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Prove: if $\sigma\in S_n$ is a factorization of $c$ disjoint cycles then $\text{sgn} (\sigma)=(-1)^{n-c}$.

We know the one cycle sign is $(-1)^{l-1}$ so $c$ of them is:

$$\text{sgn}(\sigma)=(-1)^{l-1}\cdot (-1)^{l-1}\cdot\cdot\cdot(-1)^{l-1}=(-1)^{c\cdot (l-1)}=(-1)^{c\cdot l-c}$$ and because they are disjoint cycles $n=c\cdot l$ and so $\text{sgn}(\sigma)=(-1)^{n-c}$.

Is this proof valid?

According to the proven sentence, the length of each cycle can be different, in Wikipedia there is the following fact:

"In practice, in order to determine whether a given permutation is even or odd, one writes the permutation as a product of disjoint cycles. The permutation is odd if and only if this factorization contains an odd number of even-length cycles."

Is this another way to calculate the sign? Or it is a special case of the sentence?

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  • $\begingroup$ 1) What is a "permutation of $c$ disjoint cycles"? My bets are you mean a permutation whose cycle decomposition consists of precisely $c$ cycles (where trivial cycles are counted in). When one says "permutation of [something]", it is usually this [something] that is getting switched around; but you certainly don't mean to switch around the cycles. $\endgroup$ – darij grinberg Jul 22 '15 at 16:18
  • $\begingroup$ 2) The lengths of the cycles don't have to be equal, so you cannot call them all $l$. $\endgroup$ – darij grinberg Jul 22 '15 at 16:18
  • $\begingroup$ @darijgrinberg 1) What I was meaning is cycles without a same number like (1 3)(2 4) I am translating from another language so I may using a wrong word or term. 2) yes the length do not have to be the same I used the fact that for any given cycle its sign is $(-1)^{l-1}$ $\forall l$ $\endgroup$ – gbox Jul 22 '15 at 16:25
  • $\begingroup$ So, yes, you want to say "a permutation whose cycle decomposition consists of precisely $c$ cycles". Beware, though, that these cycles might have different lengths: for example, $\left(1,2,5\right)\left(3,6\right)\left(4\right)$. $\endgroup$ – darij grinberg Jul 22 '15 at 16:26
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Your proof as it stands only works if there are $c$ disjoint cycles of an equal length, for example $(13)(26)(45)=(362542)$.

A more formal proof can be written as:

Prove: if $\sigma\in S_n$ is a factorization of $c$ disjoint cycles with $o$ odd length cycles and $e$ even length cycles, then $\text{sgn}(\sigma)=(−1)^{n−c}=(−1)^e$.

Odd length cycles always have sign $1$. After removing them, we have a permutation on $n-o$ elements, and this must be even as we only have even cycles left. So $n\equiv o\; \text{mod}2$ and we can write: $$(-1)^{n-c}=(-1)^{n-o-e}=(-1)^{n-o}\cdot (-1)^{-e}=1\cdot\dfrac{1}{(-1)^e}=(-1)^e$$

Your argument can be rescued here a bit, the parity of an even length cycle is always $-1$, so let $l=2k$ in your argument, and we can argue that:

We know the cycle sign for a cycle of even length is $(-1)^{2k-1}=-1$ so $e$ of them equals $(-1)^e$ and so $\text{sgn}(\sigma)=(-1)^e=(-1)^{n-c}$.

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First, a matter of terminology: if, like most authors, you consider cycles to be of length $>1$, then your statement is false. In order for it to be true, we shall agree to also speak of cycles of length $1$ (i.e. the fixed points of permutations).

If $\sigma_1, \sigma_2$ are disjoint cycles of lengths $l_1, l_2$, then $\sigma_1 \sigma_2$ acts on precisely $l_1 + l_2$ elements (the ones permuted by $l_1$ and $l_2$), leaving the other $n - l_1 - l_2$ fixed. If $\sigma = \sigma_1 \sigma_2 \dots \sigma_c$ then, by induction, $\sigma$ acts on precisely $l_1 + l_2 + \dots + l_c$ elements, and this number is precisely $n$ (because a decomposition into disjoint cycles exhausts a permutation). Now, $$\mathrm{sgn} \; \sigma = \mathrm{sgn} \; \sigma_1 \mathrm{sgn} \; \sigma_2 \dots \mathrm{sgn} \; \sigma_c = (-1)^{l_1 -1} (-1)^{l_2 -1} \dots (-1)^{l_c -1} = (-1)^{l_1 + l_2 \dots l_c -c} = (-1)^{n-c} ,$$ which is your desired result.

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