- The title of this article has been capitalized due to technical restrictions. The correct title should be -group.
Lemma. Let be a -group acting on a finite set ; let denote the set of fixed points of . Then
Proof. Let act on itself by conjugation. Then the set of fixed points is the center of ; thus so is not trivial.
Proof. We induct on the order of . For , the theorem is trivial. Let be the center of , and a non-identity element of . Let be the order of . Then generates a cyclic group of order ; since is contained in , it is evidently a normal subgroup of . Then is a -group of order . By inductive hypothesis, there is a sequence satisfying the theorem's requirements.
Let be the canonical homomorphism from onto , and for , let be , and let . Then for , is a normal subgroup of , and is isomorphic to ; hence it is a cyclic group of order . Also, Since is a cyclic group of order that lies in the center of , the theorem's statements are true for , as well. This completes the proof.
Corollary 1. Every -group is nilpotent.
Corollary 2. If is a -group, and is a proper subgroup of , then the normalizer of is distinct from .
This is a property of nilpotent groups in general.
Proposition. Let be a proper subgroup of a -group . Then there exists a normal subgroup of of index that contains .
Proof. Since is nilpotent there exists a normal subgroup of such that and is abelian. Let be a maximal subgroup of containing . Since is nilpotent, is normal in . Since is evidently simple, it is cyclic, and hence of order .
Corollary. Let be a -group, and a subgroup of of index . Then is a normal subgroup of .