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Math Help - No simple groups of order 48

  1. #1
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    No simple groups of order 48

    Prove there are no simple groups of order 48
    Ok, here's what i have so far.
    48=2^4*3
    n_{2} = 1 mod 2 & n_{2} | 3
    n_{2} = 1 or 3
    n_{3} = 1 mod 3 & n_{3} | 16
    n_{3} = 1, 4, or 16

    So we must eliminate these cases to get n_{2} = 1 or n_{3} = 1:
    (n_{2}, n_{3})=(3,4)
    (n_{2}, n_{3})=(3,16)

    How do I eliminate these cases?
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  2. #2
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    Quote Originally Posted by frankdent1 View Post
    Prove there are no simple groups of order 48
    Ok, here's what i have so far.
    48=2^4*3
    n_{2} = 1 mod 2 & n_{2} | 3
    n_{2} = 1 or 3
    n_{3} = 1 mod 3 & n_{3} | 16
    n_{3} = 1, 4, or 16

    So we must eliminate these cases to get n_{2} = 1 or n_{3} = 1:
    (n_{2}, n_{3})=(3,4)
    (n_{2}, n_{3})=(3,16)

    How do I eliminate these cases?
    This problem is a little involved. Consider the Sylow 2-subgroups we know that there are 1 or 3 such Sylow subgroups. If there was only one Sylow 2-subgroup then there is nothing to prove because it is invariant under conjugation and so it is normal. Thus, we will assume there are 3 Sylow 2-subgroups. Let H,K be two of these three Sylow subgroups. Notice that |H|,|K| = 16 and |HK||H\cap K| = |H||K|. Since HK\subseteq G \implies |HK| \leq 48, therefore |H\cap K| cannot be less than or equal to 4. This forces |H\cap K|=8 by applying Lagrange's theorem. Since H\cap K is of index 2 in H and K it means H\cap K \triangleleft H, H\cap K\triangleleft K. Therefore, the normalizer, N(H\cap K) must contain both H and K. By Lagrange's theorem it means |N(H\cap K)|>16 is a multiple of 16 and a divisor of |G|=48. The only possibility is N(H\cap K) = G \implies H\cap K\triangleleft G and so G is not simple.
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