Permutation as a Product of Transposition

liger123
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hi guys.. can you help me prove this theorem?
Every permutation S_n where n>1 is a product of 2 cycles..
i got a little confused with some books' proof..thnx
 
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Can you show some work you've done on this?

The proof that comes to mind for me is to write each permutation as a product of cycles (you know how to do this, right?), and then you can explicitly describe how each cycle is a product of transpositions. For example, (1234) = (14)(13)(12).
 
Yes, the thing that i don't understand is how the formula was derived. this is the formula
(a_1, a_k) (a_1, a_k-1) ... (a_1, a_2).
 
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This is the question, I understand the concept, in ##\mathbb{Z_n}## an element is a is a unit if and only if gcd( a,n) =1. My understanding of backwards substitution, ... i have using Euclidean algorithm, ##471 = 3⋅121 + 108## ##121 = 1⋅108 + 13## ##108 =8⋅13+4## ##13=3⋅4+1## ##4=4⋅1+0## using back-substitution, ##1=13-3⋅4## ##=(121-1⋅108)-3(108-8⋅13)## ... ##= 121-(471-3⋅121)-3⋅471+9⋅121+24⋅121-24(471-3⋅121## ##=121-471+3⋅121-3⋅471+9⋅121+24⋅121-24⋅471+72⋅121##...
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##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
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The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
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