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.