Three subgroup lemma

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:


 * $$trivial
 * $$trivial
 * $$trivial

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:


 * $$[[A,B],C]$$ is trivial
 * $$trivial
 * $$trivial

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.

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.

Textbook references

 * , Page 31, Theorem 2.1.2 (formal statement, with proof)