# Three subgroup lemma

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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:

• $[[A,B],C]$ is trivial
• $[[B,C],A]$ is trivial
• $[[C,A],B]$ is 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
• $[[B,C],A]$ is trivial
• $[[C,A],B]$ is 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.

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.