Return to Answer

$\DeclareMathOperator\sgn{sgn}$How about arguing by induction? I claim that there is a homomorphism map $\sgn_n : \Sigma_n \to \{\pm1\}$ that takes the value $-1$ at every transposition. This is vacuously true for $n = 0$ and $n = 1$.

In general, suppose that $n$ is greater than $1$ and we have such a map $\sgn_{n - 1}$ on $\Sigma_{n - 1}$.

Let $\tau$ be the transposition that swaps $n$ and $n - 1$. Since $\Sigma_n$ equals $\Sigma_{n - 1} \sqcup \Sigma_{n - 1}\tau\Sigma_{n - 1}$, and since two elements of $\Sigma_{n - 1}$ that are conjugate by $\tau$ are already conjugate in $\Sigma_{n - 1}$ (so that $\sgn_{n - 1}$ agrees on them), there is a unique function $\sgn_n : \Sigma_n \to \{\pm1\}$ that transforms by $\sgn_{n - 1}$ under left and right translation by $\Sigma_{n - 1}$, and that takes the value $1$ at $1$ and $-1$ at $\tau$. Since all transpositions that move $n$ are conjugate under $\Sigma_{n - 1}$, the map $\sgn_n$ is independent of the choice of $\tau$.

I originally thought that it was obvious that $\sgn_n$ was a homomorphism, and said:

It's hard to imagine your, or anyone's, liking this argument better than the one proposed by @TheoJohnsonFreyd in the comments or by @DavidESpeyer in the answers, but I think it doesn't violate any of your interdicts.

… but now I see that some computation has to be done, which was at least tacitly ruled out. Oh well!

It is clear that $\sgn_n(\sigma)^{-1}\sgn_n(\sigma\bullet)$ equals $\sgn_n$ for all $\sigma \in \Sigma_{n - 1}$; and that $\sgn_n(\tau)^{-1}\sgn_n(\tau\bullet)$ equals $1$ and $1$ and $-1$ at $\tau$, and transforms by $\sgn_{n - 1}$ under left translation by $\Sigma_{n - 2}$ and right translation by $\Sigma_{n - 1}$. It thus remains only to show that $\sgn_n(\tau)^{-1}\sgn_n(\tau\bullet)$ agrees with $\sgn_n$ at every element of the form $\sigma\tau$, where $\sigma \in \Sigma_{n - 1}$ is a transposition that moves $n - 1$. The braid relations—mentioned in one of your bullet points as a possible acceptable line of attack—say that $\tau\sigma\tau$ equals $\sigma\tau\sigma$. Thus, $\sgn_n(\tau)^{-1}\sgn_n(\tau\sigma\tau)$ equals $-\sgn_n(\sigma\tau\sigma) = \sgn_{n - 1}(\sigma)\sgn_{n - 1}(\sigma)\sgn_n(\tau)\sgn_{n - 1}(\sigma) = \sgn_n(\sigma\tau)$.

$\DeclareMathOperator\sgn{sgn}$How about arguing by induction? I claim that there is a homomorphism map $\sgn_n : \Sigma_n \to \{\pm1\}$ that takes the value $-1$ at every transposition. This is vacuously true for $n = 0$ and $n = 1$.

In general, suppose that $n$ is greater than $1$ and we have such a map $\sgn_{n - 1}$ on $\Sigma_{n - 1}$.

Let $\tau$ be the transposition that swaps $n$ and $n - 1$. Since $\Sigma_n$ equals $\Sigma_{n - 1} \sqcup \Sigma_{n - 1}\tau\Sigma_{n - 1}$, and since two elements of $\Sigma_{n - 1}$ that are conjugate by $\tau$ are already conjugate in $\Sigma_{n - 1}$ (so that $\sgn_{n - 1}$ agrees on them), there is a unique function $\sgn_n : \Sigma_n \to \{\pm1\}$ that transforms by $\sgn_{n - 1}$ under left and right translation by $\Sigma_{n - 1}$, and that takes the value $1$ at $1$ and $-1$ at $\tau$. Since all transpositions that move $n$ are conjugate under $\Sigma_{n - 1}$, the map $\sgn_n$ is independent of the choice of $\tau$.

I originally thought that it was obvious that $\sgn_n$ was a homomorphism, and said:

It's hard to imagine your, or anyone's, liking this argument better than the one proposed by @TheoJohnsonFreyd in the comments or by @DavidESpeyer in the answers, but I think it doesn't violate any of your interdicts.

