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Math Help - Permutation groups

  1. #1
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    Permutation groups

    1a) Prove that for any two 3-cycles \sigma_1, \sigma_2 \in S_n (n>4) there exists a \tau \in A_n such that \tau\sigma_1\tau^{-1}=\sigma_2
    1b) Let Z(S_n) be the center of S_n . Prove that Z(S_n)=Z(A_n)=\{e\} for all n>3 and Z(S_3)=\{e\}
    1c) Prove that the only nontrivial normal subgroup of S_n (n>4) is A_n
    1d) Give a nontrivial normal subgroup N of S_4 such that N \ne A_4
    can anyone help? thanks in advance
    Last edited by tszhin8831; November 23rd 2008 at 12:36 AM.
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  2. #2
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    Quote Originally Posted by tszhin8831 View Post
    1a) Prove that for any two 3-cycles \sigma_1, \sigma_2 \in S_n (n>4) there exists a \tau \in A_n such that \tau\sigma_1\tau^{-1}=\sigma_2
    If \sigma_1 = (a,b,c) then \tau \sigma \tau^{-1} = (\tau (a), \tau(b) , \tau (c)).
    Therefore, if \sigma_2 = (d,e,f) then let \tau be defined as a\mapsto d, b\mapsto e, c\mapsto f and fixes everything else.
    The problem is that \tau = (a,d)(b,e)(c,f)\not \in A_n.
    In order to fix that choose a'\not \in \{a,b,c\} and b'\not \in \{a',d,e,f\} and redefine \tau to send a'\mapsto b'.
    This is possible since n>4. And we see that \tau = (a,d)(b,e)(c,f)(a',b')\in A_n.
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  3. #3
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    Quote Originally Posted by tszhin8831 View Post
    1b) Let Z(S_n) be the center of S_n . Prove that Z(S_n)=Z(A_n)=\{e\} for all n>3 and Z(S_3)=\{e\}
    Let \sigma\in Z(S_n) if \sigma_1 is conjugate to \sigma then it means \tau \sigma \tau^{-1} = \sigma_1 \implies \sigma = \sigma_1.
    Therefore, elements in the center are only conjugate to themselves.
    We will show that any non-identity element in S_n ( n>2) are conjugate to some other element.
    Let \theta \in S_n be a cycle, \theta = (a_1,...,a_k), k\geq 2.
    Then \tau \theta \tau^{-1} = (\tau(a_1), ... , \tau(a_k)).
    Therefore if we pick \tau = (a_2,...,a_k,a_1) then \tau \theta \tau^{-1} \not = \theta if k>2.
    If k=2 then (a_1,a_2) and (a_2,a_1) are the same thing.
    But in this special case we can just let \tau = (a_1,a_3) where a_3 \not \in \{a_1,a_2\} and so \tau \theta \tau^{-1} \not = \theta.
    We have shown that no cycle can be in Z(S_n) for n>2.

    Let \sigma \in S_n be a non-identity permutation.
    Then we can write \sigma = \theta_1 \theta_2 ... \theta_j into a product of disjoint cycles.
    Therefore, \tau \sigma \tau^{-1} = ( \tau \theta_1 \tau^{-1})(\tau \theta_2 \tau^{-1}) ... (\tau \theta_j \tau^{-1}).
    For the moment assume there is one of \theta_1,...,\theta_j that is not a transposition, say, \theta_1 by relabing.
    Then it means, \theta_1 = (x_1,...,x_r) where r>2.
    Let \tau = (x_2,...,x_r,x_1) and we see that \tau \theta_i \tau^{-1} = \theta_i for i\not = 1.
    Therefore, \tau \sigma \tau^{-1} = \tau \theta_1 \tau^{-1} \theta_2 ... \theta_j while \sigma = \theta_1 ... \theta_j so \tau \sigma \tau^{-1} \not = \sigma since \tau \theta_1 \tau^{-1} \not = \theta_1.
    We have constructed an element conjugate and not equal to \sigma thus \sigma \not \in Z(S_n).
    But we left the case when each \theta_i is a transposition as an exceptional case.
    Try to see what you can come up with to complete the proof for Z(S_n) = \{ e \} for n>2.

    As for Z(A_n) the proof should be similar but needs to be done with more care.
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  4. #4
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    Quote Originally Posted by tszhin8831 View Post
    1c) Prove that the only nontrivial normal subgroup of S_n (n>4) is A_n
    If you happen to know that A_n is simple for n>4 then that is really helpful here. Accepting this fact we see that \{ e \} \subset A_n \subset S_n is a composition series. By Jordan-Holder theorem this is the unique composition series. Consequently, it must mean that A_n is the only proper non-trivial normal subgroup of S_n.
    1d) Give a nontrivial normal subgroup N of S_4 such that N \ne A_4
    can anyone help? thanks in advance
    Choose N = \{ \text{id}, (12)(34),(13)(24),(14)(23)\}.

    Here is another related problem can you show that N and A_4 are the only proper non-trivial normal subgroups of S_4?
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    thanks a lot
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