Results 1 to 4 of 4

Math Help - Conjugates

  1. #1
    Newbie
    Joined
    Oct 2008
    Posts
    6

    Conjugates

    How many conjugates does the permutation (123) have in the group S<sub>3</sub> of all permutations on 3 letters? Give brief reasons for your answers.

    Answers were provided for this question, but after going carefully through it, I still don't know what's going on at one point in the answer, which may prove crucial to my understanding of the problem (where I've put question marks). Below is the solution provided:

    "We can use the stabilizer-orbit relationship to see that the size of the conjugacy class of (123) is equal to the index of the stabilizer. But the stabilizer is a subgroup of S3 which contains at least <(123)> [????? why]. If it were to contain more then it must be all of S3 (by Lagrange's theorem) and so contain, for example, (12). But (12)(123)(12)-1 = (132) and so the centralizer is exactly <(123)> [????]. Since it has order 3, it also has index 2 and so (123) has 2 conjugates."

    Thank you.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by davido View Post
    How many conjugates does the permutation (123) have in the group S<sub>3</sub> of all permutations on 3 letters? Give brief reasons for your answers.

    Answers were provided for this question, but after going carefully through it, I still don't know what's going on at one point in the answer, which may prove crucial to my understanding of the problem (where I've put question marks). Below is the solution provided:

    "We can use the stabilizer-orbit relationship to see that the size of the conjugacy class of (123) is equal to the index of the stabilizer. But the stabilizer is a subgroup of S3 which contains at least <(123)> [????? why]

    Because for ANY x\in G, for ANY group G, we have that x^{-1}xx=x\Longrightarrow x centralizes itself, which simply means that any element in any group commutes with itself and, in fact, with any of its powers


    . If it were to contain more then it must be all of S3 (by Lagrange's theorem) and so contain, for example, (12). But (12)(123)(12)-1 = (132) and so the centralizer is exactly <(123)> [????]


    As seen above, (123) centralizes (123) and thus the group generated by (123), i.e. <(123>, centralizes (123)

    Tonio


    . Since it has order 3, it also has index 2 and so (123) has 2 conjugates."

    Thank you.
    .
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Swlabr's Avatar
    Joined
    May 2009
    Posts
    1,176
    Quote Originally Posted by davido View Post
    How many conjugates does the permutation (123) have in the group S<sub>3</sub> of all permutations on 3 letters? Give brief reasons for your answers.

    Answers were provided for this question, but after going carefully through it, I still don't know what's going on at one point in the answer, which may prove crucial to my understanding of the problem (where I've put question marks). Below is the solution provided:

    "We can use the stabilizer-orbit relationship to see that the size of the conjugacy class of (123) is equal to the index of the stabilizer. But the stabilizer is a subgroup of S3 which contains at least <(123)> [????? why]. If it were to contain more then it must be all of S3 (by Lagrange's theorem) and so contain, for example, (12). But (12)(123)(12)-1 = (132) and so the centralizer is exactly <(123)> [????]. Since it has order 3, it also has index 2 and so (123) has 2 conjugates."

    Thank you.
    This can actually be generalised in quite a neat way:

    Two permutations in S_n are conjugate if and only if they have the same cycle structure.

    To do this, you simply need to prove the following result, which is also very nice (and useful):

    Let i_1, i_2, \ldots i_k \in \{1, 2, \ldots, n\}. Then for \sigma \in S_n we have \sigma^{-1} (i_1 \text{ } i_2 \ldots i_k) \sigma=(i_1 \sigma \text{ } i_2 \sigma \ldots i_k \sigma).
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Oct 2008
    Posts
    6
    I see now, thanks so much. I didn't take much notice of "any element in any group commutes with itself and, in fact, with any of its powers", especially about the powers.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Galois conjugates
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: August 25th 2010, 10:43 AM
  2. G cannot be the union of conjugates
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: July 15th 2010, 03:02 PM
  3. No. of conjugates
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: February 1st 2010, 06:10 PM
  4. conjugates
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: January 8th 2010, 02:56 PM
  5. conjugates
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: April 27th 2008, 07:02 PM

Search Tags


/mathhelpforum @mathhelpforum