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Math Help - Problem: Intersection of 2-Sylow-groups

  1. #1
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    Problem: Intersection of 2-Sylow-groups

    Hi,

    I am stuck proving that a simple group G of order 60 is isomorphic to A_5.

    In particular:

    - I have shown |Syl_5(G)|=6 and |Syl_3(G)|=10, so there must be 6\cdot(5-1)=24 elements of order 5 and 10\cdot(3-1)=20 elements of order 3.

    - I could further show |Syl_2(G)|\notin\{1,3\}.

    - Now, where I get stuck: I need to show |Syl_2(G)|\neq15 and therefore |Syl_2(G)|=5.
    I know I have to use the fact that there are already 20+24=44 elements of order coprime to 2. That leaves only 60-44-1=15 elements to have order 2 or 4, which somehow has to lead to a contradiction to |Syl_2(G)|=15. But the 2-Sylow-groups have order 2^2 and can thus intersect non-trivially. So how can I argue from here?

    Any help would be much appreciated!
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  2. #2
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    Re: Problem: Intersection of 2-Sylow-groups

    Could you possibly use the fact that all sylow p subgroups are conjugate to each other (with p=2)?
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  3. #3
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    Re: Problem: Intersection of 2-Sylow-groups

    I could not yet figure out how to use that, but I will try for some hours. Thanks for the input!
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  4. #4
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    Re: Problem: Intersection of 2-Sylow-groups

    Had vague thoughts about the conjugacy class of 2-Sylow-subgroups generating a normal subgroup in G, but I couldn't see how that would lead me further. Can you help me a little?
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  5. #5
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    Re: Problem: Intersection of 2-Sylow-groups

    Hi,
    I believe this is what you need to finish the proof that G is isomorphic to A5; namely that there are 5 Sylow 2 subgroups of G. After this determination, I left the rest of the proof to you:

    Problem: Intersection of 2-Sylow-groups-mhfgroups9.png
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  6. #6
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    Re: Problem: Intersection of 2-Sylow-groups

    Thank you!
    That was all I need to conclude my proof.
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