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.