The 8-Puzzle is a famous sliding-block puzzle (a $3\times 3$ grid with one open space) that many readers will have grown up with. Here's an online implementation: https://murhafsousli.github.io/8puzzle
I am interested in representing the set $ S $ of solvable configurations of the puzzle, especially counting the number of elements of $ S $.
For some time, I was sidetracked by the red herring of thinking that $ S $ is a group. I recently cured myself of that assumption, since there is no clear way to multiply elements of $ S $. However, we can consider the group $ H $ of of moves (here we take a "move" to be any sequence of transpositions of a piece into the adjacent empty slot) which, when they act on the solved puzzle $ s_0 \in S $, leave the empty slot in-place. This is clearly a group since such moves can indeed be composed. Clearly $ H $ is a subgroup of $ S_8 $ since it acts by permuting 8 elements (the 8 tiles).
I would conjecture that $ H $ is $ A_8 $, but I am having some trouble proving it. Is there a simple argument?
If I were to prove this conjecture (and assuming it's true regardless of where we take the empty slot to be fixed), it would imply that $ |S| = 9 \times |A_8| = 9! / 2 $.