Let $\cdot$ be an action of a group $G$ on a set $X$. For $g\in G$, define the map $\phi(g):X\to X$ by $\phi(g)(x)=g\cdot x$. I'm asked to show that $\phi(g)\in {\rm Sym}(X)$ and that $\phi :G\to{\rm Sym}(X)$ is a homomorphism.   I've shown that $\phi(g)\in{\rm Sym}(X)$, but I don't understand how to show $\phi$ is a homomorphism? I know it follows from the definition of an action, but what even is the function $\phi: G\to Sym(X)$ explicity?  The solution says that $\phi$ being a homomorphism follows immediately from the fact that $\phi (gh)(x)=g\cdot (h\cdot x)$, but aren't $\phi$ and $\phi (g)$ different functions?
Checking out  and  (and some papers) was a bit difficult for understanding Group Action. The closest to understanding was this part:    I understand the definition of homomorphism (a structure-preserving map between two algebraic structures of the same type). And from what I understand the symmetric group is a group of functions (bijective functions) mapping elements of a set $X$ onto itself.  However I get confused at this line:     ...an action of $G$ on $X$ may be formally defined as a group homomorphism $\phi$ from $G$ to the symmetric group of $X$.   That seems to be the definition of group action, but I am having trouble understanding how to apply that.  What that says to me is it is a homomorphism (function) $f \colon X \rightarrow S$ where $S$ is a set of bijective functions on $X$. That itself seems weird, I thought a group action would result in a number or something related to $\{1, 2, 3\}$ (defined next).  Continuing, if we have a set $X = \{1, 2, 3\}$ and a group $(*,\{4, 5, 6\})$, then the symmetric group of $X$ would be something like $S = \{1 \rightarrow 2, 2 \rightarrow 3, \dots \}$, and $\phi$ would be something like $f : \{4, 5, 6\} \rightarrow S$. That doesn't make sense to me and I feel like I'm interpreting something wrong. I don't see how mapping to a set of functions $S$ will result in an action on the set.
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