How can I show that the number of even permutations is equal to the number of odd permutations?
You know that $\displaystyle \left[S_n:A_n\right]=2$ and so $\displaystyle S_n=A_n\cup xA_n$ for any $\displaystyle x\notin A_n$, but since all cosets have the same size this implies that $\displaystyle \displaystyle |A_n|=|xA_n|=\frac{n!}{2}$ but you can easily check that $\displaystyle xA_n$ are the odd permutations of $\displaystyle S_n$.
Another, essentially the same, idea is to prove that the map $\displaystyle \sigma\mapsto x\sigma$ is a bijection $\displaystyle A_n\to\{\text{odd permutations}\}$ where $\displaystyle x$ is any odd permutation.
Yeah, that's precisely what CAN'T happen, it is actually phrased the exact way you did, with the appropriate negation insereted, in most textbooks. A different way of looking at this is that there is a natural action of $\displaystyle S_n$ on the set $\displaystyle \left\{\pm \Delta\right\}$ where $\displaystyle \displaystyle \Delta(x)=\prod_{i<j}(x_i-x_j)$ and the action is $\displaystyle \displaystyle \sigma(\Delta)=\prod_{i<j}(x_{\sigma}(i)}-x_{\sigma(j)})$. We can then think of the set of even actions as the stabilizer of $\displaystyle \Delta$ from where everything I've told you follows.
If you haven't done group actions yet, just ignore that.
A finite set with two or more elements has equal numbers of even and odd permutations.
Proof. For each even permutation, we can obtain a unique odd permutation by transposing the first two elements. This defines a one-to-one correspondence between even and odd permutations, hence there are equal numbers of each.
I found this proof somewhere. Can you explain with an example what they mean by transposing the first two elements. This is what I was meaning...
the key fact here is that if a permutation can be expressed as a product of an odd (resp. even) number of transpositions, it can ONLY be expressed as a product of an odd (resp.even) number of transpositions.
this is a deep fact, and one which is often used to PROVE that [Sn:An] = 2. this is equivalent to proving the "signum" (or sign) function sgn:Sn-->{-1,1} is well-defined, after which it is easy to show that |An| = |Sn|/2, and that An is normal. one way of proving this is to invoke the Vandermonde polynomial:
$\displaystyle p(x_1,x_2,...,x_n) = \prod_{i<j}(x_i-x_j)$
and let a permutation σ in Sn act on p by taking:
$\displaystyle p(x_1,x_2,...,x_n) \rightarrow \prod_{i<j} (x_{\sigma(i)}-x_{\sigma(j)})$
odd permutations take p-->-p, while even permutations take p-->p.
my point is: all the difficulty is buried in ensuring that "even permutation" and "odd permutation" are well-defined terms, that we cannot have a permutation that is both. until this is done, An is not a well-defined set.
if we call our initial (even) permutation σ, and our second permutation τ, then this means:
τ(1) = σ(2)
τ(2) = σ(1)
τ(k) = σ(k), k = 3,4,...,n
this is equivalent to writing: τ = (σ(1) σ(2))σ.
note that τ has one more transpostion in its product than σ does.