In mathematics, more specifically group theory, the three subgroups lemma is a result concerning commutators. It is a consequence of Philip Hall and Ernst Witt's eponymous identity.

Notation

In what follows, the following notation will be employed:

  • If H and K are subgroups of a group G, the commutator of H and K, denoted by [H, K], is defined as the subgroup of G generated by commutators between elements in the two subgroups. If L is a third subgroup, the convention that [H,K,L] = [[H,K],L] will be followed.
  • If x and y are elements of a group G, the conjugate of x by y will be denoted by x y {\displaystyle x^{y}}.
  • If H is a subgroup of a group G, then the centralizer of H in G will be denoted by CG(H).

Statement

Let X, Y and Z be subgroups of a group G, and assume

[ X , Y , Z ] = 1 {\displaystyle [X,Y,Z]=1} and [ Y , Z , X ] = 1. {\displaystyle [Y,Z,X]=1.}

Then [ Z , X , Y ] = 1 {\displaystyle [Z,X,Y]=1}.

More generally, for a normal subgroup N {\displaystyle N} of G {\displaystyle G}, if [ X , Y , Z ] ⊆ N {\displaystyle [X,Y,Z]\subseteq N} and [ Y , Z , X ] ⊆ N {\displaystyle [Y,Z,X]\subseteq N}, then [ Z , X , Y ] ⊆ N {\displaystyle [Z,X,Y]\subseteq N}.

Proof and the Hall – Witt identity

Hall–Witt identity

If x , y , z ∈ G {\displaystyle x,y,z\in G}, then

[ x , y − 1 , z ] y ⋅ [ y , z − 1 , x ] z ⋅ [ z , x − 1 , y ] x = 1. {\displaystyle [x,y^{-1},z]^{y}\cdot [y,z^{-1},x]^{z}\cdot [z,x^{-1},y]^{x}=1.}

Proof of the three subgroups lemma

Let x ∈ X {\displaystyle x\in X}, y ∈ Y {\displaystyle y\in Y}, and z ∈ Z {\displaystyle z\in Z}. Then [ x , y − 1 , z ] = 1 = [ y , z − 1 , x ] {\displaystyle [x,y^{-1},z]=1=[y,z^{-1},x]}, and by the Hall–Witt identity above, it follows that [ z , x − 1 , y ] x = 1 {\displaystyle [z,x^{-1},y]^{x}=1} and so [ z , x − 1 , y ] = 1 {\displaystyle [z,x^{-1},y]=1}. Therefore, [ z , x − 1 ] ∈ C G ( Y ) {\displaystyle [z,x^{-1}]\in \mathbf {C} _{G}(Y)} for all z ∈ Z {\displaystyle z\in Z} and x ∈ X {\displaystyle x\in X}. Since these elements generate [ Z , X ] {\displaystyle [Z,X]}, we conclude that [ Z , X ] ⊆ C G ( Y ) {\displaystyle [Z,X]\subseteq \mathbf {C} _{G}(Y)} and hence [ Z , X , Y ] = 1 {\displaystyle [Z,X,Y]=1}.

See also

Notes

  • I. Martin Isaacs (1993). Algebra, a graduate course (1sted.). Brooks/Cole Publishing Company. ISBN0-534-19002-2.