Three subgroups lemma

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} .[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} .[2]

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

  1. ^ Isaacs, Lemma 8.27, p. 111
  2. ^ Isaacs, Corollary 8.28, p. 111

References

  • I. Martin Isaacs (1993). Algebra, a graduate course (1st ed.). Brooks/Cole Publishing Company. ISBN 0-534-19002-2.