Given two groups $H$ and $K$ such that $K$ acts on a non-empty set $\Delta$, we can compute the wreath product as $G = H \wr{_\Delta} K := Fun(\Delta,H)\rtimes K$. Now, suppose that $H$ acts on the set $\Gamma$. Following Dixon and Mortimer, and Primitive Wreath Product action, the product action of $G$ on $Fun(\Delta,\Gamma$) is defined as: $\forall (f,k) \in G = Fun(\Delta, H) \rtimes K$, $\forall \phi \in Fun(\Delta,\Gamma)$ $$\phi(f,k)(\delta):=f(k^{−1}\delta ) \cdot \phi(k^{−1} \delta)$$
I am having a hard time showing this associative. This is my attempt: $$ \begin{align} (f,k_1)((g,k_2) \cdot \phi(\delta)) &&= (f,k_1)(g(k_2^{-1}\delta) \cdot \phi(k_2^{-1}\delta) \end{align} $$
On the other hand, by multiplication in the wreath product $(f,k_1)\cdot(g, k_2) = (f(k^{-1}g, k_1k_2)$, where the action of $k^{-1}g$ is the action of K the function g such that kg = $g(k^{-1}\delta)$ So, $$ \begin{align} (fg, k_1k_2) \cdot \phi(\delta) &= fg(k_2k_1^{-1}k_1 \delta) \cdot \phi(k_2^{-1}k_1^{-1}\delta) \\ &= fg(k_2\delta)\phi(k_2^{-1}k_1^{-1}\delta) \end{align} $$
Clearly, I have gone wrong somewhere, but I don't know where. Either the action or group multiplication is wrong.