1
$\begingroup$

Let us consider the permutation $(132)$

Then $(132)=(12)(13)$

Thus $(132)$ is an even permutation because it is a product of even number of transpositions which is $2$.

Now I recently came up with another definition of a permutation which says a permutation is even if it has even number of inversions which can be found here http://mathonline.wikidot.com/even-and-odd-permutations

So here we have only one inversion which is $(32)$ and hence it is odd.Thus $(132)$ is a odd permutation

So how can a permutation be both odd and even ??

Another question is if a permutation which is a cycle has length odd then it is an even permutation. How can I say wheher a permutation is odd/even using the definition of inversion?

For example $(132)$ has length 3 but is odd permutation since it has 1 inversion whereas $(123)$ is an even permutation since it has zero inversions,both having length $3$.

I cant understand what is going on

Can someone kindly help me out??

$\endgroup$
  • 4
    $\begingroup$ $(132) = \begin{pmatrix}1&2&3\\3&1&2\end{pmatrix}$ has two inversions ($3$ is before $1$ and $3$ is before $2$). The number of inversions is counted in the second line of the two-line representation of the permutation, not in the cyclic representation of the permutation. $\endgroup$ – JMoravitz Dec 14 '18 at 5:02
  • $\begingroup$ @JMoravitz,can you kindly answer the second question too $\endgroup$ – user596656 Dec 14 '18 at 5:14
  • $\begingroup$ @JMoravitz,i understood your point,can u answer the second one please $\endgroup$ – user596656 Dec 14 '18 at 5:15
  • 1
    $\begingroup$ A permutation can only be odd or it can be even, never both simultaneously. A permutation is an odd permutation according to the definition related to inversions if and only if it is also an odd permutation according to the definition related to transpositions. The easiest proof for showing a permutation whose cyclic representation consists of a single odd-length cycle is an even permutation is through the use of the transposition form of the definition and following that with a citation or a proof that the transposition form of the definition is equivalent to the inversion one. $\endgroup$ – JMoravitz Dec 14 '18 at 5:20
  • 1
    $\begingroup$ As for "how can I say whether a permutation is odd/even using the definition (of parity pertaining to) inversions", for a given permutation $\pi$ count the number of pairs $(i,j)$ such that $i<j$ and $\pi(i)>\pi(j)$. If the number of such pairs is odd then the permutation is said to be odd, otherwise it is said to be even. $\endgroup$ – JMoravitz Dec 14 '18 at 5:23
0
$\begingroup$

No permutation is both odd and even. $(123)$ is an even permutation. It is the cycle that sends $1\mapsto 2\mapsto 3\mapsto 1$. It is not the identity permutation. This cycle notation may be a bit confusing in this way if we also use two line notation, in that we also write the two line notation with parentheses and it means something completely different. I usually see one line notation without parentheses, so $123$ is the identity permutation, but $(123)$ is a cycle with an even number of inversions.

Note that you can write $(123) = (312) = (231)$. So your method of detecting inversions is not correct. An inversion in the cycle does not correspond to an inversion in the permutation.

In general a cycle of length $2k$ is an odd permutation, and a cycle of length $2k+1$ is even. This is a pretty simple rule. If you write a permutation as a product of disjoint cycles, the parity is additive as one would expect, as is true for any product of permutations. An easy way to remember this is as follows: $$(123) = (12)(23)$$ $$(2341) = (23)(34)(41) = (23)(34)(14)$$ You can in general split a cycle into a product of transpositions this way, and the number of transpositions, while not the number of inversions, has the same parity as such.

By the definition of a cycle, it is not terribly difficult to prove this multiplication rule. You should give it a shot. Even more fun, we have $$(125347) = (125)(5347) = (125)(534)(47)$$ etc. This splitting rule is a rule I find very useful.

As a warning, if you multiply permutations in the opposite order, as in not according to function composition, the pretty splitting rule disappears. Then you'd have $$(123) = (23)(12)$$ This to me is evidence that multiplication in that order is unnatural, but it may have some advantages that I'm not aware of. I find this is sufficient reason not to use it for my purposes, but of course if it is what you use in your course or textbook, it is what it is. You can still use my splitting rule, but you have to reverse the order. You can recover the naturality of the splitting rule by interpreting cycles in the opposite order, but as far as I know this is not done.

$\endgroup$

Your Answer

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