I am trying to find the group structure $G$ of the $2\times 2\times 2$ Rubik's cube, or the "pocket cube," and I have determined that it is isomorphic to the group of permutations on $8$ numbers generated by the cycles $(1265)$, $(2376)$, and $(3487)$, each of which corresponds to the rotation of a face of the pocket cube.

If it helps, I already know that the subgroup of $S_8$ generated by $(1265)$ and $(2376)$ is isomorphic to $S_5$, though it does not consist of all permutations of the numbers $1$ through $5$ as is obvious from the nature of its generators.

Can someone please show me or give me a hint about how to determine the structure of the group generated by $(1265)$, $(2376)$, and $(3487)$, without the aid of a computer algebra system or online database of groups/subgroups of $S_8$?


  • 2
    $\begingroup$ It's almost certainly $S_8$. How transitive can you prove it? 2-transitive? 3-transitive? 4-transitive?.... $\endgroup$ – Angina Seng Nov 2 '18 at 20:47
  • 1
    $\begingroup$ Well you already know it contains $12$ distinct copies of $S_5$, for starters... $\endgroup$ – Servaes Nov 2 '18 at 20:53

Fleshing out my comments to an answer.

As Lord Shark the Unknown foresaw, your group is the full symmetric group $S_8$.

Let $G=\langle (1265),(2376),(3487)\rangle$ be the group generated by all three moves, and $K=\langle(1265),(2376)\rangle$ the subgroup generated those two moves. My suggested steps follow:

  • The group $K$ acts transitively on the set $T=\{1,2,3,5,6,7\}$. This is because the orbit of $1\in T$ contains $2,6,5$ - apply powers of the first generator to $1$. Applying powers of the second generator to $2$ shows that $3,7$ are also in the orbit.
  • The group $K$ actually acts doubly transitively on $T$. An easy way of seeing that is to calculate the product $$(1265)(2376)=(12375).$$ The presence of this 5-cycle in $K$ proves that the point-stabilizer of $6$, $\operatorname{Stab}_K(6)$, acts transitively on the set $T\setminus\{6\}$. This is known (and easily seen) to imply that $K$ acts doubly transitively on $T$.
  • Therefore the group $K$ contains an element $\alpha$ such that $\alpha(3)=7$ and $\alpha(7)=3$. Together with the third generator we can then find the following elements of $G$: $$ \begin{aligned} \alpha(3487)\alpha^{-1}&=(7483)\in G,\\ (7483)(3487)&=(384)\in G,\\ (384)(3487)&=(78)\in G. \end{aligned} $$
  • Because $K$ acts transitively on $T$, conjugating $(78)$ by suitable elements of $K$ gives us all the six 2-cycles $(x8)$, where $x\in T$. Those 2-cycles are known to generate the group $H$ of all permutations of $T\cup \{8\}$. In other words $H=\operatorname{Stab}_{S_8}(4)\le G$.
  • Because $G$ acts transitively on the set of all eight vertices, and it contains the full point stabilizer $\simeq S_7$, it must be all of $S_8$.
  • $\begingroup$ Surely there are many alternatives routes to the destination. Particularly after the observation that there is a 2-cycle in $G$. $\endgroup$ – Jyrki Lahtonen Nov 4 '18 at 10:10

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