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I am having trouble with the following passage of "Algebra and Geometry" by Alan F. Beardon in the highlighted part. I've read it several times but I do not understand what he is saying. Can you give me more explanation? Perhaps a couple of examples? Thanks.

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I am thinking of the highlighted part as:

  1. If x is an element of orbit j, then the cycle permutation j on x is equal to the original permutation of x. Meaning that cycles over elements that are parts of orbits will give the same outcome as original permutations would have given over the same element.

  2. From the answer of Shubham I think that the second part says that a cycle will only fix an element if it is disjoint from other cycles. Is "i ≠ j" meaning disjoint? I thought that disjoint and not equal were not the same notions.

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$\rho$ is original permuation .

By that way you can able to decompose $\rho$ as component function which moves distant elements

Take example $\rho=(123)(45)$

There are 2 distant cycle $\rho_1=(123),\rho_2=(45)$.

Now if you take 1 which is in first cycle $\rho_1(1)$=$\rho(1)=2$

BUT as $\rho_2$ is disjoint from $\rho_1$ it fixes 1 that is $\rho_2(1)=1$.

In this way you can decompose original permutation into component permutation

Edit:

I do not think your 2nd statement is correct

As every cycle which does not contain x , fixes it .

That is if you are having disjoint cycle then only one cycle moves other send x to x only

Function fixes x means $f(x)=x$

And for statement 1:

Cycle containing x is part of original cycle so how it could be whole cycle?

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  • $\begingroup$ I've made a mess, I re-wrote the question but I realize that I do not know where is precisely my confusion. I'll think hard and ask better. Thanks. $\endgroup$ – César D. Vázquez Nov 29 '18 at 15:18

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