# Difference between revisions of "P-group"

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'''Corollary.''' If <math>G</math> is a non-[[trivial group |trivial]] <math>p</math>-group, then the [[center (algebra) |center]] of <math>G</math> is non-trivial. | '''Corollary.''' If <math>G</math> is a non-[[trivial group |trivial]] <math>p</math>-group, then the [[center (algebra) |center]] of <math>G</math> is non-trivial. | ||

− | ''Proof.'' Let <math>G</math> act on itself. Then the set of fixed points is the center <math>Z</math> of <math>G</math>; thus | + | ''Proof.'' Let <math>G</math> act on itself by conjugation. Then the set of fixed points is the center <math>Z</math> of <math>G</math>; thus |

<cmath> \lvert Z \rvert \equiv \lvert G \rvert \equiv 0 \not\equiv 1 \pmod{p}, </cmath> | <cmath> \lvert Z \rvert \equiv \lvert G \rvert \equiv 0 \not\equiv 1 \pmod{p}, </cmath> | ||

so <math>Z</math> is not trivial. <math>\blacksquare</math> | so <math>Z</math> is not trivial. <math>\blacksquare</math> |

## Latest revision as of 17:35, 22 July 2014

*The title of this article has been capitalized due to technical restrictions. The correct title should be -group.*

A **-group** is a finite group whose order is a power of a prime .

## Properties

**Lemma.** Let be a -group acting on a finite set ; let denote the set of fixed points of . Then

*Proof.* It is enough to show that divides the cardinality of each orbit of with more than one element. This follows directly from the orbit-stabilizer theorem.

**Corollary.** If is a non-trivial -group, then the center of is non-trivial.

*Proof.* Let act on itself by conjugation. Then the set of fixed points is the center of ; thus
so is not trivial.

**Theorem.** Let be a -group of order . Then there exists a series of subgroups
such that normalizes , , and is a cyclic group of order , for all indices .

*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 .