I had come across to one theorem which says
$\pi$=$\sigma (i_1,i_2,i_3,i_4..,,i_k)\sigma ^{-1}$
where $\pi$=($\sigma(i_1),\sigma(i_2),\sigma(i_3),\sigma(i_4))......,\sigma(i_k)$ )
$\pi$$\sigma(i_1)$=$\sigma (i_1,i_2,i_3,i_4..,,i_k)\sigma ^{-1}$ $\sigma(i_1)$=$\sigma(i_2)$.Upto This I understand Now to use this theorem Here
Given Permutation x=(1,2)(3,4) and y=(5,6)(1,3).
I wanted to find 'a' such that $axa^{-1}$=y.
$axa^{-1}$(a(1))=a(2)=y(a(1)) Here is confusion what is a(1) now here ?
How to use that proof to come to answer .
Sub question : I know that 2 elements that are not of equal cycle type are not conjugate to each other .But is this argument is enough to show that following
Prove that there is no a such that $a^{-1}(1,2,3)a$=(1,3),(5,7,8) .
Or there exist some proof that I have to give for this problem.
Any help will be appreciated
Line up the permutations by their cycle length; stacking them on top of each other. Then, let $a$ be the permutation that sends the corresponding numbers to each other.
So in this case we have $$(1\ 2)(3\ 4)(5)(6),$$ $$(5\ 6)(1\ 3)(2)(4).$$ So $$a=(1\ 5\ 2\ 6\ 4\ 3).$$