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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 $A=Z$

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  • $\begingroup$ To make this a meaningful question, you need to specify the conditions you would like to impose on the pairs of actions of which you would like to have non-trivial examples. $\endgroup$ – Rasmus Nov 2 '13 at 8:48
  • $\begingroup$ The conditions could vary depending on the set you choose. Right now, I do not have a list of general conditions. In my first example of the permutation groups acting on A, the resulting action could just be a composition of actions. This might not be so in other cases. This is why I left it at "depending on suitable actions." $\endgroup$ – wannadeleteacct Nov 2 '13 at 9:02
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    $\begingroup$ What exactly is the map $G'\times(G\times A)\to G'\times A\to A$ and why are you calling it a group action? If it's $(u,v,x)\mapsto(u,vx)\mapsto uvx$, then it's just a map, not a group action, which then leaves the question of what your question is. $\endgroup$ – anon Nov 2 '13 at 9:39
  • $\begingroup$ @anon:Thanks a lot. $\endgroup$ – wannadeleteacct Nov 2 '13 at 9:49
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    $\begingroup$ Technically $G'\times(G\times A)\to A$ is not of the form $H\times X\to X$, because $G\times A\ne A$. Instead, you want to talk about $(G'\times G)\times A\to A$, which is (putatively) a group action of $G'\times G$ on $A$. Even in this case we would not use the phrase "group action of an action" or such. $\endgroup$ – anon Nov 2 '13 at 10:02
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"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 $a_1$.

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  • $\begingroup$ What is $Aut(X)$? Do you mean $Hom(G \times A ,A)$ and $Hom(G' \times A,A)$? $\endgroup$ – wannadeleteacct Nov 2 '13 at 16:14
  • $\begingroup$ How have you defined $[\rho'(G'),\rho(G)]$? $\endgroup$ – wannadeleteacct Nov 2 '13 at 16:22
  • $\begingroup$ @Manasi: ${\rm Aut}(X)$ is the set of isomorphisms $X\to X$. You should know there are actually two definitions of a group action. One is a map $G\times A\to A$ subject to certain conditions (this definition is useful because we can define topological actions with it), and the other is a map $G\to{\rm Aut}(X)$ where ${\rm Aut}(X)$ is the symmetry group of some object (e.g. the permutations of a set) - this generalizes to group actions. If $H$ and $K$ are two subgroups of $S$, then $[H,K]$ is the group generated by all commutators $[h,k]$ (it measures how much they commute with each other). $\endgroup$ – anon Nov 2 '13 at 17:29
  • $\begingroup$ Well, I know what they mean however, I am interested to know whether $G \to Aut(X)$ is meaningfully defined. What is X? $\endgroup$ – wannadeleteacct Nov 2 '13 at 17:32
  • $\begingroup$ Oh, I suppose we are using the letter $A$ aren't we. $\endgroup$ – anon Nov 2 '13 at 17:32
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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.

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  • $\begingroup$ I defined it above. (I did not define it as action of a group on a group). $\endgroup$ – wannadeleteacct Nov 2 '13 at 9:04
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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\}\}$.

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