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Since "cubes" with higher dimension than three exist I think it's natural to ask for higher dimensional Rubik's cubes. These so called hypercubes don't seem to have been described from a group theoretic point of view.

Are there any papers on this? Is the group of the $3\times 3\times 3 \times 3$ cube a subgroup of a wreath product of another wreath product?

In case you don't know about the $3\times 3\times 3$ cube. Its group is a subgroup of a product of wreath products. The wreath products describing the corner pieces and the edge pieces and representing a permutation of them with its action in the respective orientation. That is why I conjecture that in the $3\times 3\times 3 \times 3$ case we might get a wreath product of the permutation of the faces, which are now 3d cubes itself, by the the wreath product of the permutation of the 2d faces of these by the groups of their orientations.

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  • $\begingroup$ Dear @Jason Pioneer: What do you mean by "a direct group of wreath products"? $\endgroup$ – Ricardo Andrade Nov 7 '13 at 0:18
  • $\begingroup$ I mean if $A,B$ and $G,H$ are groups and $A wr B$ and $G wr H$ are wreath products of these groups then $(A wr B) \times (G wr H)$ is the direct group of these wreath groups. I should have written product of groups instead of direct group.That was a typo. Let me correct it. $\endgroup$ – Jason Pioneer Nov 7 '13 at 0:21
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    $\begingroup$ Maybe I'm missing something obvious, but it seems to me that the standard proof for ordinary Rubik's cubes works for $3^n$ cubes for any $n$ to give the cube group as a subgroup of a direct product of wreath products, with each factor corresponding to acting on cubies with $k$ stickers for $k=2,3,\ldots,n$. Is this what you want, or are you looking for something else? $\endgroup$ – Logan M Nov 7 '13 at 3:52
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The 4-dimensional, i.e. $3 \times 3 \times 3 \times 3$, equivalent of the Rubik's cube has 8 three-dimensional sides, each of which consists of $3^3 = 27$ three-dimensional colored "squares". Of these 27 "squares", the one in the center is fixed. Thus in total our 4-dimensional Rubik's cube has $8 \cdot 26 = 208$ movable "squares", and the group of its sequences of moves under composition embeds therefore into ${\rm S}_{208}$.

Now just as the usual 3-dimensional Rubik's cube has 2 kinds of movable cubies (edge stones and corner stones), our 4-dimensional Rubik's cube has 3 distinct kinds of movable cubies:

  • The corner stones, of which there are $2^4 = 16$.

    Each corner stone has 4 visible colored "squares".

  • The edge stones, of which there are $2 \cdot 12 + 8 = 32$.

    Each edge stone has 3 visible colored "squares".

  • The face stones, of which there are $2 \cdot 6 + 12 = 24$.

    Each face stone has 2 visible colored "squares".

Therefore, analogous to the usual 3-dimensional case, we can finally say that our 4-dimensional Rubik's cube group embeds into the following direct product of wreath products: $$ {\rm A}_4 \wr {\rm S}_{16} \times {\rm S}_3 \wr {\rm S}_{32} \times {\rm C}_2 \wr {\rm S}_{24}. $$

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  • $\begingroup$ Hm im pretty happy with that answer so far,because you have the calculation of the stuff i had a problem imagining. $\endgroup$ – Jason Pioneer Nov 7 '13 at 19:12
  • $\begingroup$ I agree with your argument, but if by $C_n$ in your final formula you mean the cyclic group of order $n$, then I don't see exactly where that is coming from. Naively I would think that for the corners the group should be $A_4$, and for the cubies with 3 stickers it should be $S_3$. More generally I'd expect for a $d$-dimensional cube that the corners correspond to $A_d$ and all other types of pieces correspond to $S_k$ for $k=2,\ldots,d-1$. Could you explain how you get that (or if I'm misunderstanding your claim)? $\endgroup$ – Logan M Nov 8 '13 at 19:33
  • $\begingroup$ @LoganMaingi: Sure. -- Thanks for spotting this! $\endgroup$ – Stefan Kohl Nov 8 '13 at 23:58
  • $\begingroup$ So i think the first factor of this group in the n-dimensional case will be $A_n \wr S_{2^n}$. $\endgroup$ – Jason Pioneer Nov 9 '13 at 0:07
  • $\begingroup$ @JasonPioneer That seems correct to me. The full expression should be (assuming I haven't made a typo) $$\displaystyle (A_n \wr S_{2^n}) \times \left(\prod_{i=2}^{n-1} S_i \wr S_{{n \choose i} 2^{i}} \right).$$ $\endgroup$ – Logan M Nov 9 '13 at 0:34

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