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I need some verification with regards to my understanding of permutations, disjoint cycles, transpositions, signs, permutation inverse and order:

If a permutation is defined as :

1---2---3---4---5---6---7---8---9---10---11---12---13---14

14-11--10--7----5--9---1----4--12---3----6----2----13----8

(Sorry for formatting, I am new here)

So that (1,14) is one transposition.

When writing the permutation as disjoint cycles and transpositions, should (5) and (13) be included as follows?

Disjoint cycles: (1,14,8,4,7)(2,11,6,9,12)(3,10)(5)(13)?

Transpositions: (1,14)(14,8)(8,4)(4,7)(7,1)(2,11)(11,6)(6,9)(9,12)(12,2)(3,10)(10,3)(5,5)(13,13)?

As for the order, it is the LCM of all disjoint cycles' length = 10?

And the sign is (-1)^nom of transpositions = +

And as for finding the inverse of the permutation, all I need is to write the disjoint cycles in reverse:

Inverse Disjoint cycles: (7,4,8,14,1)(12,9,6,11,2)(10,3)(5)(13)?

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When writing the permutation as disjoint cycles and transpositions, should (5) and (13) be included as follows?

It doesn't change the element you're writing down if you include them or not - both are just funny ways of writing the identity element. (They're not transpositions, though, so you probably shouldn't include them in your product of transpositions for that reason.)

Disjoint cycles: (1,14,8,4,7)(2,11,6,9,12)(3,10)(5)(13)?

Looks fine to me.

Transpositions: (1,14)(14,8)(8,4)(4,7) (7,1) (2,11)(11,6)(6,9)(9,12) (12,2) (3,10) (10,3)   (5,5)(13,13)?

Two types of errors here:

  • In bold italics: these elements don't do what you want them to. Concentrate on just a single cycle first: what happens when you apply (1, 14, 8, 4, 7) to 7? What happens when you apply (1,14)(14,8)(8,4)(4,7) to 7? What if you used (1,14)(14,8)(8,4)(4,7)(7,1) instead?
  • In bold: these aren't transpositions. They're not even cycles. Cycles are always of the form $(a_1, \dots, a_n)$ for distinct $a_i$ (otherwise bad things can happen).

As for the order, it is the LCM of all disjoint cycles' length = 10?

Yes.

And the sign is (-1)^nom of transpositions

Yes. (You should recalculate this after the above.)

And as for finding the inverse of the permutation, all I need is to write the disjoint cycles in reverse:

Inverse Disjoint cycles: (7,4,8,14,1)(12,9,6,11,2)(10,3)(5)(13)?

Yes.

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  • $\begingroup$ Regarding the (7,1) thing, I dont quite understand what you which to convey. 7 goes back to 1 so should be (7,1) be a transposition? $\endgroup$ – Mohamad Moustafa May 15 '18 at 1:38

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