A more formal but intuitive understanding (on a definition) of a group action

Question

We know that a symmetric group $S_n$ acts on the set $\{1, 2,\ldots, n\}$.

The definition of an action of a group $G$ on a set $S$ is a function $G\times S\to S$ such that:
1) $e\ast s=s$
2) $g'\ast(g\ast s)=(g'g)\ast s$

My questions are:

1. The operation $\ast$ in the context of the action of symmetric group on the set of $\{1, 2,\ldots, n\}$ is the normal operation of a function to the input. For example the transposition $(12)$ acts on the set $\{1, 2, 3\}$ and result in $\{2, 1, 3\}$.
   But when we have the second point in the definition $g'\ast g$ means a composition of function $g$ with $g'$.
   The set $G$ is a group of functions but the set $S$ is a set of elements, certainly $g\ast g'$ means differently to $g\ast s$.
   Am I correct?

2. How can we make sense of the definition?

3. In the Algebra book by Artin, it is mentioned that a group operation (or action) is more general than an automorphism, how can we understand this? How can we relate a group action to a group automorphism?

Thanks for the explanation and helps!

Answer 1

For your first question if we write the axiom like so $$g' \ast (g \ast s) = (g'\bullet g) \ast s$$ then $\ast$ is the action of $G$ on $S$, so the left operand to $\ast$ is an element of $G$, the right operand of $\ast$ is an element of $S$, and the result of the operation $\ast$ is an element of $S$. On the other hand $\bullet$ is the group operation, it's multiplication in the group $G$. So both the left and right operands as well as the result of the operation $\bullet$ are elements of the group $G$.

Basically this axiom just says that if you write down something like $g'gs$ then that expression is not ambiguous because it doesn't matter if you "multiply" $g$ and $s$ first or if you multiply $g'$ and $g$ first.

For your second question I'm afraid you'll have to be more specific, I'm not sure what you're asking.

Finally, for the third question, I haven't read Artin so I'm not sure exactly what he means, but here is one way in which actions of groups on a set are more general than just automorphisms of that set. If $\ast$ is an action $G \times S \to S$ then for any element $g \in G$ the map $S \to S$ defined by $s \mapsto g\ast s$ is an automorphism of $S$. Call this automorphism $\phi_g$, so $\phi_g(s) = g\ast s$. Now we can define a homomorphism $G \to \mathrm{Aut}(S)$ by $g \mapsto \phi_g$. So every action $G \times S \to S$ gives a homomorphism $G \to \mathrm{Aut}(S)$. Conversely, if $\Phi\colon G \to \mathrm{Aut}(S)$ is a homomorphism then we can define an action $G \times S \to S$ by $(g, s) \mapsto \Phi(g)(s)$. So actions of $G$ on $S$ are equivalent to homomorphisms $G \to \mathrm{Aut}(S)$. Automorphisms of $S$ always act on $S$ and this action $\mathrm{Aut}(S) \times S \to S$ corresponds to the identity map $\mathrm{Aut}(S) \to \mathrm{Aut}(S)$ and this is why actions are more general, it allows you to work with more groups than just automorphism groups and more homomorphisms than just identity maps.

Answer 2

1. If you look closely to 2), it is never written $g'*g$, it is written $(g'g)*s$ or $g'*(g*s)$, that is you always write "element of the group $ G$" * "element of the set $S$". When you want to multiply two elements of the group you just write $gg'$ or $g.g'$.

2. I am not sure if this what you are looking for but let's say this. Anytime a group $G$ acts on $S$ you can always acts as $g$ was an element of $Sym(S)$ and then the action just becomes the usual action of $Sym(S)$ on $S$ (that is you have a composition law for the group and an action by evaluation). We can translate this rigorously by saying that an action is exactly given by a group morphism $\rho$ :

$$\rho: G\rightarrow Sym(S) $$

$$g\mapsto [\rho(g):s\mapsto g*s]$$

3. I see a way to understand this remark (it might be wrong). In the case $S=H$ is also a group. Then we see that (by what I have already said), this is equivalent to have a group morphism :

$$\rho: G\rightarrow Sym(H) $$

If it happens that the image of $\rho$ is contained in $Aut(H)\subset Sym(H)$ then the action of $G$ on $H$ respects the law group on $H$ turning $H$ into some kind of non-commutative $G$-module. In this case we say that $G$ acts by automorphisms on $H$, in general it might happen that the action is not by automorphisms even if $H$ is not a group.

Answer 3

Consider the action of $G$ on $S$ to be a function $ f: G \times S \to S$. For each $g \in G$ we define a function $g* :S \to S$ , by $g*(s)=f(g,s)$ , but we write $g*s$ for $g*(s)$.The conditions on $f$ are that $ s=e*s=f(e,s)$ and that $g'* (g*s)= f(g',f(g,s))=f(g'g,s)=(g'g)*s$. Logically we should say that this means $g'*(g*)=(g'g)*$ but we sometimes write $g'*g=(g'g)*$......... An automorphism of a group $G$ is a bijection $ A:g \to G$ with $A(g'g)=A(g')A(g)$ for all $g',g \in G$. We may define an action $f_A : (G \times S) \to S$ by choosing $S=G$ and $f_A(g,s)=f(g)s$.(Example.Let G=Q (the rationals) with addition being the group operation. For $S=R$ (the reals) let $ f^{(1)} (q,r)=q+r$ and $f^{(2)} (q,r)= 2 q+r$. Theses are actions of $Q$ on $ R$. Let $ A(q)=2 q$ for $ q \in Q$.Then $ A $ is an automorphism and $f_A(q,s)=2 q+s$ for $q,s \in Q$.