# permutations/order of sigma

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for sigma = 1 2 3 4 5 6 7 8 9 10

3 4 10 5 7 8 2 6 9 1

i know how to find the cycles, transpositions, & inverse but not the order of sigma, will somebody pls show/explain how, thanks

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for sigma = 1 2 3 4 5 6 7 8 9 10

3 4 10 5 7 8 2 6 9 1

i know how to find the cycles, transpositions, & inverse but not the order of sigma, will somebody pls show/explain how, thanks

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the order is the least n

such that sigma^n = sigma

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the order is the least n

such that sigma^n = sigma

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if u write that permutation in disjoint cycles, the order will be the lowest common multiple of the order of the disjoing cycles. which is quite easy to find ,

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if u write that permutation in disjoint cycles, the order will be the lowest common multiple of the order of the disjoing cycles. which is quite easy to find ,

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example:

sigma = 1 2 3 4 5 6
4 1 6 2 3 5


write it down in dijoint cycles you get

sigma = (142)(365)

now the order of sigma will be lcm(3,3)=3

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example:

sigma = 1 2 3 4 5 6
4 1 6 2 3 5


write it down in dijoint cycles you get

sigma = (142)(365)

now the order of sigma will be lcm(3,3)=3

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think i got it, thanks

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think i got it, thanks

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I am studying permutations for the first time in my life. I tried to solve the problem that mathsfun's posted. I think order of sigma is 1, since the disjoint cycles of his example are ( 1 3 10)( 2 4 5 7)( 6 8 ) (9), order of each respec: 3, 4, 2, 1 and the LCM of them is 1.

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That isn;'t the lcm, that's the gcd. the only element in any group of order 1 is the identity as well.

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Will you please provide another example and show me how to find the order of sigma? Thanks in advance.

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EH? You know its jsut the LCM, as in least common multiple, of the lengths of the disjoint cycles, so are you asking me to show you how to find the LCM or are you aksing me how to write a permutation as a product of disjoint cycles?

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I want you to provide me with some examples of how to write a permutation as a product of disjoint cycles.

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well the general way is to pick an element, then see what the permutation does to that element . in cyclic notation $(a_{1}a_{2}\dotsb a_{k})$ is a k cycle and read as $a_{1}\rightarrow a_{2},a_{2}\rightarrow a_{3}\dotsb a_{k-1}\rightarrow a_{k},a_{k}\rightarrow a_{1}$

an example would be a rotation of a square. Label each vertex, i.e 1,2,3,4. (starting from top left, and going clockwise)

rotate the square by 90 degrees anti clockwise round the origin. that will take vertex 1 to 2, 2 to 3, 3 to4, and 4 to 1

written in a cycle it gives (1234)

for a cycle its order is just its lenght.

take the same square and reflect on the horizontal line through its centre. that takes 1 to 2 . 2 to 1. 3 to 4 , 4 to 3

that can be written as (12)(34)

when you have a product of disjoint cycles then its order is the least common multiple. ill just try to show you how it works.

suppose you have

sigma = (abcd)(efg)(hiklm)

then sigma^n = (abcd)^n (efg)^n (hiklm)^n since they are disjoint

and each of those cycles have order of its lenght. so to get sigma^k=sigma you need all those three to be themselves as well. when n=4 . the first cycle is same . the rest arent. when n=3 the second one is same, the rest arent, when n=5, the third one is same the rest arent. so you keep on going.

now the first cycle is same when n is a multiple of 4, second when n is multiple of 3. and third when n is multiple of 5

so the least number k that satisfies being a multiple of 4,3,5 is lcm(4,3,5)=60

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You just do it:

start with 1, where is one sent? to x, say, where is x sent? to y, say, then where is y sent to? repeat until you hit 1, then the first of the cycles is: (1xy...) then pick some element not yet hit, and start again.

For instance, to simplify (123)(345)(43)(675)

1 is missed by 675, and the 43, and the 345, but 123 sends it to 2.

2 is missed by 675, and 43 and 345, but 123 sends it to 3.

3 is missed by 675, but 43 sends it to 4, that 4 is sent to 5 by the 345, and that is left alone by the 123 so 5 is spat out.

5 is sent to 6 by 675, and that 6 is left alone by the other ones so 6 is spat out.

6 is sent to 7 by 675, 7 is left alone by the rest so 7 is spat out

7 is sent to 5 by 675, 5 is left alone by 43, but sent to 3 by 345, that 3 is sent to 1 by 123, so 1 is spat out. stop.

the first cylce is:

(123567)

that just leaves the 4. let us check 4 is left alone

675 does nothing. 43 sends it to 3, 345 sends it back to 4, and that is left alone by 123, so 4 is indeed sent to 4.

so that element has order 6.

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