I am working through Dummit and Foote and would like some help/clarification for his proof of Proposition 23 [108].
Define $\Delta\doteq\prod_{1\leq i<j\leq n}(x_i-x_j)$ and for a permutation $\pi\in S_n$, $\pi(\Delta)\doteq\prod_{1\leq i<j\leq n}(x_{\pi(i)}-x_{\pi(j)})$. Define $\epsilon(\pi)=+1$ if $\pi(\Delta)=\Delta$ and $\epsilon(\pi)=-1$ if $\pi(\Delta)=-\Delta$.
Show that $\epsilon:S_n\rightarrow\{\pm 1\}\cong C_2$ is a group homomorphism.
Let $\tau,\sigma\in S_n$. We want to show that $\epsilon(\tau\sigma)=\epsilon(\tau)\epsilon(\sigma)$. To begin, suppose that $\sigma(\Delta)$ is a polynomial with $k$ inversions. In particular, this means $\epsilon(\sigma)=(-1)^k$. When computing $(\tau\sigma)(\Delta)$, we first apply $\sigma$ to the indices of $\Delta$, we see that $(\tau\sigma)(\Delta)$ has $k$ inversions. Those terms will be of the form $(x_{\tau(j)}-x_{\tau(i)})$ where $j>i$. Next, interchange the terms, i.e., $(x_{\tau(j)}-x_{\tau(i)})\rightarrow-(x_{\tau(i)}-x_{\tau(j)})$. $(\tau\sigma)(\Delta)$ now contains only terms of the form $(x_{\tau(p)}-x_{\tau(q)})$ where $p<q$. So, $$(\tau\sigma)(\Delta)=(-1)^k\prod_{1\leq p<q\leq n}(x_{\tau(p)}-x_{\tau(q)})=\epsilon(\sigma)\prod_{1\leq p<q\leq n}(x_{\tau(p)}-x_{\tau(q)})$$ Notice that $$\prod_{1\leq p<q\leq n}(x_{\tau(p)}-x_{\tau(q)})=\epsilon(\tau)\Delta$$ Is the reason because after applying $\tau$, there may be inversions, hence the $\epsilon(\tau)$? Thus, $$(\tau\sigma)(\Delta)=\epsilon(\sigma)\epsilon(\tau)\Delta\Rightarrow\epsilon(\tau\sigma)=\epsilon(\sigma)\epsilon(\tau)=\epsilon(\tau)\epsilon(\sigma)$$ Why does that imply $\epsilon(\tau\sigma)=\epsilon(\sigma)\epsilon(\tau)$?
It might be the notation that is screwing me up because the claim makes sense.
Thanks.
I think indexing obscures the main idea. Suppose a group $G$ acts on a set $X$. Then writing $\mathbb{Z}[X]$ for the polynomial ring with elements of X as the formal variables, $G$ acts on $\mathbb{Z}[X]$ as well. Now if X is finite, pick a total order on X and consider the special element $$\Delta = \prod_{x < y \in X} (x-y) \in \mathbb{Z}[X].$$ As $\{\Delta, -\Delta\}$ is closed under the G-action, there is a group homomorphism $$G \rightarrow Sym(\{\Delta, -\Delta\}) \cong C_2.$$ Your question is the case $G = Sym(X)$.