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>
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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>.
  
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===Formulation where one is a group of automorphisms===
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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:
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* <math>[[A,B],C]</math> is trivial
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* <math>[[B,C],A]</math> is [[trivial group|trivial]]
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* <math>[[C,A],B]</math> is [[trivial group|trivial]]
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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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Statement

Two out of three formulation

Let A, B, C be three subgroups of G. Then any two of the three statements below implies the third:

Any one contained in normal closure of subgroup generated by other two

Let A, B, C be three subgroups of G. Then [[A,B],C] is contained in the normal closure of the subgroup generated by [[B,C],A] and [[C,A],B]. Equivalently, if N is a normal subgroup containing both [[B,C],A] and [[C,A],B], then N contains [[A,B],C].

Formulation where one is a group of automorphisms

Let G be a group, A,B be subgroups, and C \le \operatorname{Aut}(G). Then, using the notation of commutator of element and automorphism, any two of the three statements below implies the third:

Further, [[A,B],C] is contained in the normal closure of the subgroup generated by [[B,C],A] and [[C,A],B]. Equivalently, if N is a normal subgroup containing both [[B,C],A] and [[C,A],B], then N contains [[A,B],C].

Proof

The three subgroup lemma follows from Witt's identity.

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Corollaries

If G is a perfect group and N is a subgroup of G such that [[G,N],N] is trivial, then [G,N] 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)