Difference between revisions of "Three subgroup lemma"
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===Any one contained in normal closure of subgroup generated by other two=== | ===Any one contained in normal closure of subgroup generated by other two=== | ||
− | Let <math>A, B, C</math> be three subgroups of <math>G</math>. Then <math>[[A,B],C]</math> is contained in the [[normal closure]] of the subgroup generated by <math>[[B,C],A]</math> and <math>[[C,A],B]</math>. Equivalently, if <math>N</math> is a normal subgroup containing both <math>[[B,C],A]</math> and <math>[[C,A],B]</math>, then <math>N</math> contains <math>[[A,B],C]</math> | + | Let <math>A, B, C</math> be three subgroups of <math>G</math>. Then <math>[[A,B],C]</math> is contained in the [[normal closure]] of the subgroup generated by <math>[[B,C],A]</math> and <math>[[C,A],B]</math>. Equivalently, if <math>N</math> is a normal subgroup containing both <math>[[B,C],A]</math> and <math>[[C,A],B]</math>, then <math>N</math> contains <math>[[A,B],C]</math>. |
+ | ===Formulation where one is a group of automorphisms=== | ||
+ | |||
+ | Let <math>G</math> be a group, <math>A,B</math> be subgroups, and <math>C \le \operatorname{Aut}(G)</math>. Then, using the notation of [[commutator of element and automorphism]], any two of the three statements below implies the third: | ||
+ | |||
+ | * <math>[[A,B],C]</math> is trivial | ||
+ | * <math>[[B,C],A]</math> is [[trivial group|trivial]] | ||
+ | * <math>[[C,A],B]</math> is [[trivial group|trivial]] | ||
+ | |||
+ | Further, <math>[[A,B],C]</math> is contained in the [[normal closure]] of the subgroup generated by <math>[[B,C],A]</math> and <math>[[C,A],B]</math>. Equivalently, if <math>N</math> is a normal subgroup containing both <math>[[B,C],A]</math> and <math>[[C,A],B]</math>, then <math>N</math> contains <math>[[A,B],C]</math>. | ||
==Proof== | ==Proof== | ||
Latest revision as of 13:53, 7 July 2008
This fact is related to: commutator calculus
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Contents
Statement
Two out of three formulation
Let be three subgroups of . Then any two of the three statements below implies the third:
Any one contained in normal closure of subgroup generated by other two
Let be three subgroups of . Then is contained in the normal closure of the subgroup generated by and . Equivalently, if is a normal subgroup containing both and , then contains .
Formulation where one is a group of automorphisms
Let be a group, be subgroups, and . Then, using the notation of commutator of element and automorphism, any two of the three statements below implies the third:
Further, is contained in the normal closure of the subgroup generated by and . Equivalently, if is a normal subgroup containing both and , then contains .
Proof
The three subgroup lemma follows from Witt's identity.
PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]Corollaries
If is a perfect group and is a subgroup of such that is trivial, then is trivial.
This result has an analogue in the theory of Lie algebras.
References
Textbook references
- Nilpotent groups and their automorphisms by Evgenii I. Khukhro, ISBN 3110136724, ^{More info}, Page 31, Theorem 2.1.2 (formal statement, with proof)