 2. Finite p-groups are soluble.

Proof. We prove by induction on n\ge 0 that a group G of order p^n is soluble. The result is trivial for n=0 so assume that n\ge 1. Using the class equation one easily deduces that the centre Z of G is non-trivial. Hence Z is a normal subgroup of G. If Z=G then G is abelian and the result is clear. Otherwise Z is a proper normal subgroup of G, so that by the induction hypothesis both Z and G/Z are soluble. Thus G is soluble by Lemma 1. \square

Proof of the Theorem. If p=q we are done by Lemma 2, so assume that p\neq q. Without loss of generality assume that b<\hbox{ord}_pq. Let n_p denote the number of Sylow p-subgroups of G. By the Sylow theorems, n_p\mid q^b and n_p\equiv 1\pmod p, i.e., n_p=q^j\equiv 1\pmod p, for some 0\le j\le b. But b<\hbox{ord}_pq, which forces j=0, i.e., n_p=1. Let H be the unique Sylow p-subgroup of G. Then by the second Sylow theorem H must be self-conjugate, i.e., a normal subgroup of G.

By Lemma 2 both H (which has order p^a) and G/H (which has order q^b) are soluble. Thus G is soluble, as desired. \square


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