# Primitive Wreath Product action

I am a little confused about the primitive action of the wreath product as when to use the inverse and whether to use left or right action.

Let $H, K$ be groups and $K$ acts on $\Delta$, the wreath product of $H$ and $K$ is $H wr _{\Delta} K:= H^{\Delta} \rtimes K$.

There are two ways to interpret $H^{\Delta}$ as $\{(h_{\delta})_{\delta \in \Delta} \}$ or $Fun(\Delta ,H)$.

The action of $K$ on $H^{\Delta}$ is to permute the coordinates, so (first question:) could I use left action here and without inverse: $k \cdot (h_{\delta})_{\delta \in \Delta}=(h_{k\delta})_{\delta \in \Delta}$ or $(k \cdot f )(\delta)=f(k\delta)$, $f \in Fun(\Delta ,H)$.

Then we have $H$ acts on $\Gamma$, we can define the action of $H wr _{\Delta} K$ on $\Gamma ^{\Delta}$. Again (second question:) could I use left action here: $(((h_{\delta})_{\delta \in \Delta} ,k ) \cdot \phi )(\delta ') =h_{k^{-1}\delta '} \phi (k^{-1}\delta ')$. ( I think inverse here are necessary as otherwise it will fail the associativity of the action.)

If the above the definitions are correct, when I check the association of the action, I have:

(i) $(((g_{\delta})_{\delta \in \Delta} ,l)((h_{\delta})_{\delta \in \Delta} ,k ) \cdot \phi )(\delta )= ((g_{\delta})_{\delta \in \Delta} ,l) \cdot h_{k^{-1}\delta } \phi (k^{-1}\delta )=g_{k^{-1} l^{-1}\delta} h_{k^{-1}\delta}\phi( k^{-1} l^{-1} \delta)$

(ii) $(((g_{\delta} h_{l\delta})_{\delta \in \Delta}, lk ) \cdot \phi )(\delta ) = g_{k^{-1} l^{-1}\delta} h_{k^{-1}\delta}\phi( k^{-1} l^{-1} \delta)$

3rd: So the action of $k$ is to precompose $\delta$ with $k^{-1}$, but exactly is the action of $h$, how do we pick the subscripe $\delta$? i.e. I thought whenever $h$ acts on $\phi$, for each $\delta$, we get the corresponding $h_{\delta}$ and let it act on $\phi(\delta)$, then in (i) above, when $((g_{\delta})_{\delta \in \Delta} ,l)$ acts on $\phi (k^{-1}\delta )$, do we not have to regard $\phi (k^{-1}\delta )$ as a new map $\phi '(\delta )=\phi (k^{-1}\delta )$ and then it will be $g_{ l^{-1}\delta} \phi'( l^{-1} \delta)=g_{l^{-1}\delta} \phi( k^{-1} l^{-1} \delta)$ instead of $g_{k^{-1} l^{-1}\delta} \phi( k^{-1} l^{-1} \delta)$?

4th question: I think it would also work if we take right action and have inverse in both actions. Is there a standard way to specfify the action.

Sorry about such a long question. I would really appreciate if someone could clarify my confusion!

• You can use left or right actions according to your personal preference, but it is of course advisable to be consistent. So you should first decide whether you are going to use left or right actions and then try and figure out the details. Commented Jun 11, 2016 at 13:58

Suppose that $H$ and $K$ be groups and $K$ acts on a non-empty set $\Delta$. The action of $K$ on $Fun(\Delta ,H)$ can be defined as follows $$\forall k\in K \ \forall f\in Fun(\Delta ,H) \ f^k(\delta):=f(\delta^{k^{-1}}).$$ Also, the action of $K$ on $H^{\Delta}$, where $H^{\Delta}$ is $\{(h_{\delta})_{\delta \in \Delta} \}$, can be defined as $$\forall k\in K \ \forall (h_{\delta})_{\delta \in \Delta}\in H^{\Delta} \ ((h_{\delta})_{\delta \in \Delta})^k:=(h_{\delta^k})_{\delta \in \Delta}.$$
The wreath product of $H$ and $K$, denoted by $G$, is $H \ wr _{\Delta} K:= Fun(\Delta ,H)\rtimes K$ or $H \ wr _{\Delta} K:=H^{\Delta} \rtimes K$.
Now, suppose that $H$ acts on the set $\Gamma$. The product action of $G$ on $Fun(\Delta ,\Gamma)$ (or equivalently on $\Gamma^\Delta$) can be defined as follows $$\forall (f,k)\in G=Fun(\Delta ,H)\rtimes K \quad \forall\phi\in Fun(\Delta ,\Gamma) \quad \phi^{(f,k)}(\delta):=\phi(\delta^{k^{-1}})^{f(\delta^{k^{-1}})}$$ or equivalently as $$\forall ((h_{\delta})_{\delta \in \Delta} ,k )\in G=H^{\Delta} \rtimes K \quad \forall (\gamma_{\delta})_{\delta \in \Delta}\in \Gamma^\Delta \quad ((\gamma_{\delta})_{\delta \in \Delta})^{((h_{\delta})_{\delta \in \Delta} ,k )}:=((\gamma_{\delta^k})^{h_{\delta^k}})_{\delta \in \Delta}.$$ It is easy to see that the above actions are associative.
In your questions, the definition $k \cdot (h_{\delta})_{\delta \in \Delta}:=(h_{k\delta})_{\delta \in \Delta}$ is true, but you should define $(k \cdot f )(\delta):=f(k^{-1}\delta)$ because of the associativity of the action.
• Thanks! But I am not sure why written as a function, we have $f(\delta^{k^{-1}})$, but as a tuple, we don't have the inverse? Commented Jun 15, 2016 at 7:57
• @JeremyH: This is because of the associativity of the actions. For example with the inverse, $f^{kk'}(\delta)=f(\delta^{k'^{-1}k^{-1}})=f((\delta^{k'^{-1}})^{k^{-1}})=f^ k(\delta^{k'^{-1}})=(f^k)^{k'}(\delta)$, but without the inverse we have $f^{kk'}(\delta)=f(\delta^{kk'})=f((\delta^{k})^{k'})=f^ {k'}(\delta^{k})=(f^{k'})^{k}(\delta)$. Similarly, in writing as tuples the inverse is not needed. Commented Jun 16, 2016 at 11:31