Let $H$ be a normal $p$ -subgroup of a finite group $G$ ,then $H$ is contained into each Sylow $p$ subgroup of $G$ Any good reference for the proof of this theorem "Let $H$ be a normal $p$ -subgroup of group $G$ ,then $H$ is contained into each Sylow $p$ subgroup of $G$",i could not find after googling ?

Also a question like this here - If H is normal p-subgroup of G, then H is contained in every sylow-p subgroup.

is not clear to me.

Any help!
Let $P$ be a Sylow-$p$-subgroup of $G$,

then we need to prove that $H \subset P $ and since $P$ is any arbitrary Sylow $p$ subgroup the result will follow,

So, From Sylow's theorem we know that $H$  is contained in some Sylow -$p$-subgroup say $P'$ , now from another theorem by Sylow we also know that both $P$ and $P'$  are conjugate to each other that is $gP'g^{-1} = P$ for some $g \in G$,

Now since $H \subset P' \implies gHg^{-1} \subset gP'g^{-1} = P$ 

that is $gHg^{-1} \subset P$ , but as $H $ is normal in $G$ so $gHg^{-1} = H$ , 

so we obtained $H \subset P$ as desirable thus a normal $p$ subgroup of $G$ is contained into each Sylow $p$ subgroup.
Let $H$ normal a normal $p-$group and let $K$ a $p-$subgroup. Suppose that $|H|\leq |K|$. In particular, by Sylow second theorem, $K$ has a subgroup $L$ of cardinality $|H|$ that is conjugate to $H$, i.e. there is $g\in G$ s.t. $L=gHg^{-1}$. But since $H$ is normal, you get $L=H$. If $|K|\leq |H|$, then you take $L$ as a subgroup of $H$ and do the same.