I was given the following permutation \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 6 & 5 & 2 & 4 & 3 & 1 \\ \end{pmatrix}

and I was asked to write it as the product of disjoint cycles.

The disjoint cycles I found were $(1, 6)$ and $(2, 5, 3)$, which my textbook said was correct. However, my question is what about the $4$?

I originally had my answer as the product of 3 disjoint cycles (ie. $(1, 6)(2, 5, 3)(4)$) but my professor said that $(4)$ shouldn't be there since $4$ is fixed in the permutation. I can see that $4$ is fixed but I still don't quite understand why we can have $(4)$ as a part of the answer.

If anyone could help me understand, it would be greatly appreciated. Thanks in advance.

  • $\begingroup$ Often, "1-cycles" are considered redundant as they act the same as the identity permutation. Note that $\sigma\circ (1)=(1)\circ \sigma=\sigma$ for any permutation $\sigma$. We prefer to include only the most relevant information in order to save space. Consider the permutation in $S_{1000}$ which swaps the positions of $1$ and $2$ and doesn't change the positions of any other element., It is much easier to write this as $(1~2)$ as opposed to $(1~2)(3)(4)(5)\cdots (999)(1000)$. $\endgroup$ – JMoravitz Jun 9 '17 at 14:20

You can have $(4)$ if you want. It may be "fixed," but it's also a "$1$-cycle" disjoint from the rest, so it still fits within the definition.

It's really just a matter of notation and clarification based on what's convenient. If you are omitting $1$-cycles, however, it's usually a good idea to state the permutation's size/degree so it's clear that the omitted elements are fixed points (as opposed to not being part of the permutation in the first place).

Example: http://mathworld.wolfram.com/PermutationCycle.html

  • $\begingroup$ Ahh, I see, so it's more of a notation error than an actual error in the math? That makes sense. Thank you $\endgroup$ – Smeef Jun 9 '17 at 14:28
  • $\begingroup$ @Smeef Right -- $(1, 6)(2, 5, 3)(4)$ is technically a correct answer to your question as stated, but if your professor wants you to omit the $1$-cycles (in the context of knowing the permutation's degree), it conveys the same information. $\endgroup$ – Marcus Andrews Jun 9 '17 at 14:33
  • $\begingroup$ Dear Marcus: I think you might want to use the term "order" of a permutation, which is the lcm of the lenghts of its disjoint cycles, in this case $lcm(2, 3, 1) = lcm(2, 3) = 6$. So the one-cycle plays no roll in determing the order of the permutation. $\endgroup$ – Namaste Jun 9 '17 at 14:45

Indeed, the cycles are (1,6), (2,5,3), and (4) where (4) is a fixed point. By convention, fixed points can be left out if you know the degree of the permutation, here $n=6$.

  • $\begingroup$ That is, the order of the permutation, here $n=6$. $\endgroup$ – Namaste Jun 9 '17 at 14:45

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