# 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. $\square$