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单词 PingpongLemma
释义

ping-pong lemma


Theorem (Ping Pong Lemma).

Let k2 and let G be a group acting on a space X. Supposewe are given a classM={A1,A2,,Ak,B1,B2,,Bk} of 2kpairwise disjoint subsets of X and suppose y1,y2,,ykare elements of G such that

Bicyi(Ai)i=1,2,,k

(Bic is the complement of Bi in X). Then, the subgroupMathworldPlanetmathPlanetmathof G generated by y1,y2,,yk is free.

Before turning to prove the lemma let’s state threesimple facts:

Fact 1.

For all i=1,,k we have yi(Aic)Bi andyi-1(Bic)Ai

Proof.

Bicyi(Ai)Biyi(Ai)c=yi(Aic)

Fact 2.

If ij then AjBjAicBic.

Proof.

Ai and Aj are disjoint therefore AjAic.Similarly, AjBic so AjAicBic.In the same way, BjAicBic so AjBjAicBic.∎

Fact 3.

If R,SM then RcS

Proof.

Assume by contradictionMathworldPlanetmathPlanetmath that RcS. Then, X=RS and therefore any element of M intersects with either R orS. However, the elements of M are pairwise disjoint and thereare at least 4 elements in M so this is a contradiction.∎


Using the above 3 facts, we now turn to the proof of the Ping Pong Lemma:

Proof.

Suppose we are given w=znϵnz2ϵ2z1ϵ1 such thatz{y1,y2,,yk} and ϵ{-1,+1}.and suppose further that w is freely reduced, namely, ifzi=zi+1 then ϵi=ϵi+1. We want to showthat w1 in G. Assume by contradiction that w=1. We get acontradiction by giving R,S such thatw(Sc)R and therefore contradicting Fact 3above since Sc=w(Sc)R.

The set S is chosen as follows. Assume that z1=yi then:

S={Aiif ϵ1=1Biif ϵ1=-1

Define the following subsets P0,P1,,Pn of X:

P0=Sc;P1=z1ϵ1(P0),,Pn=znϵn(Pn-1)=w(Sc)

To completePlanetmathPlanetmathPlanetmathPlanetmathPlanetmath the proof we show by inductionMathworldPlanetmath that for=1,2,,n if z=yi then:

  1. 1.

    if ϵ=1 then PBi.

  2. 2.

    if ϵ=-1 then PAi.

For =1 the above follows from Fact 1 and thespecific choice of P0. Assume it is true for -1 andassume that z=yi. We have two cases to check:

  1. 1.

    z-1z: by the induction hypothesisP-1 is a subset of AjBj for some ji.Therefore, by Fact 2 we get that P-1 is a subset of AicBic.Consequently, we get the following:

    P=zϵ(P-1)=yiϵ(P-1)yiϵ(AicBic)

    Hence, if ϵ=1 then:

    Pyi(AicBic)yi(Aic)Bi

    And if ϵ=-1 then:

    Pyi-1(AicBic)yi-1(Bic)Ai
  2. 2.

    z-1=z: by the fact that w is freely reduced weget an equality between ϵ-1 and ϵ. Hence, ifϵ=1 then P-1BiAic andtherefore:

    P=zϵ(P-1)=yi(P-1)yi(Aic)Bi

    Similiarly, if ϵ=-1 then P-1AiBic and therefore:

    P=zϵ(P-1)=yi-1(P-1)yi-1(Bic)Ai

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