Abstract algebra: for a polynomial $p$, prove $\sigma(\tau(p))=(\sigma\tau)(p)$ for all $\sigma, \tau \in S_n$

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I'm trying to solve to following problem:

  • Part 1:

Let $\{x_1,...x_n\}$ be variables. For any polynomial $p$ in $n$ variables and for $\sigma$ $\in S_n$ define

$\sigma (p)(x_1,...,x_n)=p(x_{\sigma(1)},...,x_{\sigma(n)}$

Check that $\sigma(\tau(p))=(\sigma\tau)(p)$ for all $\sigma, \tau \in S_n$

My attempt: $\sigma(\tau(p))=\sigma(p(x_{\tau(1)},...,x_{\tau(n)}))=p(x_{\sigma(\tau(1))},...,x_{\sigma(\tau(n))})=(\sigma\tau)(p)$

  • Part 2:

Fix $n$ and define $\Delta=\prod_{1\leq i < j \leq n} (x_i-x_j)$.

Check that $\sigma(\Delta)=\pm\Delta$. Show that the map $\epsilon: \sigma\mapsto \sigma(\Delta)/\Delta$ is an homomorphism from $S_n$ to $\{-1,1\}$

My attempt: $\sigma(\Delta)/\Delta=\pm\Delta/\Delta=\pm1$, so $\epsilon: S_n\rightarrow\{-1,1\}$.

$\epsilon$ is homomorphism because $\epsilon(\sigma\tau)=(-1)^{k+l}=(-1)^k(-1)^l=\epsilon(\sigma)\epsilon(\tau)$

Is this correct?