Question:
(a) Determine sign$(\tau)$ for $$\left( \begin{array}{ccccccc} 1&2&3&4&5&6&7 \\ 2&3&5&7&1&6&4 \end{array} \right )$$.
(b) Let $A_{n} = \{\tau \in S_{n} | \mbox{sign}(\tau) = 1\}$. Show that $A_{n}$ is a sub group of $S_{n}$ and calculate the number of elements (Tip: Show that for any transposition $\tau$, $S_{n}$ is the disjoint union of $A_{n}$ and $A_{n}\circ\tau$).
My attempt so far:
(a) So I have this formula for sign$(\sigma) = \displaystyle \prod_{i < j} \frac{\sigma(j)-\sigma(i)}{j-i}$. This wouldn't be so difficult, however it's apparently ok to count the number of elements for which $i
(b) To begin with, if I want to show that something is a subgroup, I have to show that: (i)$ a,b \in U \Rightarrow a \circ b \in U$ and that (ii)$a \in U \Rightarrow a^{-1} \in U$ correct? (I'm actually a little confused at why we can just assume that the identity element is still part of a subgroup, for instance if the exercise were a little different and I was checking if $A_{n} = \{\tau \in S_{n} | \mbox{sign}(\tau) = -1\}$ was a sub group, I would say that the identity element for permutations $123...n$ has no inversions $\Rightarrow$ sign$(\tau) = 1 \Rightarrow \tau \notin A_{n}$... I mean, subgroups are groups themselves, right?
Well anyway, for (i) I tried to show it by writing:
Let $\tau, \sigma \in A_{n}$
$\Rightarrow$ sign$(\tau) = 1$ and sign$(\sigma)=1$
For a polynomial (I honestly don't totally understand this or know if I need to put this stuff with 'g' in...) $g=g(x_{1},...,x_{n})$, sign$(\tau \circ \sigma)g= (\tau \circ \sigma)(g)=\tau(\sigma(g))=\tau((\mbox{sign}\sigma)g)=(\mbox{sign}\tau)(\mbox{sign}\sigma)g$
$\Rightarrow$ sign$(\tau \circ \sigma) = (1)(1) = 1 \Rightarrow (\tau \circ \sigma) \in A_{n}$
with (ii) my attempt was:
Let $\sigma \in A_{n}$
If $\sigma \circ \sigma^{-1} = \epsilon$ (this should represent the identity element)
$\Rightarrow$ sign$(\sigma \circ \sigma^{-1}) = \mbox{sign}\epsilon$
$(\mbox{sign}\sigma)(\mbox{sign}\sigma^{-1})=1$ (Because sign$(\epsilon)=1$)
$(1)(\mbox{sign}(\sigma^{-1}))=1$ (Because $\sigma \in A_{n}\Rightarrow \mbox{sign}(\sigma) = 1$)
sign$(\sigma^{-1})=1 \Rightarrow \sigma^{-1} \in A_{n}$
Out of (i) and (ii) follows the conclusion that $A_{n}$ is a subgroup of $S_{n}$...
So hopefully that would all be on the right track so far... With the calculate the number of elements part, I don't know what to do or why. Following the tip I would try to show that $S_{n}$ is the disjoint union (not the same as showing that the intersection is empty, right?) of $A_{n}$ and $A_{n} \circ \tau$ (where $\tau$ is a transposition). I don't understand this, but directly out of the book is the statement "We call $\tau$ a transposition. If $i