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").