Proof

Throughout I shall refer to the numbering of counters as they appear on the puzzle's board. Let's introduce the three clockwise moves

• a = (1 2 6 9 8 4)
• b = (2 3 7 10 9 5)
• c = (5 6 10 12 11 8)

Let S be the group generated by a,b,c. To prove the solvability of the puzzle, I shall show that S = S₁₂. The proof proceeds in four steps.

1. S is 4-transitive.

   I'll show that there exists a combination of moves a, b, c that takes any four distinct counters into positions 1,2,4, and 8. Moving backwards, this would imply that any four counters can be moved into any four positions. Thus, let there be four distinct counters, w, x, y, and z.

   Move w into position 4. If x is in position 1, proceed to place y into 2. Otherwise, make sure x has been moved into one of the positions outside a (3, 7, 10, 11, 12 - to choose from). Apply a once so that w will be in position 1, move x into position 2 using only moves b and c. Reverse a to have w in 4 and x in 1.

   If y is already in 2, proceed placing z into 8. Otherwise, if y is in b, move it to 2. Else, first use c to move y into 5 or 10 then apply b as needed.

   If z is already in 8, or belongs to c, the first step is done. Else, it's in one of the positions 3,7,5 or 9. In any case, we have to use b to place z into c without putting y there too. Once it's done, use c once to remove z from b, reverse b to place y back into 2, and finally use c to slide z into 8.

2. d = a^-1b^-1c^-1abc = (1 7 11 4 3 12 8)(6)(2 9)(5 10).

   This is shown by direct verification. What follows is that w = d⁷ = (2 9)(5 10).

3. A₁₂S.

   Indeed from 1. we can get any two non-intersecting transpositions (k l)(m n) starting with w and using Lemma 1. Then the assertion follows from Lemma 2.

4. Finally, since any of a, b, c are odd, none of them belongs to A₁₂. Therefore, the group generated by, say, a and A₁₂ exceeds A₁₂ but is a subgroup of S. The only possibility is that S = S₁₂.