Klein Group in a combinatorics problem on checker board Came a cross problem 21 here: 
https://euclid.ucc.ie/mathenr/IMOTraining/2008%20Winter%20Camp%20-%20David%20Arthur%20-%20Invariants.pdf
Essentially, for each turn, a checker can jump diagonally over an adjacent checker and remove that adjacent checker if the other side is empty. Can we remove all and leave only one checker in the below board?

There is below hint: Consider the Klein group. This has 4 elements $\{e, a, b, c\}$ that can be multiplied according to the following rules. Let $x, y, z$ be an ordering of $a, b, c$. Then, $e · x = x · e = x, x^2 = e$, and $x · y = z$.
But I have no idea how to create a mapping from checker board to Klein group for it...
 A: That hint is most of the way to a solution. Here's a sketch, fill in the details.   


*

*Using only the non-identity elements of the Klein 4 group, label the board in a particular way. This should be obvious once you realize what the (structure-finder) conditions are. See below for more hints.       

*Define the value of a board to be the product of the elements with checkers. Show that any valid move leaves the product invariant.   

*Show that the product of the original board is $e$.   

*Since $e$ isn't represented by a single square, hence, we cannot leave only 1 checker on the board. 

*This implies that we're left with at least 2 tokens.   

Hints to help you figure out the labelling of the board   

 Each move involves 3 diagonally connected squares, which lie in 3 distinct rows and columns.   

$ $

 A sufficient condition is that each of the 3 connected squares in a move represent a different element. (The ordering doesn't matter)

$ $ 

 There are (at least) 2 labellings  that work.

A: Here's a more basic approach to solve the problem that doesn't use the Klein group, just invariants.   
Label the 1st, 4th, 7th columns X, the 2nd, 5th, 8th columns Y, the 3rd, 6th columns Z.   
Every move involves an $\{X, Y, Z\}$, and changes the parity of the number of tokens by 1.   
At the start, the number of tokens are $ \{ 10, 10, 4 \} $.
Hence, up to parity, after each move, the number of tokens toggles between $ \{ 0, 0, 0 \}$ and $ \{ 1, 1, 1 \}$. 
In particular, this implies that $ \{ 0, 0, 1 \}$ (and permutations) cannot be achieved. 
In fact, we must have at least 3 tokens left behind, each lying in distinct columns (and rows).   

