I am trying to solve a problem that involves permutations of $\{1, ..., n\}$ with all cycles even. What does this mean? Could you please give an example of such permutation?

I understand that, e.g. when $n = 4$, a permutation $\langle2,1,4,3\rangle$ involves 2 cycles, namely $(1,2)$ and $(3,4)$. Are these "even cycles"?

  • $\begingroup$ I do not want to mark this as a duplicate, but I think you can get an answer here. What does it take to convert an odd permutation to an even permutation ? $\endgroup$ – Shailesh Sep 23 '15 at 1:17
  • $\begingroup$ This means a permutation that can be written as a product of cycles of even length (i.e. of signature $+1$). Your example illustrates this correctly. $\endgroup$ – Alex M. Sep 23 '15 at 21:09
  • $\begingroup$ @AlexM. Let $p$ be an even permutation. If $p$ is a cycle, then $p$ has odd length, right? $\endgroup$ – user198044 Oct 10 '18 at 13:43
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    $\begingroup$ @JackBauer: Rereading my comment above, it should have been "number of inversions" instead of "length". And yes, you are right. $\endgroup$ – Alex M. Oct 10 '18 at 14:01
  • $\begingroup$ @AlexM. Thank you! $\endgroup$ – user198044 Oct 10 '18 at 14:05

Any permutation is a product of disjoint cycles. If we say that "all cycles even", then you might want to seek clarification. This could mean all cycles have even length or all cycles are even permutations and thus have odd length.

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    $\begingroup$ I suspect that an even cycle is meant to be even as a permutation, therefore parity probably refers to the signature. $\endgroup$ – Alex M. Oct 10 '18 at 14:57

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