Frattini subgroup is nilpotent in finite
This article gives the statement, and possibly proof, of the fact that for any Finite group (?), the subgroup obtained by applying a given subgroup-defining function always satisfies a particular subgroup property
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Contents
Statement
Verbal statement
The Frattini subgroup of any finite group is nilpotent.
Symbolic statement
Let be a finite group and denote the intersection of all maximal subgroups of (the so-called Frattini subgroup of ).
Definitions used
Frattini subgroup
Further information: Frattini subgroup
The Frattini subgroup of a (here, finite) group is the intersection of all its maximal subgroups.
Finite nilpotent group
Further information: Finite nilpotent group A finite group is nilpotent if every Sylow subgroup of it is normal.
Generalizations
The result generalizes in two important respects. First, we can prove considerably stronger results about Frattini subgroups for finite groups, and more generally, for groups in which every proper subgroup is contained in a maximal subgroup. Second, we can generalize to proving the results about any Frattini-embedded normal subgroup of an arbitrary group.
For instance:
- Frattini-embedded normal-realizable implies inner-in-automorphism-Frattini
- Frattini-embedded normal-realizable implies ACIC
Proof
Given: A finite group , and is the Frattini subgroup
To prove: For any Sylow subgroup of , is normal in .
Proof: In fact, we shall show that is normal in .
Here's the idea. By applying Frattini's argument and the fact that , we have . Now if , it is contained in a maximal subgroup of . Since is contained in every maximal subgroup, , leading to a contradiction.
References
Textbook references
- Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347, ^{More info}, Exercise 25, Page 199 (Section 6.2)