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Math Help - Simple group question

  1. #1
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    Simple group question

    I need some direction with some of the following:

    Let G be a simple group |G|=60

    1. show that G does not have sub-groups of index 3 or 4.
    2. show that if H_{1},H_{2} are 2 different 2-sylow subgroups of G such that H_{1}\cap H_{2}\neq \{e\} then N(H_{1}\cap H_{2}) is a subgroup of index 5.
    3. show that if for every 2-sylow subgroups the intersection is trivial, H_{1}\cap H_{2} = {e}. then there are exactly five 2-sylow subgroups.
    4. show that G is isomorphic to A_{5}.

    I was able to show 1,2.
    with 3, I am having trouble. using sylow's 3rd theorem I can show that the number of 2-sylow subgroups can be 1,3,5 or 15. I can eliminate the possibility of 1 (contradiction to simple group) and the possibility of 15 (too many elements considering the other p-sylow subgroups). but I can't eliminate the possibility of 3.

    I need direction as weel on how to start 4.
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  2. #2
    Senior Member roninpro's Avatar
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    To get you started, note that G acts on its Sylow 2-subgroups. If there were 3 of them, then this would induce a homomorphism \phi:G\to S_3. On one hand, \ker \phi should not be trivial, because |S_3|=6, which is too small. On the other, \ker \phi should not be G, because the action is nontrivial. This means that \ker \phi is a proper nontrivial subgroup of G. Since the kernel is always normal, this contradicts the simplicity of G.

    Try using a similar trick for part four.
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  3. #3
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    Quote Originally Posted by roninpro View Post
    To get you started, note that G acts on its Sylow 2-subgroups. If there were 3 of them, then this would induce a homomorphism \phi:G\to S_3. On one hand, \ker \phi should not be trivial, because |S_3|=6, which is too small. On the other, \ker \phi should not be G, because the action is nontrivial. This means that \ker \phi is a proper nontrivial subgroup of G. Since the kernel is always normal, this contradicts the simplicity of G.

    Try using a similar trick for part four.

    Do you mean G acts on the set of left cosets of a Sylow 2-subgroup of G? If G acts simply on its Sylow-2 subgroups, I don't think the action is well-defined for this problem.

    For OP,

    Some useful theorems for this problem are
    1) If H is a subgroup of index n in a group G and G is simple, then G is isomorphic to a subgroup of S_n. (Think about G acts on the set of all left cosets of H in G)
    2) The number of conjugates of H in G is |G:N_G(H)|
    3) Any two Sylow p-subgroup of G are conjugate.

    By (1), G does not have index 3 or 4 subgroup, since 3! or 4! is smaller than 60.
    The possible numbers of Sylow 2-subgroups are 3, 5, or 15.
    Let P be a Sylow 2-subgroup. Then |G:N_G(P)|=3, 5, or 15 by (2) and (3). 3 is not a possible number by (1). 15 is not also a possible number either by counting arguments. (Verify this).

    Thus, the number of Sylow 2-subgroups are 5. Then, |G:N_G(P)|=5. By (1), we see that G is isomorphic to the subgroup of S_5 and is simple of order 60 by assumption. Argue first that G \leq A_5 and conclude that G = A_5.
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  4. #4
    Senior Member roninpro's Avatar
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    Quote Originally Posted by TheArtofSymmetry View Post
    Do you mean G acts on the set of left cosets of a Sylow 2-subgroup of G? If G acts simply on its Sylow-2 subgroups, I don't think the action is well-defined for this problem.
    I should have been more explicit, but G acts on its Sylow p-subgroups by conjugation (which follows by #3 in your list).
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  5. #5
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    Last edited by TheArtofSymmetry; February 22nd 2011 at 08:16 AM. Reason: Deletion
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  6. #6
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    Thank you both
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