# what is the smallest symmetric group that contains the cube group as a subgroup?



## whauk (May 10, 2013)

so every finite group (with less than n+1 elements) is isomorphic to a subgroup of the symmetric group Sn. (http://en.wikipedia.org/wiki/Cayley's_theorem)
however we dont need n to be ~10^18 to find the cube group as a subgroup in it.

my current upper and lower bounds for n are:
upper bound: 48 because the cube group actually just cycles 48 stickers around. => permutation group of 48 elements suffices (=S48)
lower bound: 23 because there is an element of order 2*3*7*11(=462) namely a 11-cycle of edges with a 7 cycle of corners, a flipped edge and a twisted corner. so the smallest possible Sn to contain this must contain an 11-,7-,3- and 2-cycle simultaneously. and the sum of these numbers is 23. also 23! which is the order of S23 is already greater than the order of the cube group.

this is all i get from ~20 minutes of thinking. can anyone improve these boundaries?


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## cuBerBruce (May 11, 2013)

I believe the fact that the cube group has an element of order 720 implies a lower bound of 30.

My guess is that S48 is the smallest symmetric group that has a subgroup isomorphic to the cube group.

Of course, this assumes we're talking about the standard fixed-centers cube group ( <U,D,R,L,F,B> ). We could also ask about the smallest symmetric group having an isomorphism of <U,E,R,M,F,S> (fixed-corner cube group) or <U,E,R,L,F,S> (fixed-edge cube group) as a subgroup. These have the obvious upper bounds of 51 and 52, respectively.


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## Haru (May 17, 2013)

Consider a small cube subgroup which affects only 3 edges (UF, UL and UR). 
This subgroup is generated by a 3-edge cycle (UF -> UL -> UR -> UF) and a double edge flip (which flips UL and UR).

This subgroup is isomorphic to A4, therefore it is contained in S4.
The 3-edge cycle corresponds to (123) of S4 and the double edge flip to (12)(34).

This fact seems to suggest that a symmetric group smaller than S48 containing the Rubik's cube group might exist.


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## cuBerBruce (May 17, 2013)

There is certainly some redundancy of information in a facelet representation of the cube group. If we know the position of one facelet for each of 11 edge pieces and the position of one facelet for each of 7 corner pieces, we can determine the position of all the other facelets. However the total number of positions where these 18 facelets can reside is still 48. But it still seems possible there might be some trick to reduce the number of permutation points at least one, maybe a little more.

I noticed GAP has a function called SmallerDegreePermutationRepresentation. This seemed to be just what's needed to answer this question. However, it doesn't guarantee it will find a smaller representation even if one exists. But I might as well give it a try.


```
gap> U := ( 1, 3, 8, 6)( 2, 5, 7, 4)( 9,33,25,17)(10,34,26,18)(11,35,27,19);
(1,3,8,6)(2,5,7,4)(9,33,25,17)(10,34,26,18)(11,35,27,19)
gap> L := ( 9,11,16,14)(10,13,15,12)( 1,17,41,40)( 4,20,44,37)( 6,22,46,35);
(1,17,41,40)(4,20,44,37)(6,22,46,35)(9,11,16,14)(10,13,15,12)
gap> F := (17,19,24,22)(18,21,23,20)( 6,25,43,16)( 7,28,42,13)( 8,30,41,11);
(6,25,43,16)(7,28,42,13)(8,30,41,11)(17,19,24,22)(18,21,23,20)
gap> R := (25,27,32,30)(26,29,31,28)( 3,38,43,19)( 5,36,45,21)( 8,33,48,24);
(3,38,43,19)(5,36,45,21)(8,33,48,24)(25,27,32,30)(26,29,31,28)
gap> B := (33,35,40,38)(34,37,39,36)( 3, 9,46,32)( 2,12,47,29)( 1,14,48,27);
(1,14,48,27)(2,12,47,29)(3,9,46,32)(33,35,40,38)(34,37,39,36)
gap> D := (41,43,48,46)(42,45,47,44)(14,22,30,38)(15,23,31,39)(16,24,32,40);
(14,22,30,38)(15,23,31,39)(16,24,32,40)(41,43,48,46)(42,45,47,44)
gap> Cube := Group(U,D,L,R,F,B);
<permutation group with 6 generators>
gap> SmallerDegreePermutationRepresentation (Cube);
IdentityMapping( <permutation group of size 43252003274489856000 with 6 generators> )
```

Unfortunately it produced the identity mapping, meaning it failed to find a smaller representation. But this doesn't prove that a smaller representation doesn't exist.


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