# Three subgroup lemma

This fact is related to: commutator calculus
View other facts related to commutator calculusView terms related to commutator calculus |
This article describes a fact or result that is not basic but it still well-established and standard. The fact may involve terms that are themselves non-basic
View other semi-basic facts in group theory
VIEW FACTS USING THIS: directly | directly or indirectly, upto two steps | directly or indirectly, upto three steps|

## 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.