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?

Could anyone help me answer those questions please?

  • $\begingroup$ Yes $X$ is any set. Aut(X) is the set of all isomorphisms from $X$ to itself. @S $\endgroup$
    – user778657
    Oct 20, 2020 at 22:38
  • $\begingroup$ Then what is the difference between $\operatorname{Aut}(X)$ and $\operatorname{Sym}(X)$ for a set $X$? $\endgroup$ Oct 20, 2020 at 22:39
  • 1
    $\begingroup$ That does not make it any clearer to me. Could you provide clear definitions for both? $\endgroup$ Oct 20, 2020 at 22:40
  • 2
    $\begingroup$ 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." $\endgroup$
    – runway44
    Oct 21, 2020 at 5:33
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    $\begingroup$ 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 $x$. $\endgroup$
    – runway44
    Oct 21, 2020 at 5:38

1 Answer 1


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).

  • $\begingroup$ in your third line, do you mean ${}^fx$? $\endgroup$
    – user778657
    Oct 20, 2020 at 23:19
  • $\begingroup$ @Math Yes, fixed. $\endgroup$ Oct 21, 2020 at 7:58
  • $\begingroup$ for your first question answer, I do not know, could you help me please? $\endgroup$
    – user778657
    Nov 3, 2020 at 19:52

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