2
$\begingroup$

I was given the example as an illustration of structure of permutations in my lecture notes on algebra as shown below:

$\bigl(\begin{smallmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9\\ 2 & 1 & 3 & 4 & 9 & 6 & 5 & 8 & 7 \end{smallmatrix}\bigr)$ $= (12)(597) = (12)(57)(59) = (12)(36)(68)(36)(38)(57)(59)$

I get the first relationship was obtained by cycle decomposition, but how were 2nd and 3rd relationships obtained? If I may use some hints

$\endgroup$
1
  • $\begingroup$ Just as a general tip: this notation is a bit awkward at first, but I encourage you to try writing it out if it is confusing and see how each element is affected, then you'll gain intuition as you go. $\endgroup$
    – The Count
    Nov 6 '18 at 0:12
3
$\begingroup$

I doubt that there was any particular general method used. I should think the point is to show that a given permutation can be written as a product of cycles in many ways. To find an example you could take a given permutation $\pi$ and a random choice of cycles $\gamma_1,\ldots,\gamma_s$, write $$\pi=\gamma_1\cdots\gamma_s\sigma\ ,$$ then compute $\sigma$ (and write it as a product of cycles if you like).

The end of your example is possibly useful in that it shows how to write a $3$-cycle as a product of transpositions, $$(597)=(57)(59)\ ,$$ and the bit in the middle can be interpreted siuilarly, $$(36)(68)(36)(38)=(683)(386)=(683)(683)^{-1}=\iota\ .$$ But to sum up again, I think it's just a pretty random example.

BTW a cycle can be written as a product of transpositions in many ways, but IMHO the easiest way to remember is not the one used in your example but $$(a_1a_2a_3\cdots a_n)=(a_1a_2)(a_2a_3)(a_3a_4)\cdots(a_{n-1}a_n)\ .$$

$\endgroup$
2
$\begingroup$

I will try giving as many details as possible.

Second equality is just the equality $(a_1,\dots,a_n) = (a_1,a_n)(a_1,a_{n-1})\dots(a_1,a_2)$ (which is easy to prove) applied to the second factor. There are many other equalities of the same kind, for example $(a_1,\dots,a_n) = (a_1,a_2)(a_2,a_3)\dots(a_{n-1},a_n)$

Third equality is just multiplying in-between by the 4-terms factor $(36)(68)(36)(38)$, which can be seen to be the identity permutation the following way : $(36)(68)(36) = (38)$ as a special case of $\sigma(a_1,\dots,a_n)\sigma^{-1} = (\sigma(a_1),\dots,\sigma(a_n))$

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.