Burnside’s theorem in group theory asserts that if are prime numbers and are non-negative integers then a group of order 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 be a finite group of order . Suppose that or . Then is soluble.

Note that if then the hypothesis is trivially satisfied.

An immediate corollary is the following.

**Corollary. **Every group of order is soluble.

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

**Lemma 1. **If is a normal subgroup of then is soluble iff both and are soluble.

*Proof. *See here.

**Lemma 2. **Finite -groups are soluble.

*Proof.* We prove by induction on that a group of order is soluble. The result is trivial for so assume that . Using the class equation one easily deduces that the centre of is non-trivial. Hence is a normal subgroup of . If then is abelian and the result is clear. Otherwise is a proper normal subgroup of , so that by the induction hypothesis both and are soluble. Thus is soluble by **Lemma 1**.

*Proof of the Theorem. *If we are done by **Lemma 2**, so assume that . Without loss of generality assume that . Let denote the number of Sylow -subgroups of . By the Sylow theorems, and , i.e., , for some . But , which forces , i.e., . Let be the unique Sylow -subgroup of . Then by the second Sylow theorem must be self-conjugate, i.e., a normal subgroup of .

By **Lemma 2**** **both (which has order ) and (which has order ) are soluble. Thus is soluble, as desired.

### Like this:

Like Loading...

*Related*

Filed under Algebra

Tagged as group, prime power