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Recently, I asked a question about action of a permutation group on a set here. Let me summarize it.

Let $\mathrm{S}_{m}$ be the set of all permutations of $\{1,2,\cdots,m\}$. Then $(\mathrm{S}_{m},\circ)$ is a group where $\circ$ is function composition operation.

Show that $$\mathrm{S}_{m} \times \mathbb{N}^{m} \rightarrow \mathbb{N}^{m}, \quad(\sigma, x) \mapsto \sigma \cdot x := \left(x_{\sigma^{-1}(1)}, \ldots, x_{\sigma^{-1}(m)}\right)$$ defines an action of $\mathrm{S}_{m}$ on $\mathbb{N}^{m}$.

For $\sigma , \tau \in \mathrm{S}_{m}$ and $x \in \mathbb{N}^{m}$, I try to prove $$\sigma \cdot (\tau \cdot x) = (\sigma \circ \tau) \cdot x$$


In this answer, Wuestenfux presents the following proof:

$$\begin{aligned} \sigma \tau \cdot x &= (x_{(\sigma\tau)^{-1}(1)}, \ldots, x_{(\sigma\tau)^{-1}(n)})\\ &= (x_{\tau^{-1}\sigma^{-1}(1)}, \ldots, x_{\tau^{-1}\sigma^{-1}(n)}) \\ &= \tau\cdot (x_{\sigma^{-1}(1)}, \ldots, x_{\sigma^{-1}(n)})\\ &=\sigma\cdot(\tau \cdot x) \end{aligned}$$

Here he writes $\sigma \tau$ for $\sigma \circ \tau$.


In his proof, I think $(x_{\tau^{-1}\sigma^{-1}(1)}, \ldots, x_{\tau^{-1}\sigma^{-1}(n)}) = \tau\cdot (x_{\sigma^{-1}(1)}, \ldots, x_{\sigma^{-1}(n)})$ is wrong. Instead, it should be $(x_{\tau^{-1}\sigma^{-1}(1)}, \ldots, x_{\tau^{-1}\sigma^{-1}(n)}) = \sigma \cdot (x_{\tau^{-1}(1)}, \ldots, x_{\tau^{-1}(n)})$.

My reasoning:

  1. In $(x_{\tau^{-1}\sigma^{-1}(1)}, \ldots, x_{\tau^{-1}\sigma^{-1}(n)})$, the flow of input-output is $\sigma \longrightarrow \tau \longrightarrow x$, whereas it is $\tau \longrightarrow \sigma \longrightarrow x$ in $\tau\cdot (x_{\sigma^{-1}(1)}, \ldots, x_{\sigma^{-1}(n)})$. As such, I feel that it is counter-intuitive to have $(x_{\tau^{-1}\sigma^{-1}(1)}, \ldots, x_{\tau^{-1}\sigma^{-1}(n)}) = \tau\cdot (x_{\sigma^{-1}(1)}, \ldots, x_{\sigma^{-1}(n)})$.

  2. It's clear that $\sigma \cdot x = x \circ \sigma^{-1}$. Then $$\begin{aligned} \sigma \tau \cdot x &= x \circ (\sigma \circ \tau)^{-1}\\ &= x \circ (\tau^{-1} \circ \sigma^{-1})\\ &= (x \circ \tau^{-1}) \circ \sigma^{-1}\\ &= \sigma \cdot (x \circ \tau^{-1})\\ &= \sigma \cdot (x_{\tau^{-1}(1)}, \ldots, x_{\tau^{-1}(n)}) \end{aligned}$$

Please check if my reasoning is correct or not!

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  • $\begingroup$ See example 2.5 and 2.6 here. $\endgroup$ – Thomas Shelby Jun 1 at 4:01
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    $\begingroup$ Hi @ThomasShelby. Thank you so much for your link! It really helps me understand what is going on behind the notation. I have posted my reasoning as an answer below. Could you please have a check on it? $\endgroup$ – Le Anh Dung Jun 1 at 9:32
  • $\begingroup$ Thank you so much for your kindness @ThomasShelby :v $\endgroup$ – Le Anh Dung Jun 1 at 13:18
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I found that the blue argument is not correct in @Wuestenfux's proof.

$$\begin{aligned} \sigma \tau \cdot x &= (x_{(\sigma\tau)^{-1}(1)}, \ldots, x_{(\sigma\tau)^{-1}(n)})\\ &= \color{blue}{(x_{\tau^{-1}\sigma^{-1}(1)}, \ldots, x_{\tau^{-1}\sigma^{-1}(n)})} \\ &= \color{blue}{\tau\cdot (x_{\sigma^{-1}(1)}, \ldots, x_{\sigma^{-1}(n)})}\\ &= \sigma\cdot(\tau \cdot x) \end{aligned}$$


We have

$$\begin{aligned} \sigma \cdot (\tau \cdot x) &= \color{blue}{\sigma \cdot (x_{\tau^{-1}(1)}, \ldots, x_{\tau^{-1}(n)})} \\ &= \sigma \cdot (d_1, \ldots, d_n) \quad \text{where} \quad d_i = x_{\tau^{-1}(i)} \\ &= (d_{\sigma^{-1} (1)}, \ldots, d_{\sigma^{-1} (n)}) \\ &= (x_{\tau^{-1} (\sigma^{-1}(1))}, \ldots, x_{\tau^{-1} (\sigma^{-1} (n))})\\ &= \color{blue}{(x_{\tau^{-1}\sigma^{-1}(1)}, \ldots, x_{\tau^{-1}\sigma^{-1}(n)})} \end{aligned}$$

It follows that $$(x_{\tau^{-1}\sigma^{-1}(1)}, \ldots, x_{\tau^{-1}\sigma^{-1}(n)}) = \sigma \cdot (x_{\tau^{-1}(1)}, \ldots, x_{\tau^{-1}(n)})$$ and NOT that $$(x_{\tau^{-1}\sigma^{-1}(1)}, \ldots, x_{\tau^{-1}\sigma^{-1}(n)}) = \tau\cdot (x_{\sigma^{-1}(1)}, \ldots, x_{\sigma^{-1}(n)})$$

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  • $\begingroup$ Seems correct. I wasn't convinced either. $\endgroup$ – Wuestenfux Jun 1 at 12:34
  • $\begingroup$ Hi @Wuestenfux, I quite don't understand by "I wasn't convinced either". Did you mean that you are not convinced by my reasoning? $\endgroup$ – Le Anh Dung Jun 1 at 12:37
  • $\begingroup$ Just my reasoning. I had to do a substitution. $\endgroup$ – Wuestenfux Jun 1 at 12:39
  • $\begingroup$ Thank you so much for your interest in my question @Wuestenfux :) $\endgroup$ – Le Anh Dung Jun 1 at 14:14

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