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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:

$\operatorname{sign}(\sigma) = (-1)^{|A|-n}$?

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Does fixed points of permutation regarded as cycles of length one? –  Andrew Oct 11 '11 at 19:13
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Yes............ –  user145 Oct 11 '11 at 19:16
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2 Answers

up vote 8 down vote accepted

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 $$\operatorname{sgn}(\sigma) = \prod_{i=1}^n(-1)^{\left(\sum_{i=1}^n r_i\right) - n}.$$ 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 $$ \operatorname{sgn}(\sigma_1\sigma_2) = \operatorname{sgn}(\sigma_1)\operatorname{sgn}(\sigma_2).$$

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Your answer is much better then mine. Very organized! :-) Maybe you would like to add the observation that $\mathrm{sgn}$ is a homomorphism of groups. –  André Caldas Oct 11 '11 at 19:46
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@AndréCaldas: Done, and thanks! I had the pleasure of going over a problem like this last night, so it was rather fresh in my mind :) –  Alex Oct 11 '11 at 20:10
    
Nice way to show that sign is a homomorphism! Well, it depends on how one defines the "sign". –  André Caldas Oct 11 '11 at 20:40
    
@AndréCaldas Well it doesn't actually matter how you defined your sign function as long as you've shown that composition with a transposition negates the value. You remove $\sigma_1$ one transposition at a time to get the first equality, then you "add" each of those transposition to the identity permutation to get the second equality. I didn't actually realize this until you mentioned the definition though, but it's rather neat! –  Alex Oct 12 '11 at 15:53
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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 $\sum 1 = n$.

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Dear André: I like your answer very much! +1 –  Pierre-Yves Gaillard Oct 11 '11 at 19:52
    
@Pierre-YvesGaillard: Thanks! :-) I found Alexe's answer much better detailed... but I think it is important to emphasize the fact that $\mathrm{sgn}$ is a homomorphism of groups. That's the only advantage of my answer... ;-) –  André Caldas Oct 11 '11 at 20:03
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