Example: The two squares group

Let $H = \langle R^2,U^2\rangle$ denote the group generated by the two square moves, $R^2$ and $U^2$ or the Rubik's cube. (The reader with a cube in hand may want to try the Singmaster magic grip : the thumb and forefinger of the right hand are placed on the front and back face of the fr, br edge, the thumb and forefinger of the left hand are placed on the front and back face of the uf, ub edge; all moves in this group can be made without taking your fingers off the cube.) This group contains the useful 2-pair edge swap move $(R^2*U^2)^3$. We can find all the elements in this group fairly easily:

$$\begin{array}{c} H=\{1,R^2,R^2*U^2,R^2*U^2*R^2, (R^2*U^2)... ...(R^2*U^2)^4*R^2,(R^2*U^2)^5, (R^2*U^2)^5*R^2\}, \end{array}$$

Therefore, $\vert H\vert=12$. Note that $1=(R^2*U^2)^6$, $U^2=(R^2*U^2)^5*R^2$, and $U^2*R^2=(R^2*U^2)^5$. (By the way, this listing without repetition of $H$ by successive multiplication by $R^2$ then $U^2$ may be reformulated graphically by saying the ``the Cayley graph of $H$ with generators $R^2$, $U^2$ has a Hamiltonian circuit".) To discover more about this group, we label the vertices of the cube as follows:

The move $R^2$ acts on the set of vertices by the permutation $(1,\ 4)(2,\ 3)$ and the move $U^2$ acts on the set of vertices by the permutation $(4,\ 5)(3,\ 6)$. We label the vertices of a hexagon as follows:

The permutation $(1,\ 4)(2,\ 3)$ is simply the reflection about the line of symmetry containing both 5 and 6. The permutation $(4,\ 5)(3,\ 6)$ is simply the reflection about the line of symmetry containing both 1 and 2. By a fact stated in section 5.9.1, these two reflections generate the symmetry group of the hexagon.

Exercise 5.9.5   Work out a similar result to that in Example 5.9.1 using a knight in place of a king.

Exercise 5.9.6   We refer to Example 5.9.1. In how many ways can a king starting at $(0,0)$ reach $(1,1)$ in $5$ moves?

Exercise 5.9.7   The set of all possible knight moves also naturally forms a group which is a subgroup of King:

$$Knight:=\{x^m y^n\ \vert\ m,n\in \mathbb{Z}, \vert m\vert+\vert n\vert\equiv 0\ ({\rm mod}\, 3\},$$

(where the group operation is again multiplication). In how many ways can knight starting from $(0,0)$ reach (a) $(1,1)$ in $5$ moves? (b) $(2,1)$ in $5$ moves? [Hint: $(x^2 y + x^{-2} y + ... + x y^{-2} )^5=$ ?]

Exercise 5.9.8   We refer to Example 5.9.1 and Exercise 5.9.5. How would the group Bishop be defined? Is there a subgroup Pawn? Rook? Queen?

Exercise 5.9.9   Consider a square centered at the origin having sides parallel to the coordinate axes, superimposed on the figure in Example 5.9.4. Label the four vertices of this square starting with the upper left-hand vertex and moving around in a clockwise fashion, $1, 2, 3, 4$. Compute the symmetry group in Example 5.9.4 as a subgroup of $S_4$. In other words, write down each of the elements of $G$ explicitly in disjoint cycle notation.