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Just out of curiosity. Can the fact that the identity permutation is (only) even be proven by means of the sign function $\text{sgn}$?

In this and this post it was suggested that the simplest proof of the fact that the identity permutation can only be written as a product of an even number of transpositions is by means of $\text{sgn}$, although no proof was really specified. I've also seen proofs that use determinants such as this one.

The problem is that I do not see how one could define the sign function without first proving the identity is only an even permutation, similarly I do not know of a definition of determinants that do not appeal to this fact. I have tried, for example, to define $\text{sgn}$ as follows

Let $\rho \in \mathbb{S}_n$ and let $\sigma _1 \ldots \sigma_r$ be a product of transpositions with $r$ minimal. Then

$$\text{sgn}:=\begin{cases} 1 \ \ \ \ \text{ if } r \text{ is even} \\ -1 \ \text{ if } r \text{ is odd} \end{cases}$$

but then I seem to become unable to prove elementary properties of $\text{sgn}$ such as $\text{sgn}(\alpha \beta )=\text{sgn}(\alpha )\text{sgn}(\beta )$ which I believe are necessary to prove the main result.

Sam
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    Let $\sigma$ be an arbitrary permutation. Then $\sigma^{-1}$ has the same sign as $\sigma$, and $\mathrm{id}=\sigma\sigma^{-1}$. Hence the sign of $\mathrm{id}$ is $1$. – Arturo Magidin Jul 06 '20 at 17:46
  • @ArturoMagidin Which definition of $\text{sgn}$ are you using? By the definition I posted one could merely argue that $\text{id}\neq (ab)$ to conclude $\text{sgn}(\text{id})=1$. – Sam Jul 06 '20 at 17:49
  • It doesn't matter what definition you use: $\sigma$ will have a sign. It will be the same sign as $\sigma^{-1}$, and sign is multiplicative, so the sign of $\mathrm{id}$ is the square of the sign of $\sigma$, hence is $1$. – Arturo Magidin Jul 06 '20 at 17:50
  • One can define the sign function as the parity of the number of transpositions in any representation of $\sigma$ as a product of transpositions, and then prove the parity is independent of the representation. One proof is here. This immediately yields the multiplicativity. – Arturo Magidin Jul 06 '20 at 17:52
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    I'm saying how to use the sign function to prove $\mathrm{id}$ is even. (I honestly don't know why you keep using the qualifier "only"...). As to how to define the sign function, I just added a possibility and a proof that it is well-defined. – Arturo Magidin Jul 06 '20 at 17:54
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    The idendity permutation is the product of $0$ transpositions and $(-1)^0=1$. Hence the permutation is even. – Peter Jul 06 '20 at 17:56
  • @ArturoMagidin ah I see, I just had never seen $\text{sgn}$ being defined as in the post you linked, but yes, that answers the question. – Sam Jul 06 '20 at 18:00
  • Have you understood my answer? – user21820 Oct 03 '23 at 13:45

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Just associate each permutation $f$ with the result $r(f)$ of applying it to the terms of the finite sequence $(1,2,...,n)$, and define the sign of $f$ as the parity of the number of pairs $(x,y)$ such that $x < y$ but $x$ is after $y$ in $r(f)$. Then obviously the sign of $f$ is well-defined. And it is easy to prove that performing any swap on $r(f)$ changes the number of such pairs by an odd number, and hence the composition of $f$ with any swap has sign opposite to $f$. Done.

user21820
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