# The Rubik Square permutation groups

This post was inspired by this webpage of mathematical challenge due to Mickaël Launay (French).

Let $G_n$ be the subgroup of $S_{n^2}$ generated by the red arrow permutations as for the following picture with $n = 5$:

(Source: http://www.micmaths.com/defis/defi_07.html)

We can call $G_n$ the n-th Rubik Square permutation group.

Question: Is $G_n$ a strict subgroup of $S_{n^2}$?

#### Solutions Collecting From Web of "The Rubik Square permutation groups"

Let us look at the commutator of the top row and leftmost column operations.
We easily see that the move sequence $\leftarrow\uparrow\rightarrow\downarrow$ is a 3-cycle involving the entries at positions $(1,1), (1,2), (2,1)$. Similarly for any other pair of rows and columns. Using appropriate powers of the elementary moves we get 3-cycles on any triple of positions $(i,j), (i,k), (\ell,j)$. A lot hinges on what we can say about the subgroup generated by such 3-cycles.

I claim that such 3-cycles generate the alternating group $A_{n^2}$.
It suffices to prove that we get all the 3-cycles. The key observation is that any two 3-cycles $\in A_4$ generate all of $A_4$ unless they share a fixed point. This is because the generated subgroup is doubly transitive, and $A_4$ has no proper doubly transitive subgroups. What this means is that any pair of 3-cycles with an overlap of two positions gives us any 3-cycle among the four positions in the union.

So let $(i_t,j_t),t=1,2,3,$ be any three positions. The 3-cycles on
$(i_1,j_1),(i_2,j_2),(i_2,j_1)$ and $(i_1,j_1),(i_2,j_1),(i_2,j_3)$ thus
generate the 3-cycles on $(i_1,j_1),(i_2,j_2),(i_2,j_3)$. Together with the 3-cycles $(i_2,j_2),(i_2,j_3),(i_3,j_3)$ this then generates the desired 3-cycle.

Therefore $A_{n^2}\le G_n$.

If $n$ is odd, then all the generators of $G_n$ are in $A_{n^2}$, so we must have $G_n=A_{n^2}$. But if $n$ is even, then $G_n$ contains also odd permutations and we have $G_n=S_{n^2}$.

I won’t say anything about the Cayley graph.