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Existence of non-abelian groups of order pq watch

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    I'm having a problem with proving the following:

    Let p be a prime, and let q be a prime dividing p - 1. Then there exists a non-abelian group of order pq.

    Apparently, this can be "deduced" by considering the centralisers of elements of order p in S_p and the normalisers of Sylow p-subgroups in the same, but I don't see it. I have the following facts, which I hope are correct:
    a. There are (p - 1)! elements of order p in S_p.
    b. The centraliser of some p-cycle is the group generated by that p-cycle.
    c. There are (p - 2)! Sylow p-subgroups.
    d. The normaliser of any Sylow p-subgroup has order p(p - 1).

    (I think I can prove existence in an entirely different way by constructing a semidirect product of C_p and C_q, since there will be an element of order q in \mathrm{Aut}(C_p) = \mathbb{Z}_{p}^{*}. That said, I can't be certain that semidirect product will be non-abelian.)
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    A Sylow p-subgroup will be normal in its normalizer. ... Then we can get a subgroup of order pq. (But we already know which elements commute with p-cycles (2))
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    I'm guessing since you brought up the fact it's normal, I should use the correspondence theorem... Let H be a Sylow p-subgroup of S_p, let N be its normaliser, then N/H has order p - 1. I guess Cauchy's theorem implies N/H has an element of order q, so a subgroup of order q. Going back to N, this corresponds to a subgroup of order pq, and this is not abelian since a p-cycle commutes with only its powers. Is that right?
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    Exactly
 
 
 
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Updated: February 1, 2010
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