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Exercise F.Prove that for an isometryαthe identityασC=σCαfor all points C implies thatα=ι.
Proof.Multiplyα−1from the right side,we getασCα−1=σCαα−1=σCι=σC.
The conjugateσCαofσC isασCα−1by the definition of conjugate.SoσCα=ασCα−1.
ThusσCα=σC for all points C.
From the argument we had in Exercise C,we can conclude that Cα=C for all points C.This is the property of identityιthat it keeps every pointfixed. Thusα=ι.
Exercise e the theory for composing reflections developed in the rota-tions lesson to prove that the product of two reflections in perpendicular lines is the central reflection in their intersection point.
Proof.
Figure1:Exercise G
1
Draw line a⊥b intersecting at Q.Then by R3,α=σaσb,Q=(ab).
Let X be any point in the plane.If X=Q,σQ(Q)=Q.By R2αa(αb(Q))=
αa(Q)=Q since(Qb)and(Qa).SoσQ=σaσb.
Consider the case when X=Q,let XX b cross b at B,aX b X a b=A.X b=
σb(X)=2B−X.Subsequently,X ab=σa(X b)=2A−X b=2A−2B+X.
Since AX b⊥a,XX b⊥b,a⊥b,quadrilateral AQBX b is a rectangle.Hence,AQ X b B and AQ=X b B.But BX=X b B,so AQ=BX and AQ BX.Quadrilateral AQXB is a parallelogram.So A−B=Q−X.
So X ab=2(A−B)+X=2(Q−X)+X=2Q−X,which is also central
reflection of X about Q.
Therefore,σQ=sigma aσb.
Exercise H.Let M be a point on a line m,i.e.(Mm).Use the conjugacy theorem for reflections to show thatσM andσm commute andfind the single
reflection equal to their product.
Hint:Use super Exercise D to factorσM using mirror m and its perpendicular
h at M.Now it boils down to some algebra.But if you don’t use the conjugacy theorem in your proof,the proof is probably incomplete(hence wrong,read the Advice on proofs with gaps.)
Proof.mh=M and m⊥h.FactoringσM using Exercise D,we getσM=
σmσh=σhσm.
By CTR,σmσMσ−1
m =σM m.Since(Mm),σM m=σM.SoσmσMσ−1
m
=σM.
Multiplyσm from the right,we getσmσM=σMσm.ThusσM andσm do commute.
SubstituteσM=σmσh=σhσm into the equation we get,σm(σmσh)=(σmσm)σh. Since reflections are involutions,σmσm=ι.
σMσm=σm(σmσh)=(σmσm)σh=ισh=σh.
Exercise e the conjugacy theorems and other properties of reflections in the notes to prove that m=perbis(P,Q),the perpendicular bisector of a
2
segment,if and only if(note the iff)
σmσQσmσP=ι
Proof.IfσmσQσmσP=ιandσPσP=ι,thenσ−1
m =σm andσmσQσm(σPσP)=
σP.
σmσQσm=σP.
Similarly,σmσm=ι.Soσm=σ−1
m .ThenσmσQσ−1
m
=σP.By definition of
conjugate,this is equivalent toσσ
m(Q)
=σP.
This is equivalent toσm(Q)=P.Thus m=perbis(P,Q).
3。