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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?

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It's kind of obvious if you look at identity that it can't be $(-1)^{length of cycle}$ because then you would have $\epsilon (identity)=-1$ since identity is a cycle of length 1... –  xavierm02 Oct 2 '12 at 19:00
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3 Answers 3

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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 $(-1)^{3-1}\cdot(-1)^{6-1}=(-1)^5\cdot(-1)^2=-1$.

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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 $(a b c) = (a b) (b c)$.

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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 $6$.

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You mean $(ab)(bc)$. Or do you read them in the other direction? –  k.stm Oct 2 '12 at 18:31
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@K.Stm. I read $(ab)(ac)$ to mean that $a \to b$, $b \to a \to c$, $c \to b$, after Herstein's (admittedly backwards to perhaps the greater half of the world) tendency for permutations to act on the right. But I agree that yours works for permutations acting on the left (+1 to you too) –  mixedmath Oct 2 '12 at 18:35
    
@mixedmath: Exactly if permutations act on the right, I get $(ab)(ac)$ to mean $a\mapsto c$, $b\mapsto a$, $c\mapsto b$. Example: $(ab)(ac)a = (ab)c = c$. –  celtschk Oct 5 '12 at 8:08
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