Show that group action is homomorphism to Symmetric group I'm just barely getting my feet wet with abstract algebra, currently working on understanding group action. According to the wikipedia article, a group action $A$ of group $G$ on set $X$ is a group homomorphism from the group $G$ to the symmetric group of $X$ (i.e., the group of all permutations of $X$).
I've been able to prove to myself that for each element $g$ in group $G$ the group action $A(g,x)$ forms a a bijective mapping from $X$ onto $X$ (i.e., $A(g,\cdot)$ specifies a permutation of $X$), and so therefore the group action $h$ is a mapping from $G$ to $\mathrm{Sym}(X)$, but I haven't been able to show that this mapping is a homomorphism.
As far as I understand it, to be a homomorphism requires that $A(g,x) * A(f,x) = A(g+f,x)$ for all $g$ and $f$ in $G$ and all $x$ in $X$, where $*$ denotes the group operation of $\mathrm{Sym}(X)$ and $+$ denotes the group operation of G.
However, the only thing I've been able to do with this is rewrite the right side slightly as $A(g+f,x) = A(g, A(f,x))$.
I'm unclear about what kinds of operations I'm allowed to perform on this equation, for instance can I distribute $A(-g, \cdot)$ through the $*$ on the left side to get $A(-g, A(g,x)) * A(-g, A(f,x))$? Will that even help me?
 A: Although your notation is correct, I really dislike it -- I think it obfuscates this very clear result.  If we just denote the map $G \times X \to X$ by juxtaposition, the group action axioms become
\begin{align*}
1 x &= x\\
g_1(g_2 x) &= (g_1 g_2) x
\end{align*}
for all $x \in X$ and $g_1, g_2 \in G$.  For each $g \in G$, define
\begin{align*}
\sigma_g : X &\to X\\
x &\mapsto gx
\end{align*}
which is a bijection $X \to X$, i.e., an element of $\text{Sym(X)}$.  Define the map
\begin{align*}
\varphi : G &\to \text{Sym}(X)\\
g &\mapsto \sigma_g \, .
\end{align*}
Given $x \in X$, the second group action axiom then gives
\begin{align*}
\varphi(g_1 g_2)(x) &= \sigma_{g_1 g_2}(x) = (g_1 g_2)x = g_1(g_2 x) = g_1(\sigma_{g_2} (x)) = \sigma_{g_1}(\sigma_{g_2}(x))\\
&= (\sigma_{g_1} \circ \sigma_{g_2})(x) = (\varphi(g_1) \circ \varphi(g_2))(x) \, .
\end{align*}
Thus $\varphi(g_1 g_2) = \varphi(g_1) \circ \varphi(g_2)$, so $\varphi$ is a homomorphism.
I'm denoting both the group action and the binary operation of the group by juxtaposition, so there is some chance of confusion.  But I find it suggestive rather than confusing.  The action has to be "compatible" with the group binary operation.
A: Note that * is the group operation on $Sym(X)$. So * represents composition of functions. Replacing the notation for $A(g,.) $ by $A_g$, we see that $A_{g+f}(x)= A(g+f,x)= A(g,A(f,x))=A(g,A_f(x))=A_g*A_f(x)$ which is what we want. Note again that * represents composition of functions. 