… but now I see that some computation has to be done, which was at least tacitly ruled out. Oh well!

It is clear that $\sgn_n(\sigma)^{-1}\sgn_n(\sigma\bullet)$ equals $\sgn_n$ for all $\sigma \in \Sigma_{n - 1}$; and that $\sgn_n(\tau)^{-1}\sgn_n(\tau\bullet)$ equals $1$ at $1$ and $-1$ at $\tau$, and transforms by $\sgn_{n - 1}$ under left translation by $\Sigma_{n - 2}$ and right translation by $\Sigma_{n - 1}$. It thus remains only to show that $\sgn_n(\tau)^{-1}\sgn_n(\tau\bullet)$ agrees with $\sgn_n$ at every element of the form $\sigma\tau$, where $\sigma \in \Sigma_{n - 1}$ is a transposition that moves $n - 1$. The braid relations—mentioned in one of your bullet points as a possible acceptable line of attack—say that $\tau\sigma\tau$ equals $\sigma\tau\sigma$. Thus, $\sgn_n(\tau)^{-1}\sgn_n(\tau\sigma\tau)$ equals $-\sgn_n(\sigma\tau\sigma) = \sgn_{n - 1}(\sigma)\sgn_{n - 1}(\sigma)\sgn_n(\tau)\sgn_{n - 1}(\sigma) = \sgn_n(\sigma\tau)$.

$\DeclareMathOperator\sgn{sgn}$How about arguing by induction? I claim that there is a homomorphism map $\sgn_n : \Sigma_n \to \{\pm1\}$ that takes the value $-1$ at every transposition. This is vacuously true for $n = 0$ and $n = 1$.

In general, suppose that $n$ is greater than $1$ and we have such a map $\sgn_{n - 1}$ on $\Sigma_{n - 1}$.

Let $\tau$ be the transposition that swaps $n$ and $n - 1$. Since $\Sigma_n$ equals $\Sigma_{n - 1} \sqcup \Sigma_{n - 1}\tau\Sigma_{n - 1}$, and since two elements of $\Sigma_{n - 1}$ that are conjugate by $\tau$ are already conjugate in $\Sigma_{n - 1}$ (so that $\sgn_{n - 1}$ agrees on them), there is a unique function $\sgn_n : \Sigma_n \to \{\pm1\}$ that transforms by $\sgn_{n - 1}$ under left and right translation by $\Sigma_{n - 1}$, and that takes the value $1$ at $1$ and $-1$ at $\tau$. Since all transpositions that move $n$ are conjugate under $\Sigma_{n - 1}$, the map $\sgn_n$ is independent of the choice of $\tau$.

I originally thought that it was obvious that $\sgn_n$ was a homomorphism, and said:

It's hard to imagine your, or anyone's, liking this argument better than the one proposed by @TheoJohnsonFreyd in the comments or by @DavidESpeyer in the answers, but I think it doesn't violate any of your interdicts.

… but now I see that some computation has to be done, which was at least tacitly ruled out. Oh well!

It is clear that $\sgn_n(\sigma)^{-1}\sgn_n(\sigma\bullet)$ equals $\sgn_n$ for all $\sigma \in \Sigma_{n - 1}$; and that $\sgn_n(\tau)^{-1}\sgn_n(\tau\bullet)$ equals $\sgn_n$ on $\Sigma_{n - 1} \cup \Sigma_{n - 2}\tau\Sigma_{n - 1}$, and transforms by $\sgn_{n - 1}$ under right multiplication by $\Sigma_{n - 1}$. It thus remains only to show that $\sgn_n(\tau)^{-1}\sgn_n(\tau\bullet)$ agrees with $\sgn_n$ at every element of the form $\sigma\tau$, where $\sigma \in \Sigma_{n - 1}$ moves $n - 1$. We have that $\tau\sigma\tau$ equals $\eta\tau\sigma$, where $\eta \in \Sigma_{n - 1}$ is the transposition that swaps $n - 1$ and $\sigma(n - 1)$. (This is almost the braid relation that you mention in one of your bullet points, but I can't figure out how to make it use just the braid relation.) Thus, $\sgn_n(\tau)^{-1}\sgn_n(\tau\sigma\tau)$ equals $\sgn_{n - 1}(\eta)\sgn_{n - 1}(\sigma) = -\sgn_{n - 1}(\sigma) = \sgn_n(\sigma\tau)$.

