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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?

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  • $\begingroup$ What are the axioms that a group action must satisfy? I think if you write them down, you'll see why the mapping is a homomorphism. $\endgroup$ – André 3000 Feb 26 '14 at 19:25
  • $\begingroup$ If $h$ is a homomorphism from $G$ to any other group, what is $h(g,x)$? That seems confusing. Also, generally, $h$ is often used for a group element, since it is the letter after $g$, so it is probably bad to use $h$ here. $\endgroup$ – Thomas Andrews Feb 26 '14 at 19:27
  • $\begingroup$ @ThomasAndrews: Thanks, I've updated the question to use $A$ in place of $h$: I'm not sure why I started using $h$, must be misread something somewhere. So if $A$ is the group action, we have $A : G \times X \to X$, to it $A$ takes a two-tuple $(g,x)$ with $g \in G$ and $x \in X$. $\endgroup$ – brianmearns Feb 26 '14 at 19:33
  • $\begingroup$ @SpamIAm: Thanks, but I think you over estimate me. I have the axioms written out as $A(e,x) = x$ for all $x$ (where $e$ is the identity of $G$) and $A(g, A(f,x)) = A(g+f, x)$, but it's still not clear to me why this defined a homomorphism. $\endgroup$ – brianmearns Feb 26 '14 at 19:47
  • $\begingroup$ @sh1ftst0rm As you see in poorna's answer, the second axiom you stated immediately yields that this mapping is a homomorphism! $\endgroup$ – André 3000 Feb 27 '14 at 1:41
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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.

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  • $\begingroup$ (Thanks for updating from $h$ to $A$). So the steps to change right side, $A(g+f,x)$ into $A_g * A_f(x)$ are clear, but is that necessarily equal to $A_g(x) * A_f(x)$. I know $A_g(x)$ and $A_g$ are both permutations of X, but are they they same? $\endgroup$ – brianmearns Feb 26 '14 at 19:41
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    $\begingroup$ $A_g:X\longrightarrow X$ is a permutation of $X$. So $A_g(x)$ is an element of $X$. We can only * two maps right. So $A_g(x)*A_f(x)$ doesn't really make sense. What one means by it is $A_g(A_f(x))= A_g*A_f(x)$ $\endgroup$ – gradstudent Feb 26 '14 at 19:49
  • $\begingroup$ Actually, I take that back, $A_g(x)$ is not a permutation of $X$, it's an element of $X$. So how can you compose a permutation/function, $A_g$ with an element of a set, $A_f(x)$? $\endgroup$ – brianmearns Feb 26 '14 at 19:49
  • $\begingroup$ Oh, ok. So my original formulation for the condition of homomorphism was off. I think it makes sense now, thanks! $\endgroup$ – brianmearns Feb 26 '14 at 19:51
  • $\begingroup$ You're welcome! $\endgroup$ – gradstudent Feb 26 '14 at 19:53
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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.

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