Examples of a map involving group actions

Question

Okay, this is a trivial question but I need some non-trivial examples of a map involving group actions.

What I mean:

Let $G$ be a group acting on a set $A$. Let $G'$ be another group acting on $A$. Then $G'\times (G \times A) \to G' \times A \to A.$

(Is this valid depending on suitable actions?)

If this is valid, can one define action of action of action of.. infinitely. Consider the first example.

Some examples are

1) $ G' = G = S_n $and $A = ${$1,2,...,n : 1,2,...n$ are Natural numbers}.

2) $G'=G=Z$ and $A=Z$

Answer 1

"Then homomorphism." is not even a complete sentence. My best guess is that you're asking if

$$\begin{array}{cccccc} (G'\times G)\times A & \to & G'\times (G\times A) & \to & G'\times A &\to & A: \\ ((u,v),x) & \mapsto & (u,(v,x)) & \mapsto & (u,vx) & \mapsto & uvx \end{array}$$

defines (by the composition $(G'\times G)\times A\to A$) a group action of $G'\times G$ on $A$.

The answer is no. Observe $(a,b)(c,d)=(ac,bd)$ in $G'\times G$. But

$$\begin{cases}(a,b)(c,d)x & = & abcdx \\ (ac,bd) x &= & acbdx\end{cases} $$

and these are not generally equal. The map $(G'\times G)\times A\to A$ is a group action if and only if the actions of $G'$ and $G$ "commute" (that is, when applying an action of both of them, it doesn't matter which order you apply them in). This occurs iff $[\rho'(G'),\rho(G)]=1$ in ${\rm Aut}(A)$, where $\rho',\rho$ are the homomorphisms $G',G\to{\rm Aut}(A)$ respectively. That is, the images of $G'$ and $G$ in ${\rm Aut}(A)$ must commute with each other (everything in one commutes with everything in the other).

In general, two actions $G'\times A\to A$ and $G\times A\to A$ on a set $A$ do yield an induced action of them both on $A$, but our product group needs to be more general. Namely, these actions induce an action of the free product $G'*G$ on $A$. The free product is obtained by taking the free group on all elements of $G'$ and $G$ as letters (i.e. $G'\cup G$, where we take $G'\cap G:=\{e\}$ by fiat) and quotienting by the multiplication tables of $G'$ and $G$ as relations. Thus the elements of $G'*G$ look like words $a_1a_2\cdots a_n$ where the $a_i$ alternate between $G'$ and $G$ (possibly starting at $G$).

The action of $a_1\cdots a_n\in G'*G$ on $x\in A$ is then obvious: simply apply one letter at a time as an action from $G'$ or $G$ starting from $a_n$ and working your way to $a_1$.

Answer 2

It is impossible, generally speaking. Suppose you define "action of action": $G'\times G\to G :(g',g)\to g'*g$. Then $(g'*g)a=g'(ga)$ for $a\in A$. Let $G$ acts trivially on $A$. Then you get $a=g'a$, i.e. $G'$ must act also trivially.

Answer 3

If it is composite of actions then it is valid. It is like you wanted a set $A$ and a set $H$, such the the set $H$ consist infinitely many groups in which no two are isomorphic and each of the group are acting on the set $A$. For non-trivial example you can take $A=\mathbb{N}$ and $H= \{ H_i \mid i\in \mathbb{N} \text{, } H_i < Sym(\mathbb{N}) \text{ and $H_i$ is the pointwise stabilizer of the set }\mathbb{N}\setminus \{1,2,\dots,i\}\}$.