Sylow p-group

namelessguy · #1 $Old$ November 6th, 2009, 08:03 PM

Will anyone give me a hand on this problem? I am trying to show that if $H \vartriangleleft G$ and $H$ contains $P$ , a Sylow p-subgroup in $G$ . Then $n_p(H)=n_p(G)$ . I am not sure that this is true, but I see that since $H \vartriangleleft G$ and $P$ is contained in $H$ , then $H$ must contain all the conjugates of $P$ . By Sylow's theorem, we know all Sylow p-subgroups of $G$ are conjugates. So, $n_p(H)=n_p(G)$ . Do I need to construct a 1-1 correspondence here? I think that $n_p(H)$ and $n_p(G)$ 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 $H$ and $G$ have the same set of Sylow p-subgroups.

NonCommAlg · #2 $Old$ November 6th, 2009, 10:03 PM

Quote:

Originally Posted by namelessguy $View Post$

Will anyone give me a hand on this problem? I am trying to show that if $H \vartriangleleft G$ and $H$ contains $P$ , a Sylow p-subgroup in $G$ . Then $n_p(H)=n_p(G)$ . I am not sure that this is true, but I see that since $H \vartriangleleft G$ and $P$ is contained in $H$ , then $H$ must contain all the conjugates of $P$ . By Sylow's theorem, we know all Sylow p-subgroups of $G$ are conjugates. So, $n_p(H)=n_p(G)$ . Do I need to construct a 1-1 correspondence here? I think that $n_p(H)$ and $n_p(G)$ 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 $H$ and $G$ have the same set of Sylow p-subgroups.

if Q is a Sylow p-subgroup of G, then $Q=gPg^{-1},$ for some $g \in G.$ thus $Q =gPg^{-1} \subseteq gHg^{-1} = H,$ i.e. Q is a Sylow p-subgroup of H. the converse is trivial because $|P|$ divides $|H|.$