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permutations/order of sigma


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  • 2 weeks later...

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

 

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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