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I was going through Janet Chen's "Group Theory and the Rubik's Cube" (PDF), and I came across this question:

Let $D$ denote a clockwise quarter turn on the downward face and $R$ denote a quarter turn on the rightmost face clockwise (We have the 6 faces: front (facing us), back, down, up, left, right.) Let $\mathbb{G}$ be the group formed by all possible moves on the Rubik's cube. Let $H$ be the subgroup of $\mathbb{G}$ generated by $D^2$, $R^2$. (Here, $D^2$ is taken as the composition of two $D$ moves.) Find the number of elements of $H$.

I have listed out $D^2$, $R^2$ in disjoint cycle notation but I am stuck.

Would appreciate if anyone could help, thanks.

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  • $\begingroup$ It is easy to verify that $(D^2R^2)^3$ will return all the corner pieces back to their starting locations, but will have swapped two edge pieces on both faces (assuming 3x3x3). Therefore $(D^2R^2)$ has order six, and hence they generate the dihedral group of order twelve (by the theory of dihedral groups). $\endgroup$ – Jyrki Lahtonen Aug 29 '19 at 20:54
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$H$ is generated by $d=D^2$ and $r=R^2$. We have that $d^2=r^2=e$ is the identity, so each element of $H$ may be written on one of four forms: $$ drdr\cdots dr\\ rdrd\cdots rd\\ drdr\cdots rd\\ rdrd\cdots dr $$ ($d$ itself is of the third form, and $r$ of the fourth.)

At some point, repeating $rd$ takes you back to the identity. This puts a limit to how long these chains can be.

Finally, note that some of the chains described above, even below the "identity length", are equal. For instance, the chain $drdr\cdots rd$ that is only one $r$ short of being the identity, is actually equal to the chain $r$ (by virtue of $r=r^{-1}$).

Taking all this into account, you will have found $|H|$.

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I solved this using GAP (tag added):

gap> d := (1,3)(2,4);; r := (1,5)(2,6);;
gap> H := Group(d,r);;
gap> StructureDescription(H);
"S3"

So H is the symmetric group on three elements. How are these permutations obtained?

gap> epi := EpimorphismFromFreeGroup(H);
[ x1, x2 ] -> [ (1,3)(2,4), (1,5)(2,6) ]
gap> List(H, h->[h,PreImagesRepresentative(epi,h)]);
[ [ (), <identity ...> ], [ (3,5)(4,6), x1^-1*x2^-1*x1^-1 ],
[ (1,3,5)(2,4,6), x1^-1*x2^-1*x1^-2 ], [ (1,3)(2,4), x1^-1 ],    
[ (1,5,3)(2,6,4), (x1^-1*x2^-1)^2 ], [ (1,5)(2,6), x2^-1 ] ]

Can I express this group using a finite presentation?

gap> iso := IsomorphismFpGroup(H);
[ (3,5)(4,6), (1,5,3)(2,6,4) ] -> [ F1, F2 ]
gap> pr := PresentationFpGroup(Image(iso));
<presentation with 2 gens and 3 rels of total length 10>
gap> TzPrintRelators(pr);
#I  1. F1^2
#I  2. F1^-1*F2*F1*F2^-2
#I  3. F2^3
gap> TzGoGo(pr);
#I  there are 2 generators and 3 relators of total length 9
gap> TzPrintRelators(pr);
#I  1. F1^2
#I  2. F2^3
#I  3. (F1*F2)^2
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  • $\begingroup$ Where did you get d := (1,3)(2,4) and r := (1,5)(2,6) from? It looks like you are just permuting the big faces, and I am not convinced that this is correct. There is a natural way of viewing the Rubik's cube group as a subgroup of $S_{48}$, as you need to consider each individual face of each "cubie" (of which there are $54$), but you can ignore the central cubies of each face, so you are permuting the $55-6=48$ mini-faces. (Actually this is a worked example on the GAP website.) $\endgroup$ – user1729 Aug 29 '19 at 9:37
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    $\begingroup$ (Also, using GAP to find a presentation for $S_3$ is overkill!) $\endgroup$ – user1729 Aug 29 '19 at 9:38
  • $\begingroup$ This holds only on a 2x2x2 cube. It is not entirely clear to me whether the question is about the 2x2x2 or about a bigger cube. $\endgroup$ – Jyrki Lahtonen Aug 29 '19 at 20:56

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