Prove that sgn$(\sigma \tau) = $sgn$(\sigma)$sgn$(\tau)$ 
Prove that sign$(\sigma \tau)$ = sign$(\sigma)$sign$(\tau)$ for any permutations $\sigma, \tau \in S_n$.

I think the two thing's I'm trying to show are:


*

*If sign$(\sigma)$ = sign$(\tau) = \pm 1 \implies$ sign$(\sigma \tau)$ = $1$

*Wlog, if sign$(\sigma) = 1$, sign$(\tau) = - 1 \implies$ sign$(\sigma \tau)$ = $-1$


but I'm not sure how to start this. Can someone give me a hint?
 A: sign is defined as 1 or -1 depending on whether the number of transpositions you can write a permutation in is even or odd.
if $\sigma$ is written as $k$ transpositions and $\tau$ is written as $t$ then $\sigma \tau$ is $k+t$ transpositions.
This proves that $\text{sgn}(\sigma)\text{sgn}(\tau)=\text{sgn}(\sigma\tau)$ because even + even = even, even + odd = odd, odd + odd = even
A: We define the signature of the permutation $\sigma$:
$$\epsilon(\sigma)=(-1)^N$$
where $N$ is the number of inversion.
Let $\sigma$ and $\tau$ two permutations with $P$ and $Q$ inversions respectively. The inversions of the composition $\tau \sigma$ are:


*

*the inversions $\{i,j\}$ of $\sigma$ s.t $\{\sigma(i),\sigma(j)\}$ is not an inversion for $\tau$

*The pairs $\{i,j\}$ which is not inversions of $\sigma$ s.t $\{\sigma(i),\sigma(j)\}$ is an inversion of $\tau$.
If we add the numbers of inversions of $\sigma$ and $\tau$ we have twice the number $R$ of inversions $\{i,j\}$ of $\sigma$ s.t. $\{\sigma(i),\sigma(j)\}$ is also an inversion of $\tau$. So the total number of inversions of $\tau\sigma$ is $N=P+Q-2R$. The signature of $\tau\sigma$ is
$$\epsilon(\tau\sigma)=(-1)^{P+Q-2R}=(-1)^P(-1)^Q=\epsilon(\tau)\epsilon(\sigma)$$
A: *

*Let $\pi\in S_n$ and set $\Delta=\prod_{1\leq i\leq j\leq n}(x_i-x_j)$.

*Set $\Delta^{\pi}=\prod_{1\leq i\leq j\leq n}(x_{(i)\pi}-x_{(j)\pi})$.

*Show that $\Delta^{\pi}=\text{Sgn}(\pi)\Delta$.

*Show that for $\pi,\phi\in S_n$, $\text{Sgn}(\pi\phi)=\text{Sgn}(\pi)\text{Sgn}(\phi)$ by showing that $(\Delta^{\pi})^{\phi}=\Delta^{{\pi}{\phi}}$.

*Note that $\Delta^{{\pi}{\phi}}=\text{Sgn}(\pi\phi)\Delta$ and $(\Delta^{\pi})^{\phi}=\text{Sgn}(\pi)\text{Sgn}(\phi)\Delta$.
A: Given that you are asking this question, you may have not seen the following.  But typically, $sgn$ is defined to be a homomorphism from any symmetric group to the multiplicative group $\{1,-1\}$, with the alternating group as kernel (i.e. the even permutations map to 1).  So your formula is immediate by the definition of a homomorphism. But again, my guess is that you have not been given this characterization of $sgn$.
A: Alternatively you can show that $\operatorname{sgn}$ factors through the group of $n \times n$ permutation matrices using the determinant map. Since $\det$ is a homomorphism we get the result.
