Solving Permutations with Multiple Sets

[taylor81792]

Homework Statement

The problem says to compute the expression shown for the permutations σ, τ, and μ.
My problem in particular says to compute |{σ}| for σ= (1 2 3 4 5 6; 3 1 4 5 6 2)

The Attempt at a Solution

My attempt to solve this problem was by first trying to change σ into σ^2. And then I tried continuing to double σ until I got the identity, but at that rate it would take forever.

 

[jbunniii]

Do you know how to find the cycle decomposition of a permutation?

 

[taylor81792]

I don't think my professor ever taught me that

 

[jbunniii]

Well, here's a crash course. \sigma maps 1 to 3, 3 to 4, 4 to 5, 5 to 6, 6 to 2, and 2 to 1. We say that \sigma has a cycle consisting of these elements, and we write it as (1 3 4 5 6 2). And we have accounted for all of the elements, so in fact \sigma consists of just this cycle.

If the above makes sense to you, can you see how many times you would have to apply \sigma in order to obtain the identity?

 

[taylor81792]

So i understood that and I continued doing that. I then got (1 2 3 4 5 6; 1 4 5 6 2 3). After doing it a couple more times, I ended up getting (1 2 3 4 5 6; 1 3 4 5 6 2) again. I don't know if I did something wrong.

 

[taylor81792]

Well, here's a crash course. \sigma maps 1 to 3, 3 to 4, 4 to 5, 5 to 6, 6 to 2, and 2 to 1. We say that \sigma has a cycle consisting of these elements, and we write it as (1 3 4 5 6 2). And we have accounted for all of the elements, so in fact \sigma consists of just this cycle.

If the above makes sense to you, can you see how many times you would have to apply \sigma in order to obtain the identity?

Or would you have to keep going back to the original σ for each new cycle?

 

[jbunniii]

So i understood that and I continued doing that. I then got (1 2 3 4 5 6; 1 4 5 6 2 3).

This doesn't look right. Can you explain how you got that?

 

[jbunniii]

If you want to write out the powers of \sigma, you can do it as follows, for example to calculate \sigma^2:

\sigma maps 1 to 3 and 3 to 4. Therefore \sigma^2 maps 1 to 4.
\sigma maps 2 to 1 and 1 to 3. Therefore \sigma^2 maps 2 to 3.
\sigma maps 3 to 4 and 4 to 5. Therefore \sigma^2 maps 3 to 5.
\sigma maps 4 to 5 and 5 to 6. Therefore \sigma^2 maps 4 to 6.
\sigma maps 5 to 6 and 6 to 2. Therefore \sigma^2 maps 5 to 2.
\sigma maps 6 to 2 and 2 to 1. Therefore \sigma^2 maps 6 to 1.

This means that \sigma^2 = (1 2 3 4 5 6 ; 4 3 5 6 2 1).

 

[taylor81792]

okay i did that and then I got (1 2 3 4 5 6; 6 5 2 1 3 4), then ( 1 2 3 4 5 6; 4 3 5 6 2 1). does this look right?

 

[taylor81792]

I just tried doing σ^3 and I got the inverse because
1 maps to 4, 4 to 6 and 6 to 1
2 to 3, 3 to 5, and 5 to 2.
3 to 5, 5 to 2, and 2 to 3.

Does this look correct now?

 

[taylor81792]

Okay, i redid it a final time and I believe the answer is 6

 

[jbunniii]

I just tried doing σ^3 and I got the inverse because
1 maps to 4, 4 to 6 and 6 to 1
2 to 3, 3 to 5, and 5 to 2.
3 to 5, 5 to 2, and 2 to 3.

Does this look correct now?

No, here's what I get:

1 to 3 to 4 to 5
2 to 1 to 3 to 4
3 to 4 to 5 to 6
4 to 5 to 6 to 2
5 to 6 to 2 to 1
6 to 2 to 1 to 3

so \sigma^3 = (1 2 3 4 5 6 ; 5 4 6 2 1 3).

 

[jbunniii]

Okay, i redid it a final time and I believe the answer is 6

Yes, that's correct.

 

[Deveno]

in general, if

σ = (a₁ a₂...a_k1)(b₁ b₂...b_k2)...(t₁ t₂...t_kr)

where each cycle is disjoint from all the others,

then |σ| = lcm(k₁,k₂,...,k_r)

for example, the order of σ =

(1 2 3 4 5)
(2 1 4 5 3) = (1 2)(3 4 5) is lcm(2,3) = 6.

also, if σ is a n-cycle that maps a_j→a_j+1 (mod n),

then σ^k maps a_j→a_j+k (mod n),

that is, instead of "jumping to the next number in the cycle (circle)", we skip to the k-th following number" ("looping back around when necessary").