Question #105103
Find Z (D8) the centre of . D8 Also give the algebraic structure of D8/Z (D8.)
1
Expert's answer
2020-03-13T11:59:44-0400

Notation:

R0R_0== Rotation of 0°0^° , R90R_{90} =Rotation of 90°{90}^° , R180=R_{180}= Rotation of180°{180}^°,R270=R_{270}= Rotation of 270°{270}^° ,H=H= Flip about horizontal axis ,V=V=

Flip about vertical axis, D=D= Flip about main diagonal ,DD'== Flip about the other diagonal.

D8=D_8= { R0,R90,R180,R270,H,V,D,DR_0,R_{90},R_{180},R_{270},H,V,D,D' }

Z(D8)=Z(D_8)= { aD8:ax=xa  xD8a\in D_8:ax=xa \space \forall \space x\in D_8 }

First observe that since every rotation in D8D_8 is a power of R90R_{90} ,rotation commute with rotation. We now investigate when a rotation commutes with a reflection. Let RR be any rotation and FF be any reflection in D8D_8 .Observe that since RFRF is a reflection we have RF=(RF)1=F1R1=FR1.RF=(RF)^{-1}=F^{-1}R^{-1}=FR^{-1}. Thus it follows that R and FR\space and \space F commute if and only if FR=RF=FR1.FR=RF=FR^{-1}. By cancellation ,this hold if and only if R=R1R=R^{-1} .But R=R1R=R^{-1} only when R=R0 and R=R180R=R_0 \space and \space R=R_{180} .

Thus Z(D8)=Z(D_8)= {R0,R180R_0,R_{180} }=KK (say).

Now,

D8Z(D8)=\frac{D_8}{Z(D_8)}= { R0K,R90K,R180K,VK,DKR_0K,R_{90}K,R_{180}K,VK,DK }.


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