# Understanding what an action is?

This is a very simple question, and I am quite embarrassed to ask it! I'm trying to understand what an action is in general, and perhaps the best place to start is to try and outline my current understanding.

I understand what a group action on a set is, a map $G \times X \rightarrow X$, where $(g,x) \mapsto g\cdot x$. such that $e \cdot x = x, \forall x \in X$, and $(gh)\cdot x = g\cdot(h\cdot x)$

By the properties above each element of $G$ gives us a bijection of elements in $X$, and we have a homomorphism $G \rightarrow \textrm{Sym}(X)$

As I understand it we can also have a group acting on other objects, namely we say a group $G$ acts on an object $A$ if we have a group homomorphism $G \rightarrow \textrm{End}(A)$

E.g. in Galois theory, if we have a Galois extension $E / F$ with Galois group $G$, then $G$ acts on $E$, as we have a map $G \times E \rightarrow E$, where $(g,a) \mapsto g(a)$. By its definition $g$ is an automorphism of $E$, and the group operation is composition of functions so $(g, (h , a)) = (gh ,a)$. From my understanding, this action is not just on the underlying set of the field, we actually mean that the action respects the field structure.

So in general if we want to show that a group $G$ acts on an object $A$ we have to:

(1) associate an element $g \in G$ with a map $\phi_g: A \rightarrow A$

(2) Show that this map is a homomorphism $\phi_g \in \textrm{Hom} (A,A)$, i.e. that it respects all the operations of $A$, and

(3) show that for $g_1, g_2 \in G$ we have $\phi_{g_1g_2} = \phi_{g_1} (\phi_{g_2})$.

Hopefully I am right so far, please correct me if not. Where I am having trouble is twofold.

Firstly, I am worried that my understanding is a little off. When $X$ is a set, the action of an element of $G$ is a bijection of the set, so it seems to me that if an object $A$ has an underlying set $X$ then should we instead expect a homomorphism $G \rightarrow \textrm{Aut}(A)$? In fact, regardless of the underlying set, using an element of $G$ and its inverse then the composition property implies that the homomorphism associated with any element $g \in G$ has an inverse, namely the homomorphism associated to $g^{-1}$.

Secondly, I don't have an understanding or appreciation of the significance of group actions in a wider setting, for example if we have a Galois extension $E / F$ with Galois group $G$, and let $N$ be the fixed field of a normal subgroup $H \trianglelefteq G$, then $G$ acts on $H$ by conjugation. But I am struggling to see the interpretation of this i.e. what meaning the action has in terms of the field extension, or the group.

Thanks in advance for any pointers.

About your first question the answer is yes. Stated more in details if you are considering an object $A$ in any kind of category (it can be the category of sets, abelian groups, topological spaces or fields) an action of the group $G$ correspond exactly to an homomorphism of type $G \to \text{Aut}(A)$ where by $\text{Aut}(A)$ I mean that group of the automorphism for the given structure (they can be bijection in case of $A$ is a set, group automorphisms if $A$ is a group, group of homeomorphisms if $A$ is a topological space).

Just one remark note that the requirement is that the map of the action send every element of the group in an automorphism, which isn't just a morphism bijective: for instance in the case of topological spaces there are cases of bijective continuous function which aren't homemorphism (i.e. isomorphisms of topological spaces).

To your second question. Given a Galois extension $E/F$ with Galois group $G$ subextension $N$ such that $H$ is the corresponding subgroup of $G$ (which is the Galois group of the extension $E/N$) the fact that $H \lhd G$ is a normal subgroup is equivalent to the property of $N$ being a normal subextension of $E$ i.e. that $N/F$ is Galois extension itself. Of course there are many different proof of this fact, they depends on your definition of Galois extension.

• It should be pointed out though that $\operatorname{Aut}(S)$ is a subgroup of $\operatorname{Sym}(S)$, so both interpretations are correct, it is just that one has an "iff" associated to it. – user1729 Oct 25 '13 at 8:44

## protected by user26857Jun 19 '15 at 9:54

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