The sign of a given permutation If I have a permutation $\sigma$ on the set $A$ written by disjoint cycles. There are $n$ disjoint cycles can I then write the sign of the permutation as:
$\operatorname{sign}(\sigma) = (-1)^{|A|-n}$?
 A: You know a cycle of length $r$ can be written as $r-1$ transpositions (if you don't, it's good to convince yourself of why) so each disjoint cycle can be decomposed into transpositions.
If you can write $\sigma$ as $n$ disjoint cycles, each of length $r_i$, for $1\leq i\leq n$, you then have
$$\operatorname{sgn}(\sigma) = \prod_{i=1}^n (-1)^{r_i-1} = (-1)^{\sum_{i=1}^n (r_i - 1)}$$
But notice
$$\sum_{i=1}^n(r_i - 1) = \sum_{i=1}^n r_i - \sum_{i=1}^n 1 = \sum_{i=1}^n r_i - n.$$
So
$$\operatorname{sgn}(\sigma) = (-1)^{\left(\sum_{i=1}^n r_i\right) - n}.$$
Now what's the sum of the lengths of all the disjoint cycles?
As mentioned in the comments, it turns out the sign of a permutation is really quite a nice function. Namely, it is a homomorphism. One way to see this is to show that for any permutation $\sigma$ and transposition $\tau$ (permutations with the same underlying set of course) we have $\operatorname{sgn}(\sigma\tau) = -\operatorname{sgn}(\sigma)$. Using this we can then use the decomposition of an $r$-cycle into $r-1$ transpositions again, and one has, for any $\sigma_1,\sigma_2$ acting on $A$, of length $r_1,r_2$ respectively,
$$ \operatorname{sgn}(\sigma_1\sigma_2) = (-1)^{r_1 - 1}\operatorname{sgn}(\sigma_2).$$
But we can see $\operatorname{sgn}(\sigma_1) = (-1)^{r_1 - 1}$ and so
$$ \operatorname{sgn}(\sigma_1\sigma_2) = \operatorname{sgn}(\sigma_1)\operatorname{sgn}(\sigma_2).$$
A: Use the fact that $\mathrm{sgn}$ is a homomorphism to reduce the problem to the cycles. Then, show that for a cycle of length $\ell$, the sign for the cycle is $(-1)^{\ell-1}$.
So,
$$
  \mathrm{sgn}(\sigma)
  =
  \prod_{j=1}^n (-1)^{\ell_j - 1}
  =
  (-1)^{\sum_{j=1}^n (\ell_j - 1)}
  =
  (-1)^{|A| - n},
$$
because $\sum \ell = |A|$ and $\sum 1 = n$.
