Uniqueness of sign function on permutations.

Let $f$ be any function which assigns to each permutation an integer number, and for any two permutations $\sigma$ and $\tau$, $f(\sigma \tau) = f(\sigma)f(\tau)$. Then, $f$ is zero, or identically 1, or it is the sign function.

Solution

After reading the answer given by Hagen, I am writing the following which is what I got from his answer.

I would like to know if it is correct or not.

First, as Hagen showed, if $f(\sigma)$ is zero for any permutation, then it is zero function. So, assume that $f(\sigma)\neq 0$ for every $\sigma$.

Let $I$ be the identical permutation, and $\sigma_{ij}$ be the permutation which replaces $i$ with $j$ and vice versa. Then, we have

1 - $f(I) = 1$ (as Hagen showed).

2 - $\sigma^2_{ij}=I$, and so $f(\sigma_{ij}) = 1$ or $-1$ as $f(\sigma_{ij})$ is an integer number.

3 - Every permutation can be constructed as a sequence of $\sigma_{ij}$.

4 - If $f(\sigma_{ij}) = 1$, then for every permutation $\sigma$, $f(\sigma)$ is a multiplication of a number of $1$, and so is equal to $1$, and so $f$ is identically $1$.

5 - If $f(\sigma_{ij}) = -1$, then for every permutation $\sigma$, $f(\sigma)$ is $(-1)^m$, where $m$ is the number of simple permutations $\sigma_{ij}$ to construct $\sigma$, and therefore it is the $sgn$ function.

So, we are done.

• Hint. What are the possible values of $f(\sigma ^ 2)$? – Ethan Bolker May 20 '18 at 15:56

If $f(\sigma)=0$ for any $\sigma$, then $f(\tau)=f(\sigma\sigma^{-1}\tau)=f(\sigma)f(\sigma^{-1}\tau)=0$ for all $\tau$. So assume $f(\sigma)\ne 0$ for all $\sigma$.
From $f(I)=f(II)=f(I)f(I)$, we conclude that $f(I)=1$ (we just excluded the other possibility $f(I)=0). Let$\sigma$be a transposition. As any other transposition$\tau$is conjugate to$\sigma$, say$\tau=\phi^{-1}\sigma\phi$, we see that$f(\phi)f(\tau)=f(\phi\tau)=f(\sigma\phi)=f(\sigma)f(\phi)$ans so$f(\tau)=f(\sigma)$. From$\sigma^2=1$for transpositions, we conclude$f(\sigma)^2=f(\sigma^2)=f(I)=1$, i.e.,$f(\sigma)=\pm1$. Now use that the transpositions generate the group of permutations. As seen above, either$f(\sigma)=1$for all transpositions and then by induction$f(\phi)=1$for all$\phi$. Or$f(\sigma)=-1$for all transpositions and then by induction$f(\phi)=\operatorname{sgn}(\phi)$for all$\phi\$.