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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!

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    I think the group $G$ should be finite. If this is the case, then the statement is clear from Sylow's theorem (and hence it is not likely to be stated explicitly as a theorem; you might find somewhere as an exercise though). – 2017-02-18
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    Yes,did an edit. – 2017-02-18
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    Also this - http://math.stackexchange.com/questions/1709031/a-normal-p-group-of-g-is-contained-in-each-sylow-subgroup?rq=1 might help me! – 2017-02-18
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    Hint: Combine the followings (i) Every p-subgroup is contained in a Sylow p-subgroup, and (ii) Every Sylow p-subgroups are conjugate to each other. Try to write and post a proof by yourself! – 2017-02-18
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    Ok..Will boost myself ... – 2017-02-18
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    Is the answer correct @Orat ? – 2017-02-18
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    You’ve done well! – 2017-02-18

2 Answers 2

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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.

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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.