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Math Help - Klein four-group is normal group of A4

  1. #1
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    Klein four-group is normal group of A4

    I have shown that in strait forward way .
    gVg^(-1) = V

    But is there some other way to show that ?

    Sylow dint worked for me here .

    Thanks for help .
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  2. #2
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    Quote Originally Posted by s.lateralus View Post
    I have shown that in strait forward way .
    gVg^(-1) = V

    But is there some other way to show that ?

    Sylow dint worked for me here .

    Thanks for help .
    What you have done looks like what I would have suggested!
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  3. #3
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    some other ideas ?
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  4. #4
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    Quote Originally Posted by s.lateralus View Post
    some other ideas ?


    If you already know that two permutations in S_n are conjugate iff they have the very same cycle decomposition, then it is straightforward that V is a normal subgroup of S_4 and, thus, also of A_4 since V\subset A_4 .

    Tonio
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  5. #5
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    Quote Originally Posted by tonio View Post
    If you already know that two permutations in S_n are conjugate iff they have the very same cycle decomposition, then it is straightforward that V is a normal subgroup of S_4 and, thus, also of A_4 since V\subset A_4 .

    Tonio
    can you please explain first part of your statement plz ,
    some example will really help.
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  6. #6
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    Quote Originally Posted by s.lateralus View Post
    can you please explain first part of your statement plz ,
    some example will really help.

    Well, for example (12)(345), (14)(235) are conjugate permutations in S_n\,,\,n\ge 5 , since they both are the product of a 2-cycle and a 3-cycle, but none of them is conjugate to (13)(24) since this last is the product of two 2-cycles (transpositions), and etc.
    Of course, one must know first that ANY permutation is expressable as the product of DISJOINT cycles...

    Tonio
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  7. #7
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    If you want a geometric interpretation, use the fact that A4 is the group of rotations of a regular tetrahedron. The tetrahedron has three axes, namely the lines joining the midpoints of opposite edges. A rotation of the tetrahedron takes axes to axes, and thus induces a homomorphism from A4 to S3, whose kernel is the Klein group (consisting of the identity map together with rotations through 180 about each axis).
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  8. #8
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    Also, Klein is the unique subgroup of A_4 such as |H|=4, so it's normal.
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