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Conjugacy in A_n

  1. Apr 4, 2004 #1
    What is a rule/method for determining whether two elements in the alternating group A_n are conjugate?
    Or, given an element sigma in A_n, how do you determine the conjugacy class of sigma in A_n?
    References/additional resources on the topic appreciated.
     
  2. jcsd
  3. Apr 5, 2004 #2

    matt grime

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    Hmm, it can be tricky can't it? Well, simplest is to just memorize the conjugecy classes for small n, and ignore large n.

    But if that's not acceptable, here's a start on how to think about it:

    (12345) is not conjugate to (21345) in A_5 or A_6 but is in larger A_n.

    Swapping over two entries is an odd permutation, and if there's not enough room left over to correct that oddness by conjugating with a disjoint cycle then they can't be conjugate, otherwise they are. So in that example, I'd need to conjugate with (12), in A_7 I can use the cycle (67) and then conjugation by (12)(67) is a conjugacy in A_7
    Basically, if I recall correctly, then the conjugacy classes are those of S_n possibly split in two depending on which conjugacies can only be obtained by odd elements.
     
  4. Apr 5, 2004 #3
    Can this idea be generalized?
    For instance, would (12345) and (32145) be conjugate only through an even permutation, since I've swapped over 3 entries? I'm thinking
    (32145)-->(23145)-->(21345)-->(12345) requires three swaps, and since these cylces are conjugate by (123)=(13)(12), I feel like I'm on my way to a simple generalization.
    How about (12345) and (52341)?
    (52341)-->(25341)-->(23541)-->(23451)=(12345) so again I needed three swaps but we find that these two cycles are conjugate by (12)... hmmm... maybe my generalization needs work. I guess I could have looked at this as (51234) and (15234) which is essentially your example.
     
  5. Apr 5, 2004 #4

    matt grime

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    12345 and 32145 - you've only swapped two entries - 321 is an odd permutatation of 123
     
  6. Apr 5, 2004 #5
    ..............ok, how about this, what's an odd permutation?
     
  7. Apr 5, 2004 #6

    matt grime

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    A transposition is when you swap two objects. If you need to do an odd number of transpositions the permutation is odd, if you do an even number of transpositions the permutation is even.

    take 123, transpose 12 to get 213, so 213 is an odd perm of 123. as is 321 and 132.

    the even perms of 123 are: swap 12 then 23 to get 312, and so on these are the cyclic permutations

    in summary

    123, 231, 312 are the even perms of 123, and 213, 132, 321 are the odd ones.

    Inside A_3 these odd and even perms are not all conjugate. Inside A_5 they are.
     
  8. Apr 5, 2004 #7
    Is it correct to say that if you conjugate a k-cycle by a 2-cycle, then the entries in that are in the 2-cycle are swapped in the k-cycle? For example,
    (34)(2345)(34)=(2435)
    (13)(1436)(13)=(3416)
    (13)(4365)(13)=(4165)
    This is a very useful observation! Now I know if I conjugate a k-cycle by an even permutation, there will be an even number of transpositions -- hence the term "even"? We learned that an even permutation is one that can be written as an even number of two cycles. I never realized the term actually had more significance.
    So, (12345) and (14325) are NOT conjugate in A_5 but (12345) and (34125) are? The latter case is because I can swap the 3 with the 1 by conjugating with (13) and I can swap the 4 and the 2 by conjugating with (42). Hence, (12345)=(42)(13)(34125)(13)(42). This is really beautiful stuff here!
     
    Last edited: Apr 5, 2004
  9. Apr 5, 2004 #8

    matt grime

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    You're on the right track but remember the redundancy that might exist in larger groups.

    123 and 213 are not conjugate in A_3 but they are in A_5 by say, (12)(45) using the spare 4 and 5.
     
  10. Apr 5, 2004 #9
    Hence if we have any two permutations (with the same cycle type) whose "total length" is less than n-1, these two cycles are conjugate in A_n. However, it is quite discouraging to try to compute the size of the conjugacy class for permutations other than these types. I tried doing it for (12345) in S_5 by reasoning that there are n(n-1)/2=10 ways to conjugate this by a two cycle (thus, these 10 conjugates are not conjugate to (12345) in A_5). The actual size of this conjugacy class in A_5 is 12 (while the size of the conjugacy class of (21345) is also 12), so I'm overlooking two conjugates. I imagine this is due to the fact that we may get new permutations if we conjugate (12345) by a 4-cylce, but I don't have any idea how to count how many such conjugates exist, and how many repeat with the 10 I've already found.
     
  11. Apr 5, 2004 #10

    matt grime

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    I've got a formula for sizes of conjugacy classes in S_n (and hence A_n) lying around somewhere. I'll give them to you tomorrow when I've found them in one of the folders by my desk. Either the classes are the same or they split into two equal sized parts.
     
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