Okay, I only just found out what it is. All groups of odd order are soluble (or solvable).

And this implies that all finite groups of odd composite order are not simple? I can prove it.

Let

be a finite group of composite odd order (so it’s soluble by Feit–Thompson). We may assume that

is not Abelian, since we all know that any Abelian group of composite order is not simple. Since it is not Abelian, its derived subgroup

is not trvial. Also

cannot be all of

otherwise

for all

whereas being soluble means that

must be trivial after a finite number of derivations. Hence the commutator subgroup of

is a nontrivial and proper normal subgroup, proving that

is not simple.

There. That was not really heavy machinery, was it?