# What is generated by $(1265)$, $(2376)$, and $(3487)$? (Pocket cube group)

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 $$S_8$$?

Thanks!

• It's almost certainly $S_8$. How transitive can you prove it? 2-transitive? 3-transitive? 4-transitive?.... – Lord Shark the Unknown Nov 2 '18 at 20:47
• Well you already know it contains $12$ distinct copies of $S_5$, for starters... – Inactive - avoiding CoC Nov 2 '18 at 20:53

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 $$S_8$$.
• Surely there are many alternatives routes to the destination. Particularly after the observation that there is a 2-cycle in $G$. – Jyrki Lahtonen Nov 4 '18 at 10:10