The alternating group

The following remarkable result about the alternating group will not be needed to understand the structure of the Rubik's cube. However, the theorem below is interesting because of its connection with the fact (due to N. Abel and E. Galois) that you cannot solve the general polynomial of degree 5 or higher using radicals, i.e., that there is no analog of the quadratic formula for polynomials of degree 5 or higher. Explaining this connection would take us too far from our main topic. The interested reader is referred to Artin [Ar], chapter 14.

Theorem 5.16.7   If $X$ has $5$ elements or greater then $A_X$ has no non-trivial proper normal subgroups. In other words, if $H \triangleleft A_X$ is a normal subgroup then either $H = \{1\}$ or $H = A_X$.

This will not be proven here. (For an algebraic proof, see for example [R].) For a visual, geometric ``proof'', that $A_5$ is simple, see [K]. The discussion there hinges on the fact that the symmetry group of the icosehedron can be identified with $A_5$, much like the symmetry group of the cube can be identified with $S_4$. See Exercise 5.12.13. Unlike the above theorem, the next fact about the alternating group will be needed later. It will be used in our determination of the structure of the Rubik's cube group. This fact also arose in connection with our discussion of the ``legal positions'' of the 15 puzzle in a previous chapter.

Proposition 5.16.8   Let $H$ be the subgroup of $S_n$ generated by all the 3-cycles in $S_n$ then $H=A_n$.

proof: Since $sgn:S_n\rightarrow \{\pm 1\}$ is a homomorphism, and since any 3-cycle is even, any product of 3-cycles must also be even. Therefore, $H\subset A_n$. If $g\in A_n$ then $g$ must swap an even number of the inequalities $1<2<...<n-1<n$, by Definition 4.2.2. Therefore, (since any permuation may be written as a product of $2$-cycles, Theorem 4.5.1) $g$ must be composed of permutations of the form $(i,\ j)(k,\ l)$ or $(i,\ j)(j,\ k)$. But $(i,\ j)(k,\ l)=(i,\ j,\ k)(j,\ k,\ l)$ and $(i,\ j)(j,\ k)=(i,\ j,\ k)$. Therefore, $g\in H$. This implies $A_n\subset H$, so $A_n=H$. $\Box$ As an application to the Rubik's cube, suppose you have all the corners in the correct position. They might be twisted, that's okay. The point is that they don't need to be swapped or permuted. Suppose that the edges are not in the correct position. It turns out that the permuation which puts the edges in the correct position must be an even permutation. There are simple moves for a $3$-cycle on edges: $M_2^2U^{-1}M_R^{-1}U^2M_RU^{-1}M_R^2$.