This is exercise 9.6a from Section. The Rational Numbers in textbook Analysis I by Amann/Escher.
On the symmetric group $\mathrm{S}_n$, define the sign function by $$\operatorname{sign} \sigma := \prod_{1 \leq j<k \leq n} \frac{\sigma(j)-\sigma(k)}{j-k}, \qquad \sigma \in \mathrm{S}_{n}$$
Prove that
$\operatorname{sign}\left(\mathrm{S}_n\right) \subseteq\{ \pm 1\}$.
$\operatorname{sign}(\sigma \circ \tau)=(\operatorname{sign} \sigma)(\operatorname{sign} \tau)$ for $\sigma, \tau \in \mathrm{S}_{n}$.
Could you please verify if my attempt is fine or contains logical gaps/errors?
My attempt:
- We have
$$\begin{alignat}{2} |\operatorname{sign} \sigma| &= \left |\prod_{1 \leq j<k \leq n} \frac{\sigma(j)-\sigma(k)}{j-k} \right | &&= \frac{\prod_{1 \leq j<k \leq n} |\sigma(j)-\sigma(k)|}{\prod_{1 \leq j<k \leq n}|j-k|}\\ &= \frac{\prod_{1 \leq j'<k '\leq n} |j' - k'|}{\prod_{1 \leq j<k \leq n}|j-k|} &&= 1 \end{alignat}$$
The conclusion then follows.
- We have
$$\begin{align}\operatorname{sign} (\sigma \circ \tau) &= \prod_{1 \leq j<k \leq n} \frac{\sigma \circ \tau (j)-\sigma \circ \tau (k)}{j-k} \\ &= \prod_{1 \leq j<k \leq n} \frac{(\sigma \circ \tau (j)-\sigma \circ \tau (k))(\tau(j)-\tau(k))}{(j-k)(\tau(j)-\tau(k))} \\ &= \left (\prod_{1 \leq j<k \leq n} \frac{\sigma \circ \tau (j)-\sigma \circ \tau (k)}{\tau(j)-\tau(k)} \right ) \left ( \prod_{1 \leq j<k \leq n} \frac{\tau(j)-\tau(k)}{j-k} \right ) \\ &= \left (\prod_{1 \leq j'<k' \leq n} \frac{\sigma (j')-\sigma (k')}{j'-k'} \right ) \left ( \prod_{1 \leq j<k \leq n} \frac{\tau(j)-\tau(k)}{j-k} \right ) \\ &= (\operatorname{sign} \sigma) (\operatorname{sign} \tau) \end{align}$$
Update: On the basis of @Bernard's comment, I added an update to make my proof more rigorous.
In 1., for a pair $(j,k)$ with $j < k$: If $\sigma(j) < \sigma(k)$ then I define $j':= \sigma(j)$ and $k':= \sigma(k)$, otherwise I define $j':= \sigma(k)$ and $k':= \sigma(j)$. In either case, $|\sigma(j)-\sigma(k)| = |j'-k'|$.