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Math Help - conjugacy classes

  1. #1
    Senior Member abhishekkgp's Avatar
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    conjugacy classes

    Let g_1, g_2, \text{...},g_r be the representatives of the conjugacy classes of the finite group G and assume that these elements pairwise commute. Prove that G is abelian.
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    Senior Member Tinyboss's Avatar
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    Did you try anything or have any ideas yet? How far did you get?
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  3. #3
    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by Tinyboss View Post
    Did you try anything or have any ideas yet? How far did you get?
    i took arbitrary elements from the conjugacy classes represented by g_i \text{ and } g_j, viz, r g_i r^{-1} \text{ and } s g_j s^{-1} \text{ where } r, s \in G
    these two should commute given that g_i \text{ and } g_j commute.
    when i work it out it comes down to saying that r g_i r^{-1} \text{ and } s g_j s^{-1} commute iff [(s^{-1} r) g_i (s^{-1} r})^{-1}]g_j = g_j[(s^{-1} r) g_i (s^{-1} r})^{-1}]
    Calling s^{-1}r=h we have that  h g_i h^{-1}\text{ and } g_j should commute. here i am stuck.
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    Hmm.. well that's not the way to go. Just because representatives of classes commute, doesn't mean that all elements between the classes commute. For example, consider the classes [(1\,2)],[(3\,4\,5)]\subseteq S_5. There must be another way..

    EDIT: Try this.

    First, we need the fact that, for any finite group G and subgroup H, G\neq \bigcup_{a\in G}a^{-1}Ha.

    Now, we let G act on itself by conjugation. Since it is assumed that the g_i commute pairwise, we have g_i^{-1}g_kg_i=g_k for all 1\leq i,k \leq r. That is, for each k, g_1,\ldots ,g_r\in G_{g_k) (the stabilizer of g_k under this group action).

    Let g\in G be arbitrary. Then there exist i\in \{1,\ldots,r\},x\in G such that g=x^{-1}g_ix. Hence g=x^{-1}g_ix\in x^{-1}G_{g_k}x for each k.

    We have shown that, for any g\in G, we can find x\in G so that g\in x^{-1}G_{g_k}x. It follows that G=\bigcup_{x\in G}x^{-1}G_{g_k}x (for each k).

    By the claim above, since G is finite, this forces G=G_{g_k} for each k. That is, the conjugacy class of g_k is \{g_k\}. We know that this means g_k\in Z(G) and, since these elements form a complete set of representatives for the conjugacy classes of G, G=\{g_1,\ldots,g_r\}, and the group is abelian.
    Last edited by topspin1617; March 7th 2011 at 11:52 AM.
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    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by topspin1617 View Post
    Hmm.. well that's not the way to go. Just because representatives of classes commute, doesn't mean that all elements between the classes commute. For example, consider the classes [(1\,2)],[(3\,4\,5)]\subseteq S_5. There must be another way..

    EDIT: Try this.

    First, we need the fact that, for any finite group G and subgroup H, G\neq \bigcup_{a\in G}a^{-1}Ha.

    Now, we let G act on itself by conjugation. Since it is assumed that the g_i commute pairwise, we have g_i^{-1}g_kg_i=g_k for all 1\leq i,k \leq r. That is, for each k, g_1,\ldots ,g_r\in G_{g_k) (the stabilizer of g_k under this group action).

    Let g\in G be arbitrary. Then there exist i\in \{1,\ldots,r\},x\in G such that g=x^{-1}g_ix. Hence g=x^{-1}g_ix\in x^{-1}G_{g_k}x for each k.

    We have shown that, for any g\in G, we can find x\in G so that g\in x^{-1}G_{g_k}x. It follows that G=\bigcup_{x\in G}x^{-1}G_{g_k}x (for each k).

    By the claim above, since G is finite, this forces G=G_{g_k} for each k. That is, the conjugacy class of g_k is \{g_k\}. We know that this means g_k\in Z(G) and, since these elements form a complete set of representatives for the conjugacy classes of G, G=\{g_1,\ldots,g_r\}, and the group is abelian.
    now that's quite something!
    i agree, the proof is complete.
    but i can understand the proof. there was no way that i would think that way.
    how does all that come to ones mind?

    EDIT: AND THIS TOO IS A SERIOUS QUESTION, AS SERIOUS AS THE THREAD ITSELF!!
    Last edited by abhishekkgp; March 7th 2011 at 06:15 PM. Reason: left out something important.
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  6. #6
    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by topspin1617 View Post
    First, we need the fact that, for any finite group G and subgroup H, G\neq \bigcup_{a\in G}a^{-1}Ha.
    to prove this please check whether this is correct or not.
    g_1 H g_1^{-1} = g_2 H g_2^{-1}

    \iff g_2^{-1} g_1 H (g_2^{-1} g_1)^{-1}=H

    \iff g_2^{-1} g_1 \in N_G(H)

    \iff g_1 N_G(H)=g_2 N_G(H)

    so maximum number of distinct non-identity elements which could be there in \bigcup_{a\in G}a^{-1}Ha is (|H|-1)|G:N_G(H)|
    now since H \leq N_G(H) the proposition follows.
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