# The Unapologetic Mathematician

## Rubik’s Cube Wrapup

I want to tie up a few loose ends about Rubik’s group today.

We can fit Rubik’s group into a sequence that more clearly shows all the structure I’m talking about. Specifically, it’s a subgroup of the bigger group I mentioned back at the beginning. We can restate the three restrictions as saying the maneuvers in Rubik’s group are those in the kernel of a certain homomorphism. So, first let’s write down the big group.

The unrestricted edge and corner groups are just wreath products, which I’ll write out as semidirect products. Without restrictions, these two groups are independent, so we just have a direct product to give the unrestricted Rubik’s group. $\bar{G}=\left(\mathbb{Z}_2^{12}\rtimes S_{12}\right)\times\left(\mathbb{Z}_3^8\rtimes S_8\right)$
I’ll write $(((e_1,e_2,...,e_{12}),\sigma_e),((c_1,c_2,...,c_8),\sigma_c))$ for a generic element of this group. Each part of this list corresponds to part of the expression for $\bar{G}$ above.

Now we want to add up all the edge flips and make them come out to zero. We can write this sum as a homomorphism: $e:\bar{G}\rightarrow\mathbb{Z}_2$ $e(((e_1,e_2,...,e_{12}),\sigma_e),((c_1,c_2,...,c_8),\sigma_c))=e_1+e_2+...+e_{12}$
where the sum is taken in the group $\mathbb{Z}_2$. You should be able to verify that this actually is a homomorphism. Similarly, we want the sum of the total twists as a homomorphism: $c:\bar{G}\rightarrow\mathbb{Z}_3$ $c(((e_1,e_2,...,e_{12}),\sigma_e),((c_1,c_2,...,c_8),\sigma_c))=c_1+c_2+...+c_8$
where the sum is taken in $\mathbb{Z}_3$.

Finally, the permutation condition uses the “signum” homomorphism from a symmetric group to $\mathbb{Z}_2$. It assigns the value ${}0$ to even permutations and the value $1$ to odd ones. We use it to write the last restriction as a homomorphism: $p:\bar{G}\rightarrow\mathbb{Z}_2$ $p(((e_1,e_2,...,e_{12}),\sigma_e),((c_1,c_2,...,c_8),\sigma_c))={\rm sgn}(\sigma_e)+{\rm sgn}(\sigma_c)$

Now we assemble our overall restriction homomorphism as the direct product of these three: $f=e\times c\times p:\bar{G}\rightarrow\mathbb{Z}_2\times\mathbb{Z}_3\times\mathbb{Z}_2$
and get the short exact sequence: $\mathbf{1}\rightarrow G\rightarrow\bar{G}\rightarrow^{f}\mathbb{Z}_2\times\mathbb{Z}_3\times\mathbb{Z}_2\rightarrow\mathbf{1}$

Commenter Dan Hoey brought up where my fundamental operations come from. To be honest, these four are just ones I remember off the top of my head. He’s right, though, that there are systematic ways of coming up with maneuvers that perform double-flips, double-twists, and $3$-cycles. I’ll leave you to read his comment and work out yourself that you can realize four such basic maneuvers as commutators — products of elements of the form $m_1m_2m_1^{-1}m_2^{-1}$. This means that the commutator subgroup $\left[G,G\right]$ of Rubik’s group is almost all of $G$ itself. It just misses a single twist. In fact, $G/\left[G,G\right]\cong\mathbb{Z}_2$ — Rubik’s group is highly non-abelian.

Incidentally, this approach to the cube is not the first one I worked out, but it’s far more elegant than my pastiche of particular tools. I picked it up back when I was at the University of Maryland from a guy who had worked it out while he was at Yale as a graduate student back when the cube first came out: Jeff Adams.

March 28, 2007 - Posted by | Group theory, Rubik\'s Cube

## 2 Comments »

1. Maybe this is a little late for this entry but you might like to know that 5 students at Carinthia University of Applied Sciences in Austria built a completely automated system that solves the Rubik Cube puzzle in fewer than 26 moves in less than 2 minutes. The system looks at the cube with a camera, processes data, runs solving algorithm, and uses mechanical grippers to make the moves. Comment by Steve Myers | March 30, 2007 | Reply

2. […] important representation I haven’t mentioned. Well, actually, I mentioned it in passing while talking about Rubik’s group, but not very explicitly. And it’s a very important […]

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