Show that group action is homomorphism to Symmetric group

Question

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?

Answer 1

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.

Answer 2

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.