wreath product

Let $A$ and $B$ be groups, and let $B$ act on the set $\mathrm{\Gamma }$.Define the action of $B$ on the direct product $A^{\mathrm{\Gamma }}$ by

$$bf(\gamma ):=f(b^{-1}\gamma ),$$

for any $f\in A^{\mathrm{\Gamma }}$ and $\gamma \in \mathrm{\Gamma }$.The wreath product of $A$ and $B$ according to the action of $B$ on $\mathrm{\Gamma }$, denoted $A{\wr }_{\mathrm{\Gamma }}B$, is the semidirect product of groups $A^{\mathrm{\Gamma }}⋊B$.

Let us pause to unwind this definition. The elements of $A{\wr }_{\mathrm{\Gamma }}B$ are ordered pairs $(f,b)$, where $f\in A^{\mathrm{\Gamma }}$ and $b\in B$. The group operation is given by

$$(f,b)(f^{\prime },b^{\prime })=(fb{f}^{\prime },b{b}^{\prime }).$$

Note that by definition of the action of $B$ on $A^{\mathrm{\Gamma }}$,

$$(fb{f}^{\prime })(\gamma )=f(\gamma )f^{\prime }(b^{-1}\gamma ).$$

The action of $B$ on $\mathrm{\Gamma }$ in the semidirect product permutes the elements of a tuple $f\in A^{\mathrm{\Gamma }}$, and the group operation defined on $A^{\mathrm{\Gamma }}$ gives pointwise multiplication. To be explicit, suppose $\mathrm{\Gamma }$ is an $n$-tuple, and let $(f,b),(f^{\prime },b^{\prime })\in A{\wr }_{\mathrm{\Gamma }}B$.Let $b_i$ denote $b^{-1}(i)$. Then

$(f,b)(f^{\prime },b^{\prime })$$=$$((f(1),f(2),\mathrm{\dots },f(n)),b)((f^{\prime }(1),f^{\prime }(2),\mathrm{\dots },f^{\prime }(n)),b^{\prime })$

$=$$((f(1),f(2),\mathrm{\dots },f(n))(f^{\prime }(b_1),f^{\prime }(b_2),\mathrm{\dots },f^{\prime }(b_n)),b{b}^{\prime })\text{(*)}$

$=$$((f(1)f^{\prime }(b_1),f(2)f^{\prime }(b_2),\mathrm{\dots },f(n)f^{\prime }(b_n)),b{b}^{\prime }).$

Notice the permutation of the indices in (*).

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