Group action candidate on a set and necessity of the inverse permutation

Question

I've translated a text, originally in french :

"Let $n\geq 1$, and $E$ a set. Then $\mathfrak{S}_n$ acts on $E^n$ by :

$\mathfrak{S}_n\times E^n\rightarrow E^n$

$(\sigma,(x_1,\ldots,x_n)\mapsto(x_{\sigma^{-1}(1)},\ldots,x_{\sigma^{-1}(n)})$

The conscientious reader will check that taking $\sigma^{-1}$ is necessary to have a group action."

What I've tried :

I've named the group action candidate $f$ such that :

$f:\mathfrak{S}_n\times E^n\rightarrow E^n$

$(\sigma,(x_1,\ldots,x_n)\mapsto(x_{\sigma^{-1}(1)},\ldots,x_{\sigma^{-1}(n)})$

To prove that $f$ is a group action, given an arbitrary n-tuple $x=(x_1,\ldots,x_n)\in E^n$, and $(\sigma_1,\sigma_2)\in\mathfrak{S}_n^2$, you must have :

• Identity : $f(\textrm{Id},x)=x$
• Compatibility : $f(\sigma_1,f(\sigma_2,x))=f(\sigma_1\circ\sigma_2)(x)$

The identity works, but the compatibility seems to fail. I derive the left member and the right member of the equality :

• $f(\sigma_1,f(\sigma_2,x))\\ =f(\sigma_1,(x_{\sigma_2^{-1}(1)},\ldots,x_{\sigma_2^{-1}(n)}))\\ =(x_{\sigma_1^{-1}(\sigma_2^{-1}(1))},\ldots,x_{\sigma_1^{-1}(\sigma_2^{-1}(n))})$
• $f(\sigma_1\circ\sigma_2)(x)\\ =(x_{(\sigma_1\circ\sigma_2)^{-1}(1)},\ldots,x_{(\sigma_1\circ\sigma_2)^{-1}(n)})\\ =(x_{(\sigma_2^{-1}\circ\sigma_1^{-1})(1)},\ldots,x_{(\sigma_2^{-1}\circ\sigma_1^{-1})(n)})\\ =(x_{\sigma_2^{-1}(\sigma_1^{-1}(1))},\ldots,x_{\sigma_2^{-1}(\sigma_1^{-1}(n))})$

For the equality to hold, you must have :

$$\sigma_1^{-1}\circ\sigma_2^{-1}=\sigma_2^{-1}\circ\sigma_1^{-1}$$

Since $\sigma_1$ and $\sigma_2$ are arbitrary, their inverse must necessary commute for $f$ to be a group action.

The problem:

I do not understand why "$\sigma^{-1}$ is necessary to have a group action", since it implies commutativity, and even if it was an hypothesis, I am not sure if the set of permutations whose inverse commute is still a group (maybe leading to a contradiction).

Actually, I think the group action can be defined without the inverse. It seems to me the inverse cannot work since by taking the inverse of a composition of permutations, it will swap the order of their inverse, causing the commutative property to be necessary.

Moreover by searching on the web and in books, I can see that there is a "natural group action" with the permutations which look like this one, and indeed it seems rather natural. Furthermore they do not mention constructions with permutation inverses on the the documents I read.

Additional readings :

According to these 2 posts (in this order, the second one thinks there is a mistake in the first one) :

• Show that $\mathrm{S}_{m} \times \mathbb{N}^{m} \rightarrow \mathbb{N}^{m}$ defines an action of $\mathrm{S}_{m}$ on $\mathbb{N}^{m}$
• Is this a typo in a proof regarding action of a permutation group on a set?

It says that $f(\sigma_1,(x_{\sigma_2^{-1}(1)},\ldots,x_{\sigma_2^{-1}(n)}))$ would not be equal to : $$(x_{\sigma_1^{-1}(\sigma_2^{-1}(1))},\ldots,x_{\sigma_1^{-1}(\sigma_2^{-1}(n))})$$ But instead : $$(x_{\sigma_2^{-1}(\sigma_1^{-1}(1))},\ldots,x_{\sigma_2^{-1}(\sigma_1^{-1}(n))})$$

The reasoning is the following, by letting $d_i=x_{\sigma_2^{-1}(i)}$ :

$f(\sigma_1,(x_{\sigma_2^{-1}(1)},\ldots,x_{\sigma_2^{-1}(n)}))\\ =f(\sigma_1,(d_1,\ldots,d_n)) =(d_{\sigma_1^{-1}(1)},\ldots,d_{\sigma_1^{-1}(n)})\\ =(x_{\sigma_2^{-1}(\sigma_1^{-1}(1))},\ldots,x_{\sigma_2^{-1}(\sigma_1^{-1}(n))})$

Alternative definition :

But I can in the same way define $h_i=x_{\sigma_1^{-1}(i)}$. We have the following property :

$$h_{\sigma_2^{-1}(i)}=x_{\sigma_1^{-1}(\sigma_2^{-1}(i))}$$

Then :

$f(\sigma_1,x)=f(\sigma_1,(x_1,\ldots,x_n))=(x_{\sigma_1^{-1}(1)},\ldots,x_{\sigma_1^{-1}(n)})=(h_1,\ldots,h_n)$

$f(\sigma_1,f(\sigma_2,x))=f(\sigma_1,(x_{\sigma_2^{-1}(1)},\ldots,x_{\sigma_2^{-1}(n)})=(h_{\sigma_2^{-1}(1)},\ldots,h_{\sigma_2^{-1}(n)})=(x_{\sigma_1^{-1}(\sigma_2^{-1}(1))},\ldots,x_{\sigma_1^{-1}(\sigma_2^{-1}(n))})$

Suspicions :

So it seems to me there is something going on, I suspect there are at least two ways to define $f(\sigma_1,f(\sigma_2,x))$ :

• $f(\sigma_1,f(\sigma_2,x))=(x_{\sigma_2^{-1}(\sigma_1^{-1}(1))},\ldots,x_{\sigma_2^{-1}(\sigma_1^{-1}(n))})$
• $f(\sigma_1,f(\sigma_2,x))=(x_{\sigma_1^{-1}(\sigma_2^{-1}(1))},\ldots,x_{\sigma_1^{-1}(\sigma_2^{-1}(n))})$

Letting $d$ or $h$ would in fact provide a choice between the constructions, which are incompatible simultaneously. A subtlety would reside in the way for $f$ to apply its first argument to the indices of the its second one.

Moreover, the author of the translated text would assume a "$d$"-construction.

Questions :

1. Are there no errors in what I've written ?
2. Is my conclusion correct ? I.e. it is a problem of definition and hidden assumption about the construction of the group action candidate.
3. Do the author really meant the "$d$"-construct ?