Results 1 to 2 of 2

Thread: Sylow p-group

  1. #1
    Member
    Joined
    Nov 2007
    Posts
    108

    Sylow p-group

    Will anyone give me a hand on this problem? I am trying to show that if $\displaystyle H \vartriangleleft G$ and $\displaystyle H$ contains $\displaystyle P$, a Sylow p-subgroup in $\displaystyle G$. Then $\displaystyle n_p(H)=n_p(G)$. I am not sure that this is true, but I see that since $\displaystyle H \vartriangleleft G$ and $\displaystyle P$ is contained in $\displaystyle H$, then $\displaystyle H$ must contain all the conjugates of $\displaystyle P$. By Sylow's theorem, we know all Sylow p-subgroups of $\displaystyle G$ are conjugates. So, $\displaystyle n_p(H)=n_p(G)$. Do I need to construct a 1-1 correspondence here? I think that $\displaystyle n_p(H)$ and $\displaystyle n_p(G)$ are just the number of Sylow p-subgroups, how can I really construct a 1-1 correspondence if they are not sets? I don't claim that $\displaystyle H$ and $\displaystyle G$ have the same set of Sylow p-subgroups.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by namelessguy View Post
    Will anyone give me a hand on this problem? I am trying to show that if $\displaystyle H \vartriangleleft G$ and $\displaystyle H$ contains $\displaystyle P$, a Sylow p-subgroup in $\displaystyle G$. Then $\displaystyle n_p(H)=n_p(G)$. I am not sure that this is true, but I see that since $\displaystyle H \vartriangleleft G$ and $\displaystyle P$ is contained in $\displaystyle H$, then $\displaystyle H$ must contain all the conjugates of $\displaystyle P$. By Sylow's theorem, we know all Sylow p-subgroups of $\displaystyle G$ are conjugates. So, $\displaystyle n_p(H)=n_p(G)$. Do I need to construct a 1-1 correspondence here? I think that $\displaystyle n_p(H)$ and $\displaystyle n_p(G)$ are just the number of Sylow p-subgroups, how can I really construct a 1-1 correspondence if they are not sets? I don't claim that $\displaystyle H$ and $\displaystyle G$ have the same set of Sylow p-subgroups.
    if Q is a Sylow p-subgroup of G, then $\displaystyle Q=gPg^{-1},$ for some $\displaystyle g \in G.$ thus $\displaystyle Q =gPg^{-1} \subseteq gHg^{-1} = H,$ i.e. Q is a Sylow p-subgroup of H. the converse is trivial because $\displaystyle |P|$ divides $\displaystyle |H|.$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. p-sylow group abelian
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Feb 22nd 2011, 08:24 PM
  2. Replies: 1
    Last Post: Dec 10th 2008, 07:20 AM
  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