Justifying definitions of a group action.

Question

Definition: Let $G$ be a group and $A$ a set. We say that $G$ acts on $A$ if there is a map $G\times A \rightarrow A$ denoted by $g \ast a$ satisfying $1 \ast a = a$ for all $a \in A$ and $g_1 \ast (g_2 \ast a) =(g_1g_2)\ast a$ for all $g_1,g_2 \in G$ and $a \in A$.

I've also seen that you can define a group action of $G$ on $A$ to be any homomorphism from $G \rightarrow S_{A}$; i.e a homomorphism from $G$ into the set of bijections of $A$. We also know that for each $g$ in $G$ we can specify $\sigma(g):A \rightarrow A$ by $a \mapsto g \ast a$, I wanted to understand why both definitions are the same, and I believe it follows from the following claim.

Claim: Let $G$ act on $A$. The map $\varphi:G \rightarrow S_A$ defined by $g \mapsto \sigma_g$ is a homomorphism. Conversely, given any homomorphism $\phi:G \rightarrow S_A$, the map $G \times A \rightarrow A$ by $g \ast a = \phi(g)(a)$ is a group action.

The proof is straightforward,

Want to show that $\varphi:G \rightarrow S_A$; where recall $S_A$ is the set of all bijections of $A$, defined by $g \mapsto \sigma_g$ is a homomorphism. Let $g_1,g_2 \in G$, then for any $a \in A$, \begin{align*} \varphi(g_1g_2)(a) & = \sigma_{g_1g_2}(a) \\ & = (g_1g_2) \ast a \\ & = g_1 \ast (g_2 \ast a) \\ & = g_1 \ast (\sigma_{g_2}(a)) \\ & = \sigma_{g_1}(\sigma_{g_2}(a)) \\ & = (\sigma_{g_1} \circ \sigma_{g_2})(a) \\ & = (\varphi(g_1) \circ \varphi(g_2))(a). \end{align*} Therefore the map $\varphi:G \rightarrow S_A$ by $g \mapsto \sigma_g$ is a homomorphism. Next let $\phi:G \rightarrow S_A$ be a homomorphism, we want to show that $g \ast a = \phi(g)(a)$ is an action of $G$ on $A$. Because $\phi$ is a homomorphism we know that $\phi(1_G) = 1_{S_A}$ and so $1 \ast a = 1(a) = a$, next let $g_1,g_2 \in G$ and \begin{align*} (g_1g_2)\ast a &= \varphi(g_1g_2)(a)\\ &=(\varphi(g_1) \circ \varphi(g_2))(a) \\ & = \varphi(g_1)(\varphi(g_2)(a)) \\ & = g_1 \ast (\varphi(g_2)(a)) \\ & = g_1 \ast(g_2 \ast a). \end{align*} So we conclude that any homomorphism from $G \rightarrow S_A$ gives rise to a valid action of $G$ on $A$ by $g \ast a = \varphi(g) (a)$.

So what is this really saying? I understand that if we have a homomorphism from $G$ into the symmetric group on $A$ we have a valid action, but don't we also need that any action is a homomorphism? I don't see why the map $\varphi:G \rightarrow S_A$ by $\varphi(g) = \sigma_g$ is the group action. I suppose my question is why is this claim enough to conclude that both definitions are equivalent. Thanks in advance for the clarification.

Answer 1

I think that the problem is simply a flaw in the formulation of your definition and of your claim.

The phrase $G$ acts on $A$ is not well-defined; it should only be used in a very clear context where the action itself is already given, either explicitly or implicitly.

What do I mean by the action itself? I mean that map $G \times A \to A$ in your definition. Let me reformulate your definition:

Definition Let $G$ be a group and $A$ a set. An action of $G$ on $A$ is a map $G \times A \to A$ denoted by $g*a$ satisfying... [now copy the rest of the definition as stated].

Your claim is similarly flawed. I would suggest breaking it into two separate parts.

Claim Let $G$ be a group and $A$ a set.

1. For any action $G \times A \to A$ denoted by $g*a$ the map $\varphi : G \to S_A$ defined by $g \mapsto \sigma_g$, where $\sigma_g(a)=g*a$, is a group homomorphism.
2. For any group homomorphism $\varphi : G \to S_A$ the map $G \times A \to A$, defined by $(g,a) \mapsto \varphi(g)(a)$, is an action of $G$ on $A$.

And with that done, your proof can be similarly fixed, and you might even allow yourself to use that tricky phrase $G$ acts on $A$ if the context is clear.

Answer 2

What is going on is this:

The group action permutes the elements of $A$

The symmetric group on $A$ comprises all the permutations on A.

We can identify the elements of the symmetric group which correspond to the elements of $G$ - we just identify the permutation corresponding to $g$.

The key step is that the identification respects the group structure: but this is obvious - in each case we are composing permutations of the same set in the same way. (We have to be a little careful, because two different elements of $G$ can lead to the same permutation of $A$).

The proof specifies the "identification" as the map $\varphi$.

Then, if we have a homomorphism from $G\to S_A$ we can associate each element of $G$ with a permutation of $A$ (that is what the homomorphism does does) - and we use the fact that the map is a homomorphism to show that these permutations give rise to an Action on A (ie the satisfy the conditions for an Action).

Group Actions are really important and worth understanding, and the relationship with the Symmetric Group is present every time an Action is involved - again this is significant.