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LuisRodg

Real Member

Registered: 2007-10-23

Posts: 322

Quick question on conjugate permutations.

Last edited by LuisRodg (2009-03-28 13:19:37)

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Quick question on conjugate permutations.

Two elements in Sn are conjugate if and only if they have the same cycle types.  So provided this is so, arrange their cycle types to line up, say:

(123)(57)(46) and (145)(23)(67)

Now send the first element of the first permutation to the first element of the second permutation, and so on:

1 -> 1
2 -> 4
3 -> 5
5 -> 2
7 -> 3
4 -> 6
6 -> 7

This gives (246735).  This will conjugate the first into the second.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

LuisRodg

Real Member

Registered: 2007-10-23

Posts: 322

Re: Quick question on conjugate permutations.

Excellent. Just tried it out and it works. So basically:

Is this correct?

Now, I have one more question. Lets say you have two arbitrary 3-cycles in A5. These 3-cycles are always conjugate (which is what im trying to prove). If I define gamma by the above definition then I have that those arbitrary 3-cycles are conjugate. However, my question is, does the conjugating permutation (gamma) also have to be in A5?

Last edited by LuisRodg (2009-03-29 04:35:46)

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Quick question on conjugate permutations.

Is this correct?

Yes.

I'm not entirely sure how much group theory you know, so if this doesn't make sense there is an easier but less general way to explain it.

Given a group G and an element g in G, we define:

C_g is called the conjugacy class of g.  It is called this because for any group, the conjugacy classes form equivalence classes.  So in C_g you will find all the elements which are conjugate to g.  Now your question:

However, my question is, does the conjugating permutation (gamma) also have to be in A5?

Becomes a bit trivial, if you look at it upside down.  Instead of asking "does their exist", let's just conjugate an element of A_5 by an element not in A_5:

(12)(123)(21) = (132)

So the answer is no.  But a more interesting question is: can we always take the conjugating element to be in A5?  In the previous example, if we conjugate by (12)(45) instead:

(12)(45)(123)(54)(21) = (132)

We get the same result.  The answer in general is no, which can be seen by taking any group with with a nontrivial center.  For example, in the integers modulo 2, the conjugacy class of 1 is just 1 itself.  However, to get from 1 to 1 you need to add and then subtract 0, which is not in the conjugacy class.

To see if this is the case with A_5, the only way I know is to do the computations by hand.

---

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

LuisRodg

Real Member

Registered: 2007-10-23

Posts: 322

Re: Quick question on conjugate permutations.

Yes it made sense.

So then if I want to prove the following:

Avon

Member

Registered: 2007-06-28

Posts: 80

Re: Quick question on conjugate permutations.

Your question does make sense and you are right.

To prove that

are conjugate you must find

such that

.

You already have an element

such that

.

Suppose

. If you could find an element

that commutes with

then you might consider the element

.

If you are interested in a similar problem, you might like to think about the number of conjugacy classes of 5-cycles in

.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Quick question on conjugate permutations.

Correct, you have not proved the statement.  What you want to do instead of finding the elements that conjugate, is just to count.  You should know of some ways to count the number of conjugates in a group (think about centralizers).  You also have the theorem:

Two elements of S_n are conjugate if and only if they have the same cycle type.  Specifically, all 3 cycles are conjugate in S_5.

A more explicit hint: Compute the number of 3 cycles in S_5, then compute the centralizer of (123) in A_5.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Guest2344

Guest

Re: Quick question on conjugate permutations.

Is there a way to show uniqueness for gamme or is it enough that it's implied?