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Let the permutation group $S_3$ act on $\mathbb{R}^3$ by permuting the elements.

How can we show that this action can be written as $\sigma (x_1, x_2, x_3) = ( x_{\sigma^{-1}(1)}, x_{\sigma^{-1}(2)}, x_{\sigma^{-1}(3)})$ for $\sigma \in S_3$, $(x_1, x_2, x_3) \in \mathbb{R}^3$.

For instance, if $\sigma = (123)$, we have $\sigma (x_1, x_2, x_3) = (123)(x_1, x_2, x_3) = (x_3, x_1, x_2)$.

Just looking at some examples, this seems so obvious, but I'm not sure how to go about showing it.

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    $\begingroup$ You cannot "show" them, because you didn't define any action of $S_3$ on $\mathbb{R}^3$. The given formula might even be used as definition, and you cannot show a definition from nothing $\endgroup$ – user160738 Mar 1 '17 at 4:32
  • $\begingroup$ Sorry, $\sigma$ is meant to be a permutation of $\{ 1, 2, 3 \}$, so it's just a cycle. For instance, if $\sigma = (123)$, we have $\sigma (x_1, x_2, x_3) = (123)(x_1, x_2, x_3) = (x_3, x_1, x_2)$. Is that clearer? $\endgroup$ – zxmkn Mar 1 '17 at 4:44
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    $\begingroup$ It seems obvious because you're taking as a definition exactly what you are trying to prove. Your question as stated does not quite make sense. What you COULD do is prove that your "definition" actually satisfies all of the defining properties of a group action. $\endgroup$ – mathematician Mar 1 '17 at 4:48
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AFAIK, we define the action of group $G$ on set $X$ like this. Note that they denote $\varphi(g, x)$ as $g.x$

We can think of action of group $G$ on an object $R$ as a homomorphism from $G$ to $Aut(R)$. Or, as the the first definition, you have $\varphi: G\times R \rightarrow R$, where $\varphi(g,x) = \phi_g(x)$. Please check this satisfy the first definition.

Here we have $\varphi: S_3 \rightarrow Aut(\mathbb{R}^3)$, where $\varphi(\sigma) : \mathbb{R}^3 \rightarrow \mathbb{R}^3$ and $\varphi(\sigma)(x_1.x_2,x_3) = (x_{\sigma^{-1}(1)},x_{\sigma^{-1}(2)},x_{\sigma^{-1}(3)})$.

Then, $\sigma(x_1,x_2,x_3) = \varphi(\sigma(x_1,x_2,x_3)) = \varphi(\sigma)(x_1,x_2,x_3) = (x_{\sigma^{-1}(1)},x_{\sigma^{-1}(2)},x_{\sigma^{-1}(3)})$.

Now we check that $\varphi$ is homomorphism, i.e $\varphi(\sigma\circ \theta) =\varphi(\sigma)\circ \varphi(\theta)$.

Let $(x_1,x_2,x_3) \in \mathbb{R}^3$, we have $$\varphi(\sigma\circ \theta)(x_1,x_2,x_3) = (x_{(\sigma\circ\theta)^{-1}(1)},x_{(\sigma\circ\theta)^{-1}(2)},x_{(\sigma\circ\theta)^{-1}(3)})$$ $$= (x_{\theta^{-1}\circ\sigma^{-1}(1)},x_{\theta^{-1}\circ\sigma^{-1}(2)},x_{\theta^{-1}\circ\sigma^{-1}(3)}) = \varphi(\theta)(x_{\sigma^{-1}(1)},x_{\sigma^{-1}(2)},x_{\sigma^{-1}(3)}) = \varphi(\sigma)\circ \varphi(\theta).$$

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I think I've figured this out.

We know that $S_3$ acts on $\mathbb{R}^3$ by permuting the coordinates, and we want to show that this action can be written as $\sigma (x_1, x_2, x_3) = ( x_{\sigma^{-1}(1)}, x_{\sigma^{-1}(2)}, x_{\sigma^{-1}(3)})$ for $\sigma \in S_3$, $(x_1, x_2, x_3) \in \mathbb{R}^3$.

Let $(x_{n_1}, x_{n_2}, x_{n_3}) := \sigma (x_1, x_2, x_3)$ and note that $(n_1, n_2, n_3)$ is a permutation of $\{1,2,3\}$.

Let $\sigma = (ijk)$ or $(ij) \in S_3$.

WLOG, consider the $j$th entry of $\sigma (x_1, x_2, x_3)$, i.e. $x_{n_j}$.

$\sigma(i) = j$, so $x_{n_j} = x_i$.

$\sigma^{-1}\sigma(i) = i = \sigma^{-1}(j)$, so $x_{n_j} = x_i = x_{\sigma^{-1}(j)}$.

Therefore, $\sigma (x_1, x_2, x_3) = ( x_{\sigma^{-1}(1)}, x_{\sigma^{-1}(2)}, x_{\sigma^{-1}(3)})$.

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Write $e_1=(1,0,0)$, $e_2=(0,1,0)$ and $e_3=(0,0,1)$. Then, $(x_1,x_2,x_3)=x_1e_1+x_2e_x+x_3e_3$.

Now, $S_3$ acts linearly on $\mathbb{R}^3$ by $\sigma.e_i=e_{\sigma(i)}$ since $1.e_i=e_i$ and $$ \sigma.(\tau.e_i)=\sigma(e_{\tau(i)})=e_{\sigma(\tau(i))}=e_{(\sigma\tau)(i)}=(\sigma\tau).e_i.$$ Now, \begin{align} \sigma.(x_1,x_2,x_3)&=\sigma.(x_1e_1+x_2e_2+x_3e_3)\\ &=x_1e_{\sigma(1)}+x_2e_{\sigma(2)}+x_3e_{\sigma(3)}\\ &=x_{\sigma^{-1}(1)}e_1+x_{\sigma^{-1}(2)}e_2+x_{\sigma^{-1}(3)}e_3\\ &=(x_{\sigma^{-1}(1)},x_{\sigma^{-1}(2)},x_{\sigma^{-1}(3)}). \end{align}

Example: Take $\sigma=(123)$, then $$ (123).(x_1,x_2,x_3)=(123).(x_1e_1+x_2e_2+x_3e_3)=x_1e_2+x_2e_3+x_3e_1=(x_3,x_1,x_2).$$

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