Let $A$ and $B$ be groups, and let $B$ act on the set $\Gamma$ Define the action of $B$ on the direct product $A^{\Gamma}$ by $$ b f(\gamma) := f(b^{-1}\gamma), $$ for any $f\in A^{\Gamma}$ and $\gamma\in\Gamma$ The wreath product of $A$ and $B$ according to the action of $B$ on $\Gamma$ denoted $A\wr_{\Gamma} B$ is the semidirect product of groups $A^{\Gamma}\rtimes B$

Let us pause to unwind this definition. The elements of $A\wr_{\Gamma}B$ are ordered pairs $(f,b)$ where $f\in A^{\Gamma}$ and $b\in B$ The group operation is given by $$(f,b)(f',b') = (fbf', bb').$$ Note that by definition of the action of $B$ on $A^{\Gamma}$ $$(fbf')(\gamma) = f(\gamma)f'(b^{-1}\gamma).$$

The action of $B$ on $\Gamma$ in the semidirect product permutes the elements of a tuple $f\in A^{\Gamma}$ and the group operation defined on $A^{\Gamma}$ gives pointwise multiplication. To be explicit, suppose $\Gamma$ is an $n$ tuple, and let $(f,b),~(f',b')\in A\wr_{\Gamma} B$ Let $b_i$ denote $b^{-1}(i)$ Then \begin{eqnarray*} (f,b)(f',b') &=& \bigl((f(1),~f(2),~\ldots,~f(n)),~b\bigr) \bigl((f'(1),~f'(2),~\ldots,~f'(n)),~b'\bigr) \\ &=& \bigl((f(1),~f(2),~\ldots,~f(n)\bigr)\bigl(f'(b_1),~f'(b_2),~\ldots,~f'(b_n)\bigr),~bb') \text{(*)} \\ &=& \bigl((f(1)f'(b_1),~f(2)f'(b_2),~\ldots,~f(n)f'(b_n)),~bb'\bigr). \end{eqnarray*}Notice the permutation of the indices in (*).

A moment's thought to understand this slightly messy notation will be illuminating, and might also shed some light on the choice of terminology.