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Let $G$ a group, $X$ a set. An action of $G$ on $X$ is given by a mapping $(g,x) \to g \cdot x$, which satisfies $g \cdot (h \cdot x) = (gh) \cdot x$ and $e_G \cdot x = x$. In other words, to determine an action one must know how every element of the group acts on every element of the set.

Can an action be determined by means other then giving explicitly the action of every element of $G$ on every element of $X$?

What if I have several equations (depending on elements of X and G, and the action of G) that give some information of the action - when are the equations strong enough so they induce an action on the entire set? Furthermore, are there cases where I can fully recover the action?

For example, suppose that $X$ is a (given) set of elements of a field extension $K/k$, and $G \subseteq Gal(K / k)$. We would like to extend that action of $G$ to $K(X)$, such that the extended action satisfies several equations.

Is there any theory in mathematics that asks and answers this sort of questions?

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  • $\begingroup$ You only need to specify the action for a set of generators of $G$, because that enables you to calculate it for all elements of $G$. $\endgroup$
    – Derek Holt
    Mar 24, 2015 at 14:29
  • $\begingroup$ Remember that giving an action of $G$ on $X$ is equivalent to give a group morphism from $G$ to $Bij(X)$ (the set of bijections of $X$). Now, a morphism is only defined by the images of the generators of $G$ (by the way if you have a set of relations for a particular set of generators of $G$ then you have conditions to know if some particular images of the generators of $G$ define an action or not). $\endgroup$ May 15, 2016 at 15:07

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Let $F_n$ denote the free group of rank $n$. You can characterise the action of $\operatorname{Aut}(F_2)$ on $F_2$ by an equation as follows:

Theorem (Theorem 3.9, Magnus, Karrass and Solitar, Combinatorial group theory, 1975 (2nd Ed.))

A map $f: a\mapsto x$, $b\mapsto y$, where $x, y\in F_2$, is an automorphism of $F_2$ if and only if there exists an element $z\in F_2$ such that $$z^{-1}[x, y]z=[a, b]~\text{or}~z^{-1}[x, y]z=[a, b]^{-1}.$$

Hence, automorphisms somehow correspond to solutions to the equations $z^{-1}[x, y]z=[a, b]^{\pm1}$. On the other hand, automorphisms of $F_n$ for $n>2$ cannot be described in this way. In fact, I understand that they cannot be characterised precisely using first-order sentences (although I cannot find a reference for this).

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