Thread: Permutations and Conjugacy classes

1) "Explain what is meant by is a permutation of . The permutation is 123456 -> 635124, write in cycle notation. Find . How many permutations of are there in the conjugacy class of "

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. 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?

Originally Posted by TeaWithoutMussolini

1) "Explain what is meant by is a permutation of . The permutation is 123456 -> 635124, write in cycle notation. Find . How many permutations of are there in the conjugacy class of "

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. 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 then

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...

Originally Posted by TeaWithoutMussolini

1) "Explain what is meant by is a permutation of .

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

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

Correct.

Pi^-1, = 452631.

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

How many permutations of are there in the conjugacy class of

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.

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 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.