Verifying a Permutation: Even or Odd?

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1. I think my professor gave me credit for this problem when he shouldn't have! I just want to understand for the test, so I want someone to verify for me.
2. How do you find out whether a given permutation is even or odd without factoring it into transpositions?
3. My answer was:
Factor it into disjoint cycles (not necessarily 2-cyles, or transpositions).
Count the number of even cycles (those with an odd r).
If this count is even, the permutation is even. If this count is odd, then the permutation is odd.

but in looking back over my work to study for the test, I think it should have been:
Factor it into disjoint cycles (not necessarily 2-cyles, or transpositions).
Count the number of odd cycles (those with an even r).
If this count is even, the permutation is even. If this count is odd, then the permutation is odd.

Am I right in doubting my previous answer?
 
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Hi catherinenanc! :smile:

Nope. Your previous answer is correct.
An even cycle is an odd permutation.
With odd and odd you can make even, but with even and even you can only make even.
 


I like Serena said:
Hi catherinenanc! :smile:

Nope. Your previous answer is correct.
An even cycle is an odd permutation.
With odd and odd you can make even, but with even and even you can only make even.

You mean to say that that an "even cycle" is an "odd number of transpositions", right?
(BTW, odd r means (x1,x2,x3,...,xr) like (2,3,5) or (1,5,6,8,9).)

Also, I agree that "With odd and odd you can make even, but with even and even you can only make even" - that's precisely why I think my answer was wrong.
For example, Take (abc)(def)(ghij)(klmn) and (abc)(def)(ghij)(klmn)(opqs).
The first would be even+even+odd+odd=even
The second would be even+even+odd+odd+odd=odd.
 


catherinenanc said:
an "even cycle" is an "odd number of transpositions"

Or, rather, that an "even cycle" can be expressed as an "odd number of transpositions"
 


Ok the problem here is that our terminology is off. I used a term in my answer "even cycle" that threw us off. My bad.
My point is, shouldn't this be correct instead of my original?:

Factor it into disjoint cycles (not necessarily 2-cyles, or transpositions).
Count the number of odd permutations (those with an even r=number of values shown in a row).
If this count is even, the permutation is even. If this count is odd, then the permutation is odd.
 


Formulating it like that removes the ambiguity, which is better. :)

Btw, note that the decomposition into disjoint cycles is unique except for order.
 
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