Until today, I was convinced that we needed at least 6 generators to generate the Rubik's Cube group. However, consider the following...

$$a = (0,5,7,2)(1,3,6,4)(8,16,24,32)(9,17,25,33)(10,18,26,34)$$

$$b = (5,15,42,24)(6,12,41,27)(7,10,40,29)(16,21,23,18)(17,19,22,20)$$

$$c = (0,26,47,13)(1,28,46,11)(2,31,45,8)(32,37,39,34)(33,35,38,36)$$

$$d = (0,39,40,16)(3,36,43,19)(5,34,45,21)(8,13,15,10)(9,11,14,12)$$

$$e = (2,18,42,37)(4,20,44,35)(7,23,47,32)(24,29,31,26)(25,27,30,28)$$

d^{-1}c^{-1}da^{-1}e^{-1}aeb^{2}d^{-1}b^{-1}ab^{-1}e^{-1}b^{-1}d^{-1}baba^{-1}d^{-1}a^{-1}ca^{-1}e^{-1}aeab^{2}db^{-2}e^{-1}d^{-1}c^{-1}d^{-1}cd^{-2}b^{-1} dbd^{-1}ec^{-1}e^{-1}d^{-1}ada^{-1}dcd^{-1}c^{-1}eb^{2}c^{2}d^{-1}b^{-1}c^{-2}ebc^{-1}a^{-1}b^{-2}ca^{-1}e^{-1}c^{-1}eca^{2}b^{-1}d^{-1}ba^{-1}b^{-1}dbd^{-1}ada^{-1}dcd^{-1}c^{-1}e^{-1}c^{-1}d^{-1}eada^{-1}de^{-1}cd^{-1}eacea^{-1}e^{-1}a^{-2}c^{-1}e^{-1}ceac^{-1}ba^{-2}b^{-1}ce^{-2}c^{2}e^{2}d^{2}c^{2}a^{-1}b^{2}c^{2}d^{-2}e^{2}ac^{2}abd^{-1}c^{-2}b^{-2}ec^{2}ba^{-1}c^{-2}a^{-1}ed^{-1}b^{2}e^{-1}da^{-2}eabab^{-1}e^{-1}a^{2}b^{-1}e^{-1}beae^{-1}a^{2}cea^{-1}e^{-1}a^{-2}c^{-1}e^{-1}ceaeba^{-1}c^{-1}bdcb^{-2}e^{-1}c^{-1}bd^{-1}cb^{-1}ae^{2}db^{-1}e^{2}d^{-2}ceb^{2}de^{-1}a^{-2}b^{2}e^{-2}d^{-2}c^{-2}d^{-2}e^{-2}ce^{-1}c^{-1}bab^{-1}a^{-1}eacec^{-1}ba^{-1}b^{-1}a^{-1}e^{-1}a^{-1}b^{-1}ce^{2}bc^{-1}ec^{-2}ed^{-1}a^{-2}de^{-2}a^{-1}c^{2}a^{-2}c^{-2}a^{2}c^{2}d^{-1}a^{-1}de^{-1}cdc^{-1}d^{-1}e = $(13,21,29,37)(14,22,30,38)(15,23,31,39)(40,42,47,45)(41,44,46,43)$

This is a lot of information, but bear with me. The permutations $a$, $b$, $c$, $d$ and $e$ correspond to 5 faces of a Rubik's Cube. I then write an expression in these 5 generators that gives me the inverse of what is usually cited as the 6th generator of the group.

This expression is large, but can anyone verify it?

Note: If someone cares, I am willing to post a You-Tube video that animates the Rubik's Cube and shows that the above sequence actually does rotate the remaining face using only the other faces of the cube.

  • $\begingroup$ These ones may also help: math.stackexchange.com/questions/1836898/… math.stackexchange.com/questions/249476/… $\endgroup$ – Jean-Claude Arbaut Mar 31 '17 at 5:58
  • $\begingroup$ It has been known for a long time that any five of the six standard generators of the cube group are sufficient to generate the group. $\endgroup$ – Derek Holt Mar 31 '17 at 8:32
  • $\begingroup$ Thanks. It has only been known for a short time in my brain that five of the six standard generators of the cube are sufficient. Singmaster's result makes things even more interesting. Even if a generating set can't lose a member, that doesn't mean a smaller set can't be found. $\endgroup$ – Spencer Parkin Mar 31 '17 at 18:26
  • 1
    $\begingroup$ @SpencerParkin There is a trivial example: consider the cartsian product of two cyclic groups, $(\Bbb Z/2) \times (\Bbb Z/3)$. You can give a set of two generators ($(1,0)$ and $(0,1)$), that you can't reduce to one, but $(1,1)$ alone is also a generator. $\endgroup$ – Jean-Claude Arbaut Apr 1 '17 at 17:29
  • $\begingroup$ Thank you for the example, Jean. I suppose it just wasn't obvious to me, though clearly, there are obvious examples. I suck at group theory. $\endgroup$ – Spencer Parkin Apr 2 '17 at 23:32