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Math Help - If |G|=42, and the 3-Sylow is not normal, for all elements s.t. |x|=3, |y|=7, xy!=yx

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    If |G|=42, and the 3-Sylow is not normal, for all elements s.t. |x|=3, |y|=7, xy!=yx

    Here's what I've thought thus far.

    There is only one 7-Sylow (easy to check). Then, the number of 3-Sylows is 7. We know that they are conjugates of each other. Let P be the 3-Sylow and Q be the 7-Sylow. Then consider Q acting on P via conjugation. The orbit stabilizer theorem implies that

    |\text{orbit}| = \frac{|Q|}{|\text{stabilizer}|}

    It follows that the orbit must divide 7, so either 1 or 7. If the orbit is 7, we are in luck: Conjugating by Q gives us all possible 3-Sylows and each of them is disjoint apart from the identity, so that means elements of the 3-Sylows can't possibly commute with elements of the 7-Sylow.

    I have no idea what to do in the case where the orbit is just 1 element: i.e. the 7-Sylow is in the normalizer of the 3-Sylow. It doesn't imply that the 3-Sylow is normal, so there is no contradiction there. What more must I do?
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  2. #2
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    Quote Originally Posted by eeyore View Post
    Here's what I've thought thus far.

    There is only one 7-Sylow (easy to check). Then, the number of 3-Sylows is 7. We know that they are conjugates of each other. Let P be the 3-Sylow and Q be the 7-Sylow. Then consider Q acting on P via conjugation. The orbit stabilizer theorem implies that

    |\text{orbit}| = \frac{|Q|}{|\text{stabilizer}|}

    It follows that the orbit must divide 7, so either 1 or 7. If the orbit is 7, we are in luck: Conjugating by Q gives us all possible 3-Sylows and each of them is disjoint apart from the identity, so that means elements of the 3-Sylows can't possibly commute with elements of the 7-Sylow.

    I have no idea what to do in the case where the orbit is just 1 element: i.e. the 7-Sylow is in the normalizer of the 3-Sylow. It doesn't imply that the 3-Sylow is normal, so there is no contradiction there. What more must I do?

    Since there are 7 Sylow 3-sbgps. of G, this means that the normalizer of any of them is a sbgp. of index 7

    in G, which means the normalizer of any Sylow 3-sbgp. of G has order \frac{42}{7}=6 .

    In particular, no element of order 7 fixes an element of order 3 and we're done...

    Tonio
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