Three subgroup lemma

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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)