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Math Help - p-sylow group abelian

  1. #1
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    p-sylow group abelian

    I am having some difficulties here and would like some guidance.

    Let G be a group such that |G|=p^2q^2 where p,q are primes and p>q and q is not a divisor of p\pm 1. Show that G is abelian.

    What I can determine so far is that G has p-sylow group of order p^2 and a q-sylow group of order q^2 and that both must be abelian.

    Additionally I can try to eliminate the possibilities for s_p,s_q (the number of p,q-sylow groups).

    What I got is that the options for s_q are 1, pq and for s_p they are 1,q^2. but I can't get anywhere from here.

    Any help appreciated.
    Last edited by skyking; February 22nd 2011 at 04:00 PM. Reason: typos
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  2. #2
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    Quote Originally Posted by skyking View Post
    I am having some difficulties here and would like some guidance.

    Let G be a group such that |G|=p^2q^2 where p,q are primes and p>q and q is not a divisor of p\pm 1. Show that G is abelian.

    What I can determine so far is that G has p-sylow group of order p^2 and a q-sylow group of order q^2 and that both must be abelian.

    Additionally I can try to eliminate the possibilities for s_p,s_q (the number of p,q-sylow groups).

    What I got is that the options for s_q are 1, pq and for s_p they are 1,q^2. but I can't get anywhere from here.

    Any help appreciated.

    The given conditions give us that there are exactly one Sylow p-sbgp. and one Sylow

    q-sbgp. which are thus normal.

    Since they're intersection is obviously trivial their product is their own direct product, and since

    the product of their orders is the order of G then G is the direct product of

    abelian groups...

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    The given conditions give us that there are exactly one Sylow p-sbgp. and one Sylow

    q-sbgp.
    Tonio
    Thanks for the reply, I understand it all besides how you get the above conclusion.

    The divisors of p^2q^2 are 1,p,q,p^2,q^2,pq,p^2q,pq^2.

    For S_q we need to have s_q=1(modq), s_q\leq p^2 so p,q,q^2,p^2,p^2q,pq^2 are all out, which leaves us with 1,pq

    Similarly for s_p we need s_p=1(modp), s_p\leq q^2 which leaves us with 1,q^2

    How do we eliminate those other options?
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