Help explaining the sign of a permutation I have a permutation that has been expressed in disjoint cycles (this isn't my actual question, this is an example done in lectures which I'm trying to understand):
(a b c)(d e f g h i)
Now the procedure taught to calculate the sign of the permutation is to multiply the signs of the disjoint cycles, but in the example it is written:
$ (-1)^2 \cdot (-1)^5 = -1 $
Why is it being raised to the powers of 2 and 5? Should it not be 3 and 6 because that is the length of the cycles?
Because when I try my question, I calculate the inversions of the permutation and using that gives me the sign to be negative, but the disjoint cycle has 8 "elements" so $ (-1)^8 = 1 $, which is not negative.
So does that mean you always do $(-1)^{\mathrm{(Length\, \, of\, \, cycle\, \, - 1)}} $ and then multiply like this?
 A: The sign of a cycle of length $n$ is indeed $(-1)^{n-1}$. That is because transpositions have sign $-1$ and permutations of length $n$ can easily be rewritten as the product of $n-1$ transpositions by just splitting them up. For example $(a b c) = (a b) (b c)$.
A: It turns out that $(a b c)$ can be written as $(ab)(ac)$, and in general an $n$-cycle can be written with $(n-1)$ transpositions.
This is why it's $2$ and $5$ as opposed to $3$ and $6$.
A: The sign of a permutation says whether you need an even or an odd number ow two-element swaps to achieve that permutation. Each swap gets a factor $-1$, so $k$ swaps get a factor of $(-1)^k$. Now how many swaps do you need to do a cycle of length $n$?
Well, for $n=2$, it's obvious: The cycle is swapping two elements. Therefore you need only one swap.
To implement the $n$-cycle, first cycle the first $n-1$ elements. This will bring almost all elements into the right place, except that the $(n-1)$-st element will end up at position $1$ instead of position $n$, and the $n$-th element will remain where it is instead of moving to place $1$. However both problems are fixed by a single additional swap, of the first with the last element. Therefore the $n$-cycle needs one swap more than the $n-1$-cycle, thus has the opposite sign.
From those two facts you get by simple induction that the $n$-cycle needs $n-1$ swaps and thus has sign $(-1)^{n-1}$.
Now your permutation is a product of a $3$-cycle and a $6$-cycle, therefore its sign is $(-1)^{3-1}\cdot(-1)^{6-1}=(-1)^5\cdot(-1)^2=-1$.