$\DeclareMathOperator\sgn{sgn}$How about arguing by induction? I claim that there is a homomorphism map $\sgn_n : \Sigma_n \to \{\pm1\}$ that takes the value $-1$ at every transposition. This is vacuously true for $n = 0$ and $n = 1$.

In general, suppose that $n$ is greater than $1$ and we have such a map $\sgn_{n - 1}$ on $\Sigma_{n - 1}$.

Let $\tau$ be the transposition that swaps $n$ and $n - 1$. Since $\Sigma_n$ equals $\Sigma_{n - 1} \sqcup \Sigma_{n - 1}\tau\Sigma_{n - 1}$, and since two elements of $\Sigma_{n - 1}$ that are conjugate by $\tau$ are already conjugate in $\Sigma_{n - 1}$ (so that $\sgn_{n - 1}$ agrees on them), there is a unique function $\sgn_n : \Sigma_n \to \{\pm1\}$ that transforms by $\sgn_{n - 1}$ under left and right translation by $\Sigma_{n - 1}$, and that takes the value $1$ at $1$ and $-1$ at $\tau$. Since all transpositions that move $n$ are conjugate under $\Sigma_{n - 1}$, the map $\sgn_n$ is independent of the choice of $\tau$.

I originally thought that it was obvious that $\sgn_n$ was a homomorphism, and said:

It's hard to imagine your, or anyone's, liking this argument better than the one proposed by @TheoJohnsonFreyd in the comments or by @DavidESpeyer in the answers, but I think it doesn't violate any of your interdicts.

… but now I see that some computation has to be done, which was at least tacitly ruled out. Oh well!

It is clear that $\sgn_n(\sigma)^{-1}\sgn_n(\sigma\bullet)$ equals $\sgn_n$ for all $\sigma \in \Sigma_{n - 1}$; and that $\sgn_n(\tau)^{-1}\sgn_n(\tau\bullet)$ equals $1$ and $1$ and $-1$ at $\tau$, and transforms by $\sgn_{n - 1}$ under left translation by $\Sigma_{n - 2}$ and right translation by $\Sigma_{n - 1}$. It thus remains only to show that $\sgn_n(\tau)^{-1}\sgn_n(\tau\bullet)$ agrees with $\sgn_n$ at every element of the form $\sigma\tau$, where $\sigma \in \Sigma_{n - 1}$ is a transposition that moves $n - 1$. The braid relations—mentioned in one of your bullet points as a possible acceptable line of attack—say that $\tau\sigma\tau$ equals $\sigma\tau\sigma$. Thus, $\sgn_n(\tau)^{-1}\sgn_n(\tau\sigma\tau)$ equals $-\sgn_n(\sigma\tau\sigma) = \sgn_{n - 1}(\sigma)\sgn_{n - 1}(\sigma)\sgn_n(\tau)\sgn_{n - 1}(\sigma) = \sgn_n(\sigma\tau)$.

$\DeclareMathOperator\sgn{sgn}$How about arguing by induction? I claim that there is a homomorphism map $\sgn_n : \Sigma_n \to \{\pm1\}$ that takes the value $-1$ at every transposition. This is vacuously true for $n = 0$ and $n = 1$.

In general, suppose that $n$ is greater than $1$ and we have such a map $\sgn_{n - 1}$ on $\Sigma_{n - 1}$.

Let $\tau$ be the transposition that swaps $n$ and $n - 1$. Since $\Sigma_n$ equals $\Sigma_{n - 1} \sqcup \Sigma_{n - 1}\tau\Sigma_{n - 1}$, and since two elements of $\Sigma_{n - 1}$ that are conjugate by $\tau$ are already conjugate in $\Sigma_{n - 1}$ (so that $\sgn_{n - 1}$ agrees on them), there is a unique function $\sgn_n : \Sigma_n \to \{\pm1\}$ that transforms by $\sgn_{n - 1}$ under left and right translation by $\Sigma_{n - 1}$, and that takes the value $1$ at $1$ and $-1$ at $\tau$. Since all transpositions that move $n$ are conjugate under $\Sigma_{n - 1}$, the map $\sgn_n$ is independent of the choice of $\tau$.

I originally thought that it was obvious that $\sgn_n$ was a homomorphism, and said:

It's hard to imagine your, or anyone's, liking this argument better than the one proposed by @TheoJohnsonFreyd in the comments or by @DavidESpeyer in the answers, but I think it doesn't violate any of your interdicts.

