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?
Best Answer
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.