Let $H$ be a normal $p$-subgroup of a group $G$. Proof $H$ is contained in $S$ for all $S$ Sylow $p$-subgroup of $G$.
I'm having trouble trying to understand what is that I have to proof. I don't fully understand the problem so this is what I've got so far.
On my notes I have that: Every p-subgroup is contained in a $S_{p}$-subgroup.
An $S_{p}$-subgroup is the same as a Sylow $p$-subgroup?
Also there's a Lemma that states: If $S$ is a $S_{p}$-subgroup of $G$ and $H$ is a p-subgroup then $H\subset N_{G}(S)$ if and only if $H\subset S$.
I'm pretty lost so any help would be very appreciated.
If $S \in Syl_p(G)$ and $H$ is a $p$-subgroup of $G$ normalizing $S$, which is $H \subseteq N_G(S)$, then $HS$ is a subgroup. In addition, $|HS|=\frac{|H| \cdot |S|}{|H \cap S|}$, which is a $p$-power. Since $S$ is a maximal $p$-subgroup, and $S \subseteq HS$ it follows that $S=HS$, or equivalently, $H \subseteq S$. The converse is trivial.
Different proof: observe $S \unlhd N_G(S)$, so, applying Sylow theory inside $N_G(S)$, this latter group has a unique Sylow $p$-subgroup, namely $S$. If the $p$-subgroup $H \subseteq N_G(S)$, then by your first note ("every $p$-subgroup is contained in some Sylow $p$-subgroup"), again you get $H \subseteq S$, since there is no other!