# elaboration of group action definition.

My professor wrote at the beginning of speaking about group actions this:

In general, Aut$$(X) \subset$$ Sym$$(X)$$ acts on $$X$$. If $$G \subset Aut(X)$$ is a subgroup, we say that "G acts on $$X$$ by appropriate automorphism."

Then he gave us a first definition for Group action which is: If $$G$$ a group, $$X$$ a set. a group action by $$G$$ on $$X$$ is a function : $$G \times X \rightarrow X$$ defined by $$(g,x) \mapsto {}^gx$$ such that $${}^g({}^hx) = {}^{(gh)}x$$ for all $$g, h \in G.$$

Then he gave us a second definition which is: any group homomorphism $$G \rightarrow Aut(X).$$

Then he gave us examples as follows:

If $$V$$ is a vector space over $$k$$ of dim. $$n < \infty.$$

1- $$GL_{n}(k) = GL(V)$$ acts on $$V$$ by linear transformations. $$SL_{n}(k)$$ acts on $$V$$ by restriction.

My questions are:

1- I do not understand how $$Aut(X)$$ acts on $$X.$$ what is the implied operation in that case?

2- I do not understand how is the second definition is also a group action definition? what is the implied operation in that case?

3- How can I prove that the example given is really a group action?

• Yes $X$ is any set. Aut(X) is the set of all isomorphisms from $X$ to itself. @S
– user778657
Oct 20, 2020 at 22:38
• Then what is the difference between $\operatorname{Aut}(X)$ and $\operatorname{Sym}(X)$ for a set $X$? Oct 20, 2020 at 22:39
• That does not make it any clearer to me. Could you provide clear definitions for both? Oct 20, 2020 at 22:40
• The author seems to be speaking in a generality that students may not be able to appreciate. Here, ${\rm Aut}(X)$ depends on what kind of thing $X$ is (i.e. what "category" it is an object of), e.g. topological space, poset, ring, etc. The automorphisms preserve the structure, e.g. the topology, relations and operations that are defined on $X$. Thus, ${\rm Aut}(X)$ is in general smaller than the full permutation group on a concrete object $X$. For instance if $X$ is the vertices of a cube, we may say there are $8!$ permutations of the vertices, but only $48$ "automorphisms." Oct 21, 2020 at 5:33
• So (1) the action of ${\rm Aut}(X)$ depends on what kind of thing $X$ is, but generally it's just by functions we plug stuff into. (2) To apply $g\in G$ to $x\in X$, first turn $g$ into an element of ${\rm Aut}(X)$ (according to $G\to{\rm Aut}(X)$) and then apply that to $x$. Oct 21, 2020 at 5:38

For your first question: Going by the first definition, an action of $$\operatorname{Aut}(X)$$ on $$X$$ should be a function $$\operatorname{Aut}(X)\times X\ \longrightarrow\ X:\ (f,x)\ \longmapsto\ {}^fx,$$ satisfying $${}^g({}^hx)={}^{(gh)}x$$. So to every pair $$(f,x)\in\operatorname{Aut}(X)\times X$$, we associate a new element $${}^fx\in X$$. What is the obvious choice of element to associate with the pair $$(f,x)$$? Can you show that this association satisfies $${}^g({}^hx)={}^{(gh)}x$$?

For your second question: Given a group homomorphism $$\varphi:\ G\ \longrightarrow\ \operatorname{Aut}(X),$$ for every $$g\in G$$ its image $$\varphi(g)\in\operatorname{Aut}(X)$$ is an automorphism of $$X$$, so in particular a map from $$X$$ to $$X$$. For notational clarity define $$\varphi_g:=\varphi(g)$$ for all $$g\in G$$, which is an automorphism of $$X$$ for each $$g\in G$$. Then $$\psi:\ G\times X\ \longrightarrow\ X:\ (g,x)\ \longmapsto\ \varphi_g(x),$$ defines a group action of $$G$$ on $$X$$. You should verify this from the definitions of group homomorphism and group action.

Once you have verified and understood these two answers, I think you should be able to answer the third question by yourself; it would at least be a good exercise to try it (again).

• in your third line, do you mean ${}^fx$?
– user778657
Oct 20, 2020 at 23:19
• @Math Yes, fixed. Oct 21, 2020 at 7:58