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Math Help - Groups-Permutation groups

  1. #1
    Member Jason Bourne's Avatar
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    Groups-Permutation groups

    I have the following questions but I don't totally understand them, so im not sure about it all:

    Let H be the subgroup of S_8 generated by the permutations

    \alpha = (1 2 3 4)(5 6 7 8), \beta = (1 5 3 7)(2 8 4 6)<br />

    (1) Find the order of \alpha, \alpha\beta. What's the order of \beta^{-1}\alpha\beta ?

    (2) Find the order of H.

    (3) Find the conjugacy classes of H.

    (4) Show that Z=\{e, \alpha^2 \} is a normal subgroup of H

    (5) To which well-known group is the quotient H/Z isomorphic?
    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
    For (1) I know the orders are all 4 I think.

    For (2) Is the fastest way of doing this to actually go through and compute every new element? Or is there an easier way of finding the order?

    For (3) I think the conjugacy classes are the sets in which the permutions all have the same cycle structure?

    For (4) Z is normal iff Z is a union of conjugacy classes?

    For (5) I think this is isomorphic to the Klein Four Group. Find the orders of elements in H/Z ?

    Thanks for any help.
    Last edited by Jason Bourne; May 12th 2009 at 04:18 AM.
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  2. #2
    Newbie adit38's Avatar
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    i am nebie....., but i enjoy grup theory, can i follow this thread?
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  3. #3
    Member Jason Bourne's Avatar
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    Quote Originally Posted by adit38 View Post
    i am nebie....., but i enjoy grup theory, can i follow this thread?
    Yes you can follow this thread and feel free to join in.
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  4. #4
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    Quote Originally Posted by Jason Bourne View Post
    I have the following questions but I don't totally understand them, so im not sure about it all:

    Let H be the subgroup of S_8 generated by the permutations

    \alpha = (1 2 3 4)(5 6 7 8), \beta = (1 5 3 7)(2 8 4 6)<br />

    (1) Find the order of \alpha, \alpha\beta. What's the order of \beta^{-1}\alpha\beta ?

    (2) Find the order of H.

    (3) Find the conjugacy classes of H.

    (4) Show that Z=\{e, \alpha^2 \} is a normal subgroup of H

    (5) To which well-known group is the quotient H/Z isomorphic?
    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
    For (1) I know the orders are all 4 I think.

    For (2) Is the fastest way of doing this to actually go through and compute every new element? Or is there an easier way of finding the order?

    For (3) I think the conjugacy classes are the sets in which the permutions all have the same cycle structure?

    For (4) Z is normal iff Z is a union of conjugacy classes?

    For (5) I think this is isomorphic to the Klein Four Group. Find the orders of elements in H/Z ?

    Thanks for any help.
    this is a long exercise ... here are some hints:

    1. In any group, the order of an element and any of its conjugates are equal because (y^{-1}xy)^k=y^{-1}x^ky. so o(\beta^{-1} \alpha \beta)=o(\alpha)=4.

    2. show that \alpha^2=\beta^2 and \alpha \beta \alpha = \beta and therefore H \cong Q_8, the quaternion group of order 8.

    3. H has 5 conjugacy classes: \{1 \}, \ \{\alpha, \alpha^3 \}, \ \{\alpha^2 \}, \ \{\beta, \beta^3 \}, \ \{\alpha \beta, \alpha \beta^3 \}.
    Last edited by NonCommAlg; May 12th 2009 at 12:22 PM. Reason: 5 not 4.
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  5. #5
    Member Jason Bourne's Avatar
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    2. show that \alpha^2=\beta^2 and \alpha \beta \alpha = \beta and therefore H \cong Q_8, the quaternion group of order 8.
    I don't understand how this implies H is isomorphic to Q8

    3. H has 4 conjugacy classes: \{1 \}, \ \{\alpha, \alpha^3 \}, \ \{\alpha^2 \}, \ \{\beta, \beta^3 \}, \ \{\alpha \beta, \alpha \beta^3 \}.
    I would have though that the conjugacy classes are:

    \{1 \},\ \{\alpha^2 \}, \ \{\alpha, \alpha^3 , \beta, \beta^3, \alpha \beta, \alpha \beta^3 \}

    because the elements in each set have the same disjoint cycle structure?
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  6. #6
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    Quote Originally Posted by Jason Bourne View Post

    I don't understand how this implies H is isomorphic to Q8
    because they have the same presentations.


    I would have though that the conjugacy classes are:

    \{1 \},\ \{\alpha^2 \}, \ \{\alpha, \alpha^3 , \beta, \beta^3, \alpha \beta, \alpha \beta^3 \}

    because the elements in each set have the same disjoint cycle structure?
    those are conjugates in S_8 not in H. don't forget that we want to find the conjugacy classes of H.
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  7. #7
    Member Jason Bourne's Avatar
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    because they have the same presentations.
    Presentations????

    those are conjugates in S_8 not in H. don't forget that we want to find the conjugacy classes of H.
    Now im confused. So how do I find the conjugacy classes of H? I'm clearly not understanding this stuff.
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