ADVANCED TOPIC: Realizing PGL(2,F5) inside the Rubik's cube group The material in this chapter was communicated to me by Dan Bump. This chapter is relatively advanced in that it requires more mathematical background from the reader than the previous chapters. We have seen in an earlier chapter how to "realize" the group of quaternions {1,-1,i,-i,j,-j,k,-k} inside the Rubik's cube group. This chapter is devoted to a similar but more complicated realization. Let F5 denote the finite field with 5 elements, so F5 is, as a set, F5 = {0,1,2,3,4}, with addition and multiplication being performed mod 5. Let G denote the Rubik's cube group. Let H be the subgroup generated by F and U: H = . We describe how to label the six vertices on the "up" and "front" faces of the cube, fru, flu, dfl, dfr, bru, blu, with the elements "projective plane" P^2(F5) = {0,1,2,3,4,infinity} in a certain way. This so-called projective plane may be identified with the set of lines through the origin in the Cartesian plane F5^2 by associating each number (including infinity) with the slope of the corresponding line. Exercise: The group H acts on the set of vertices above. Exercise: The group [a b] PGL(2,F5) = { [ ] | a,b,c,d in F5 [c d] ad-bc <> 0 } acts on the set P^2(F5) by means of the linear fractional transformations: [a b] ax + b [ ] : x |---> -------- [c d] cx + d [a b] [ ] : P^2(F5) ---> P^2(F5). [c d] We shall label the 6 vertices above with elements of P^2(F5) in such a way that Lemma 1: There are a0,a1,b0,b1,c0,c1,d0,d1 in F5 (given explicitly below) such that (a) the action of F on these vertices is the same as the action of some linear fractional transformation a0x + b0 f_F(x) = ---------, c0x + d0 (b) the action of U on these vertices is the same as the action of some linear fractional transformation a1x + b1 f_U(x) = --------- . c1x + d1 Lemma 2: PGL(2,F5) = . The Labeling Label the up and front vertices as 3 ______________2 /______/u_____/ | /______/______/| | 0 | | | 4 | | | | /| |______f______ /| | | | | | | | | | |/ |______|______|/ 1 infinity Let x - 1 f_F(x) = --------, x + 1 f_U(x) = 3x + 3 . The map phi : F |--> f_F, U |--> F_U, extends to an isomorphism of groups phi : ---> subset PGL(2,F5). Exercise: Verify Lemma 1 above. Proof of Lemma 2 Let [a b] [ ] in PGL(2,F5) [c d]* denote the image of [a b] [ ] in GL(2,F5) [c d] under the natural map GL(2,F5) --> PGL(2,F5), g |--> F5^x*g. Since [0 -1] f_U^2 = [ ] [1 0]* we have [0 -1] [ ] in . [1 0]* Since [-1 0] f_F*f_U^5 = [ ] [-1 -1]* it follows that [1 0] [ ] in . [1 1]* Conjugating this matrix by f_U^2, we find that [1 -1] [ ] in . [0 1]* Recall that SL(2,F5) is generated by elementary transvections (see [R]). Therefore, PSL(2,F5) subset subset PGL(2,F5). It is also known (see [R]) that |PSL(2,F5)| = 60 and |PGL(2,F5)| = 120. It remains to show that there is an element of which does not belong to PSL(2,F5). We claim that such an element is f_U. Note that det(f_U) belongs to the set 3(F5^x)^2 = {3x^2 | x in F5^x}. But an element of PSL(2,F5) must have determinant 1. Since 3^(-1) = 2 (mod 5) is not a square mod 5, there is no element of F5 which satisfies 1 = 3x^2. Thus f_U does not belong to PSL(2,F5). QED 