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.
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 $\displaystyle |H|.$