Permutations and Conjugacy classes

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June 13th 2011, 08:43 AM
TeaWithoutMussolini

Permutations and Conjugacy classes

1) "Explain what is meant by $\pi$ is a permutation of $\bar{n} =(1,2...n)$ . The permutation $\pi$ is 123456 -> 635124, write in cycle notation. Find ${\pi}^{-1}$ . How many permutations of ${S}_{6 }$ are there in the conjugacy class of $\pi$ "

Ok for the first part I don't know how to word it. I know what a permutation does, but I don't know how to call it. $\pi$ sends each element to another? I don't know how to define this. Cycle notation I make it (1 6 4)(2 3 5). Just need a confirmation on this one. Pi^-1, = 452631. The conjugacy classes part I have no idea. I think it means with the same cycle types? But I don't know how to do this?

Also, how many permutations of 6 are there in total? Is it 6!?

2) "The following are permutations of 9. a=(12)(345)(78), b=(1234)(6789), c=(197)(34)(58). Calculate bc. Calculate the conjugate c.b.c^-1.

Does there exist a permutaiton oao^-1 = b, or oao^-1 = c."

Now I remember a rule which says the conjugate sends the part of one to the other, the corresponding part. So a conjugate of something has to have the same cycle type? So for this there would be one for c but NOT a? And for oao^-1 = c, o must be... I forget how to do it?
June 13th 2011, 10:50 AM
Also sprach Zarathustra

Re: Permutations and Conjugacy classes

Quote:

Originally Posted by TeaWithoutMussolini

1) "Explain what is meant by $\pi$ is a permutation of $\bar{n} =(1,2...n)$ . The permutation $\pi$ is 123456 -> 635124, write in cycle notation. Find ${\pi}^{-1}$ . How many permutations of ${S}_{6 }$ are there in the conjugacy class of $\pi$ "

Ok for the first part I don't know how to word it. I know what a permutation does, but I don't know how to call it. $\pi$ sends each element to another? I don't know how to define this. Cycle notation I make it (1 6 4)(2 3 5). Just need a confirmation on this one. Pi^-1, = 452631. The conjugacy classes part I have no idea. I think it means with the same cycle types? But I don't know how to do this?

Also, how many permutations of 6 are there in total? Is it 6!?

2) "The following are permutations of 9. a=(12)(345)(78), b=(1234)(6789), c=(197)(34)(58). Calculate bc. Calculate the conjugate c.b.c^-1.

Does there exist a permutaiton oao^-1 = b, or oao^-1 = c."

Now I remember a rule which says the conjugate sends the part of one to the other, the corresponding part. So a conjugate of something has to have the same cycle type? So for this there would be one for c but NOT a? And for oao^-1 = c, o must be... I forget how to do it?

1. If $\pi(j)=k$ then $\pi^{-1}(k)=j$

For: "Also, how many permutations of 6 are there in total? Is it 6!?"
The Answer is yes.

I don't understand the second question.

But if you need to find the composition of permutations try to put the permutations on matrices...
June 13th 2011, 02:43 PM
emakarov

Re: Permutations and Conjugacy classes

Quote:

Originally Posted by TeaWithoutMussolini

1) "Explain what is meant by $\pi$ is a permutation of $\bar{n} =(1,2...n)$ .

A permutation of a set A is a bijection from A to A.

Quote:

Cycle notation I make it (1 6 4)(2 3 5).

Correct.

Quote:

Pi^-1, = 452631.

Yes, if this is understood as the second row (under 123456) of a permutation, not as a cycle.

Quote:

How many permutations of ${S}_{6 }$ are there in the conjugacy class of $\pi$

Wikipedia says that conjugating preserves the cycle structure. So I think that the conjugates of (1 6 4)(2 3 5) are all possible products of two cycles of three elements.

You may get a better response about conjugating in the Algebra forum.
June 15th 2011, 02:28 AM
Deveno

Re: Permutations and Conjugacy classes

as emakarov points out, the term you are looking for is bijection, or as they are sometimes called "1-1 correspondences" (functions that are 1-1 AND onto). with a small finite set, it is sometimes easier to explicitly list the correspondence than to enumerate the "functional rule".

another way to write the inverse $\pi^{-1}$ is to "write the cycles in reverse order", so we would have (4 6 1)(5 3 2) = (1 4 6)(2 5 3).

it is not that hard to prove that if (a b c)(d e f) is any pair of disjoint 3-cycles, (a b c)(d e f) is conjugate to (1 6 4)(2 3 5). let σ be the permutation:

1-->a
2-->d
3-->e
4-->c
5-->f
6-->b, then (a b c)(d e f) = σ(1 6 4)(2 3 5)σ^-1. on the other hand σ(1 6 4)(2 3 5)σ^-1 = (σ(1) σ(6) σ(4))(σ(2) σ(3) σ(5))

for any permutation σ, so those are ALL the conjugates of (1 6 4)(2 3 5).

so, how many 3-cycles are there in S6? we can pick a in 6 ways, and then b in 5 ways, and then c in 4 ways: 6(5)(4) = 120.

but (a b c) = (b c a) = (c a b), so we must divide that total by 3, for 40 possible 3-cycles in S6.

but given any 3-cycle (a b c), we can only make the remaining elements {d,e,f} into a 3-cycle 2 ways: (d e f) and (d f e).

however, this does not give us 80 pairs of disjoint 3-cycles, because (d e f)(a b c) and (a b c)(d e f) are the same permutation.

so we have 40 pairs of disjoint 3-cycles.

for your second question, if you look closely i gave you a way to find a permutation σ such that σaσ^-1 = c, and since a and b have

diferent cycle structure, there is no way they can be conjugate.