I'm trying to make sense of the primitive Wreath product action by looking at an example. I took $S_3$ acting on a triangle $\Delta=\{1,2,3\}$, $C_2$ acting on $\Gamma=\{1,2\}$ and constructed the Wreath product $S_3 wr C_2$. I looked at the action of $S_3 wr C_2$ on $\Delta\times \Gamma$ (two copies of the triangle) and found generators $\langle(1\ 2\ 3),(2\ 3),(4\ 5\ 6),(5\ 6),(1\ 4)(2\ 5)(3\ 6) \rangle$ in $S_6$ for this action on the set $\Delta\times \Gamma$ of cardinality 6.
The problem I found is that when I considered the action of $S_3 wr C_2$ on $\Delta^{\Gamma}$ (three copies of the triangle) I found the generators $$\langle (1\ 2\ 3),(2\ 3),(4\ 5\ 6),(5\ 6),(7,8,9),(8,9),(1\ 4\ 7)(2\ 5\ 8)(3\ 6\ 9),(4\ 7)(5\ 8)(6\ 9),(2\ 4)(3\ 7)(6\ 8) \rangle$$ in $S_9$ of the action on the set $\Delta^{\Gamma}$, (this is simply the action of $S_3$ on each coordinate and swapping coordinates with $C_2$).
These generators generate the whole group $S_9$, whereas generators in $S_6$ generate a group of order 72, which is the right order for the Wreath product. What got me even more confused is that $\langle (1\ 2\ 3),(4\ 5\ 6),(7,8,9),(4\ 7)(5\ 8)(6\ 9)\rangle = S_9$, so if I used $C_3$ instead of $S_3$ I apparently get a primitive action, and this should be impossible.
I am clearly not understanding this action correctly, but I don't know where I am mistaken exactly.
It should be
$$\langle (1\ 2\ 3)(4\ 5\ 6)(7\ 8\ 9),(2\ 3)(5\ 6)(8\ 9),(1\ 4\ 7)(2\ 5\ 8)(3\ 6\ 9),(4\ 7)(5\ 8)(6\ 9),(2\ 4)(3\ 7)(6\ 8) \rangle.$$
You got the final three generators correct, but you must have got confused and split the first two generators up into separate cycles.