wreath product

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

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

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

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

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

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

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 $(f,b),~{}(f^{\prime},b^{\prime})\in A\wr_{\Gamma}B$. Let $b_{i}$ denote $b^{-1}(i)$. Then

 $\displaystyle(f,b)(f^{\prime},b^{\prime})$ $\displaystyle=$ $\displaystyle\bigl{(}(f(1),~{}f(2),~{}\ldots,~{}f(n)),~{}b\bigr{)}\bigl{(}(f^{% \prime}(1),~{}f^{\prime}(2),~{}\ldots,~{}f^{\prime}(n)),~{}b^{\prime}\bigr{)}$ $\displaystyle=$ $\displaystyle\bigl{(}(f(1),~{}f(2),~{}\ldots,~{}f(n)\bigr{)}\bigl{(}f^{\prime}% (b_{1}),~{}f^{\prime}(b_{2}),~{}\ldots,~{}f^{\prime}(b_{n})\bigr{)},~{}bb^{% \prime})\text{(*)}$ $\displaystyle=$ $\displaystyle\bigl{(}(f(1)f^{\prime}(b_{1}),~{}f(2)f^{\prime}(b_{2}),~{}\ldots% ,~{}f(n)f^{\prime}(b_{n})),~{}bb^{\prime}\bigr{)}.$

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.

Title wreath product WreathProduct 2014-12-30 11:23:20 2014-12-30 11:23:20 mps (409) juanman (12619) 21 mps (12619) Definition msc 20E22 wreath product