# Justifying definitions of a group action.

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.

• I'm not sure I understand your question. In the proof it shows that a homomorphism gives rise to an action, and an action gives rise to a homomorphism. Is your question asking about these translations being inverse to each other?
– Dave
Oct 11, 2022 at 15:33
• Yes @Dave. It seems somehow circular in my mind; i.e a group action can be defined as a homomorphism from $G \rightarrow S_A$ because any action gives rise to such a homomorphism, and any homomorphism is a group action? In this way do we mean that the action is the map $\varphi:G \rightarrow S_A$ and also $g\ast a$? Oct 11, 2022 at 15:57
• A homomorphism $\varphi:G\to S_A$ defines an action on $A$ via $g*a:=\varphi(g)(a)$ since $\varphi(g)$ is a bijection on $A$; the fact that $\varphi$ is a homomorphism makes this satisfy the axioms of an action. Conversely, an action $G\times A\to A$ defines a homomorphism $\varphi:G\to S_A$ via $\varphi(g)$ being the map $A\to A$ sending $a\mapsto g*a$; the fact that $g*a$ is an action makes this map a homomorphism.
– Dave
Oct 11, 2022 at 16:31
• These translations are inverse to each other because, starting with a homomorphism $\varphi:G\to S_A$, we get an action $g*a:=\varphi(g)(a)$, and then if we convert this back to a homomorphism using our recipe, the homomorphism would send $g\mapsto (a\mapsto g*a)=(a\mapsto \varphi(g)(a)$, which is just the map $\varphi$. You can check that the other composition also gives the identity.
– Dave
Oct 11, 2022 at 16:31
• Thanks @Dave, that's very clear! Oct 11, 2022 at 17:30

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.

• So just to confirm, what the second definition i.e an action is a hom. from $G$ into $S_A$ is saying is that we can interchange $g \ast a$ and $\varphi(g)$? Sorry if I'm not seeing something evident, the ideas just seem circular to me i.e a group action can be defined as a homomorphism from $G \rightarrow S_A$ because any action gives rise to such a homomorphism, and any homomorphism is a group action? And in that way the group action is both $g \ast a$ and $\varphi(g): G \rightarrow S_A$. Oct 11, 2022 at 15:59
• You cannot just exchange $g*a$ and $\phi(g)$, because they are not objects of the same type: $g*a$ is an element of $A$; whereas $\phi(g)$ is a function from $A$ to itself, in particular a permutation of $A$. But what you can do is exchange $g*a$ and $\phi(g)(a)$, which is exactly what you wrote in your post. Oct 11, 2022 at 16:09
• Also, when you write about group actions, you have to decide what you adopt as the definition of a action. In your post, you adopted the $g*a$ concept as the definition of a group action, which is fine. Oct 11, 2022 at 16:11
• Also, after you have completed the proof of equivalence (not before) you are justified in using circular language. That's what proofs of equivalence let you do!! Oct 11, 2022 at 16:11
• Ah! Thanks so much, I don't know why but that last comment hit the nail on the head. Very clear and helpful! Oct 11, 2022 at 16:19

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.

• Note - I have seen Leo Mosher's answer and like the clarity of that. The Group Action is the fundamental thing (ie the group G together with the map $G \times A \to A$, which tells us -in informal language - how the elements of $G$ permute $A$). A group can act on the same set in different ways - for example, G can act in itself by multiplication or by conjugation. Oct 11, 2022 at 15:41