Results 1 to 3 of 3

Thread: p-sylow group abelian

  1. #1
    Junior Member
    Joined
    Oct 2010
    From
    Zulu-5
    Posts
    60

    p-sylow group abelian

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

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

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

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

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

    Any help appreciated.
    Last edited by skyking; Feb 22nd 2011 at 04:00 PM. Reason: typos
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    3
    Quote Originally Posted by skyking View Post
    I am having some difficulties here and would like some guidance.

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

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

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

    What I got is that the options for $\displaystyle s_q$ are $\displaystyle 1, pq$ and for $\displaystyle s_p$ they are $\displaystyle 1,q^2$. 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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2010
    From
    Zulu-5
    Posts
    60
    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 $\displaystyle p^2q^2$ are $\displaystyle 1,p,q,p^2,q^2,pq,p^2q,pq^2$.

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

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

    How do we eliminate those other options?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Is the subgroup of an abelian group always abelian?
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: Dec 6th 2009, 11:38 PM
  2. Sylow p-group
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Nov 6th 2009, 08:03 PM
  3. Sylow group
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Dec 6th 2008, 12:08 PM
  4. Sylow group question
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Aug 24th 2008, 09:49 AM
  5. Sylow p-group
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: May 9th 2008, 09:33 AM

Search Tags


/mathhelpforum @mathhelpforum