… but now I see that some computation has to be done, which was at least tacitly ruled out. Oh well!

It is clear that $\sgn_n(\sigma_{n - 1})^{-1}\sgn_n(\sigma_{n - 1}\bullet)$ equals $\sgn_n$; and that $\sgn_n(\tau)^{-1}\sgn_n(\tau\bullet)$ equals $\sgn_n$ on $\Sigma_{n - 1} \cup \Sigma_{n - 2}\tau\Sigma_{n - 1}$, and transforms by $\sgn_{n - 1}$ under right multiplication by $\Sigma_{n - 1}$. It thus remains only to show that $\sgn_n(\tau)^{-1}\sgn_n(\tau\bullet)$ agrees with $\sgn_n$ at every element of the form $\sigma\tau$, where $\sigma \in \Sigma_{n - 1}$ moves $n - 1$. We have that $\tau\sigma\tau$ equals $\eta\tau\sigma$, where $\eta \in \Sigma_{n - 1}$ is the transposition that swaps $n - 1$ and $\sigma(n - 1)$. (This is almost the braid relation that you mention in one of your bullet points, but I can't figure out how to make it use just the braid relation.) Thus, $\sgn_n(\tau)^{-1}\sgn_n(\tau\sigma\tau)$ equals $\sgn_{n - 1}(\eta)\sgn_{n - 1}(\sigma) = -\sgn_{n - 1}(\sigma) = \sgn_n(\sigma\tau)$.

$\DeclareMathOperator\sgn{sgn}$How about arguing by induction? I claim that there is a homomorphism map $\sgn_n : \Sigma_n \to \{\pm1\}$ that takes the value $-1$ at every transposition. This is vacuously true for $n = 0$ and $n = 1$.

In general, suppose that $n$ is greater than $1$ and we have such a map $\sgn_{n - 1}$ on $\Sigma_{n - 1}$.

Let $\tau$ be the transposition that swaps $n$ and $n - 1$. Since $\Sigma_n$ equals $\Sigma_{n - 1} \sqcup \Sigma_{n - 1}\tau\Sigma_{n - 1}$, and since two elements of $\Sigma_{n - 1}$ that are conjugate by $\tau$ are already conjugate in $\Sigma_{n - 1}$ (so that $\sgn_{n - 1}$ agrees on them), there is a unique function $\sgn_n : \Sigma_n \to \{\pm1\}$ that transforms by $\sgn_{n - 1}$ under left and right translation by $\Sigma_{n - 1}$, and that takes the value $1$ at $1$ and $-1$ at $\tau$. Since all transpositions that move $n$ are conjugate under $\Sigma_{n - 1}$, the map $\sgn_n$ is independent of the choice of $\tau$.

I originally thought that it was obvious that $\sgn_n$ was a homomorphism, and said:

It's hard to imagine your, or anyone's, liking this argument better than the one proposed by @TheoJohnsonFreyd in the comments or by @DavidESpeyer in the answers, but I think it doesn't violate any of your interdicts.

… but now I see that some computation has to be done, which was at least tacitly ruled out. Oh well!

It is clear that $\sgn_n(\sigma)^{-1}\sgn_n(\sigma\bullet)$ equals $\sgn_n$ for all $\sigma \in \Sigma_{n - 1}$; and that $\sgn_n(\tau)^{-1}\sgn_n(\tau\bullet)$ equals $\sgn_n$ on $\Sigma_{n - 1} \cup \Sigma_{n - 2}\tau\Sigma_{n - 1}$, and transforms by $\sgn_{n - 1}$ under right multiplication by $\Sigma_{n - 1}$. It thus remains only to show that $\sgn_n(\tau)^{-1}\sgn_n(\tau\bullet)$ agrees with $\sgn_n$ at every element of the form $\sigma\tau$, where $\sigma \in \Sigma_{n - 1}$ moves $n - 1$. We have that $\tau\sigma\tau$ equals $\eta\tau\sigma$, where $\eta \in \Sigma_{n - 1}$ is the transposition that swaps $n - 1$ and $\sigma(n - 1)$. (This is almost the braid relation that you mention in one of your bullet points, but I can't figure out how to make it use just the braid relation.) Thus, $\sgn_n(\tau)^{-1}\sgn_n(\tau\sigma\tau)$ equals $\sgn_{n - 1}(\eta)\sgn_{n - 1}(\sigma) = -\sgn_{n - 1}(\sigma) = \sgn_n(\sigma\tau)$.