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?

  • $\begingroup$ 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

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 $\text{sgn}(\sigma)=(-1)^e=(-1)^{n-c}$.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.