$S_3$ permutes the elements of $\mathbb{R}^3$, show that $\sigma (x_1, x_2, x_3) = ( x_{\sigma^{-1}(1)}, x_{\sigma^{-1}(2)}, x_{\sigma^{-1}(3)})$.

Question

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.

Answer 1

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

Answer 2

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

Answer 3

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 $$ (123).(x_1,x_2,x_3)=(123).(x_1e_1+x_2e_2+x_3e_3)=x_1e_2+x_2e_3+x_3e_1=(x_3,x_1,x_2).$$