# The wreath product $K ~\mbox{wr}_{\Gamma} ~ H$ acts faithful on $\Delta \times \Gamma$ iff $K$ acts faithful on $\Delta$

Let $K$ and $H$ be groups, and let $H$ act on $\Gamma$. Also let $\operatorname{Fun}(\Gamma, K) = \{ f : f : \Gamma \to K \}$ be the set of all functions from $\Gamma$ to $K$. This set is a group under pointwise multiplication $$(fg)(\gamma) := f(\gamma)g(\gamma)$$ and $H$ acts on $\operatorname{Fun}(\Gamma, K)$ by setting for $f : \Gamma \to K$ and $x \in H$ the function $f^x$ to be $$f^x(\gamma) := f(\gamma^{x^{-1}})$$ for all $\gamma\in \Gamma$. Now the semidirect product $$\operatorname{Fun}(\Gamma, K) \rtimes H$$ is called the wreath product of $K$ and $H$ and denoted by $K ~\mbox{wr}_{\Gamma} ~ H$. Its elements are tupels $(f, x)$ of functions and elements from $H$. If $K$ also acts on some set $\Delta$, then the wreath product acts in a natural way on $\Delta \times \Gamma$, by setting for $(\delta, \gamma)$ $$(\delta, \gamma)^{(f,x)} := (\delta^{f(\gamma)}, \gamma^x).$$ (This action is called the imprimitive action of the wreath product on wikipedia).

Now I want to prove, in the case we have an action of $K$ on $\Delta$ and $H$ on $\Gamma$, and therefore an action of the wreath product on $\Delta \times \Gamma$, that:

The wreath product group $G = K ~\mbox{wr}_{\Gamma}~ H$ acts faithfully on $\Delta \times \Gamma$ if and only if $K$ acts faithfully on $\Delta$.

(This is exercise 2.6.6. from the book Permutation Groups by Dixon & Mortimer).

My Attempt: Let $G$ act faithfully on $\Delta \times \Gamma$ and suppose for each $\delta \in \Delta$ we have $\delta^k = \delta$. Denote by $\overline k : \Gamma \to K$ the constant function with $\overline k(\gamma) = k$. The we have for each $(\delta, \gamma)$ that $$(\delta, \gamma)^{(\overline k, 1)} = (\delta^{\overline k(\gamma)}, \gamma^1) = (\delta^k, \gamma) = (\delta, \gamma).$$ Hence as $G$ acts faithful $(\overline k, 1) = (\overline 1, 1)$ as $(\overline 1, 1)$ is the identity in $G$, but this shows that $\overline k = \overline 1$ or $k = 1$.

Now conversely let $K$ act faithfully on $\Delta$ and suppose we have $$(\delta, \gamma)^{(f,x)} = (\delta, \gamma)$$ for each $(\delta, \gamma) \in \Delta \times \Gamma$. By this for each $\delta \in \Delta$ and each $\gamma \in \Gamma$ we have $$\delta^{f(\gamma)} = \delta \quad \mbox{ and } \quad \gamma^{x} = \gamma.$$ As $K$ acts faithful we have $f(\gamma) = 1$ for each $\gamma$, i.e. $f$ is the constant function which equals $1$ (denoted $\overline 1$ above). Now we just need that $x = 1$ and we are done, but I do not see that this is implied by the above (it would follow if also $H$ acts faithful, but this is not part of the exercise, nor is it included in the general definition of the wreath product)?

So is there any way to show that $(f, x) = (\overline 1, 1)$?

Maybe the exercise is wrong and it should read:

The wreath product acts faithful on $\Delta \times \Gamma$ if and only if $K$ and $H$ act both faithful on $\Delta$ and $\Gamma$.

but the errata does not mentions this, so I am not sure?

• You are correct, $H$ must act faithfully on $\Gamma$. Look, if $|\Gamma|=1$, then every element $x\in H$ satisfies $\gamma^x=\gamma$. – David Hill Sep 15 '15 at 17:16
• @DavidHill: Yes, thank you for your anwer. Would you mind making your comment into an answer so I can mark this post as answered? – StefanH Sep 15 '15 at 17:34

## 1 Answer

You definitely need $H$ to act faithfully on $\Gamma$. If $\Gamma=\{\gamma\}$ has one element, then $\gamma^x=\gamma$ for all $x\in H$.