I get step 1, that's easy enough. Step 2 I understand, aside from the fact that e=d^7. Why? As far as I can see (2 9)(5 10) is not the identity at all. And why do I need step 2 at all?

Step 3: I understand that lemma 1 tells us that we can transform any cycle (and therefore any transposition) into any other of equal length by simply conjugating with a suitable permutation g in S_n. Step 3 wants to conclude that A_12 is a subset of S because A_12 is generated by pairs of disjoint transpositions. Maybe I'm being stupid, but I don't see how we can use Lemma 1 (which is about cycles) on pairs of transpositions.

If we can, then we know that the permutation g that we want always is in S (the group generated by the three legal moves of the game) because step one shows us that we can always move any four counters into any four positions.

Thus, A_12 is a subset of S, and the rest is easy.

Please help me! I feel pretty close to closing all the gaps, but it's frustrating me that I don't see the significance of step 2 ( (2 9)(5 10) is a disjoint pair of transpositions, which obviously has something to do with things, but... what?)

The action of d can be described with 4 cycles:

(1 7 11 4 3 12 8)
(6)
(2 9)
(5 10).

Non-intersecting cycles commute, and we can consider them individually.

The first is a 7-cycle. It is identical after 7 applications:

(1 7 11 4 3 12 8)⁷ = Id.

The third and the fourth are 2-cycles, so that

(2 9)² = (5 10)² = Id.

Therefore

(2 9)⁷ = (2 9) and
(5 10)⁷ = (5 10).

> As far as I can see (2 9)(5 10)
> is not the identity at all.

No, of course it is not an identity.

> And why do I need step 2 at all?

I needed a sequence of moves that affects exactly four counters.

>Step 3: I understand that lemma 1 tells us that we can
>transform any cycle (and therefore any transposition) into
>any other of equal length by simply conjugating with a
>suitable permutation g in S_n. Step 3 wants to conclude that
>A_12 is a subset of S because A_12 is generated by pairs of
>disjoint transpositions. Maybe I'm being stupid, but I don't
>see how we can use Lemma 1 (which is about cycles) on pairs
>of transpositions.

For, say, two non-intersecting cycles c₁ and c₂,

g^-1c₁c₂g = (g^-1c₁g)(g^-1c₂g)

so it is possible to apply Lemma 1 to each of the cycles separately. In particular, it works for a pair of non-intersecting transpositions.

>If we can, then we know that the permutation g that we want
>always is in S (the group generated by the three legal moves
>of the game) because step one shows us that we can always
>move any four counters into any four positions.
>
>Thus, A_12 is a subset of S, and the rest is easy.

Right.

>Please help me!

Is it any clearer now?

(I am glad you asked.)