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I'm attempting to prove the following: Let $P$ be a normal Sylow $p$-subgroup of $G$ and let $H$ be any subgroup of $G$. Prove that $P \cap H$ is the unique Sylow $p$-subgroup of $H$.

My Attempt: I've already proven that $P \cap H \unlhd H$ so that if $P \cap H$ is a Sylow $p$-subgroup then it will be unique in $H$. My problem is proving how $P \cap H$ is a Sylow $p$-subgroup of $H$ at all. For instance if $H = \{1\}$ then $P \cap H = \{1\}$ and $H$ wouldn't have any Sylow subgroups at all would it, let alone $p$-subgroups with $p > 1$. As an additional note, does every finite group contain some Sylow $p$-subgroups due to the prime decomposition theorem?

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    $\begingroup$ Hint: $|PH|/|P| = |H|/|P \cap H|$. $\endgroup$
    – user169852
    Commented Nov 7, 2017 at 2:00
  • $\begingroup$ P.S. Presumably D&F are using the convention that if $|H|$ is not divisible by $p$, then $\{1\}$ is a $p$-Sylow subgroup of $H$ (of order $p^0 = 1$). $\endgroup$
    – user169852
    Commented Nov 7, 2017 at 2:04
  • $\begingroup$ Thank you for the hint! $\endgroup$
    – user427380
    Commented Nov 7, 2017 at 2:55

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Every non trivial finite group contains a Sylow p-subgroup by Sylow's Theorem. This is became any number can be written as $p^rm$ where $p$ does not divide $m$.

Secondly, the fact that $P \cap H$ is a Sylow subgroup of H follows from the fact that $P \cap H$ is a maximal $p$ subgroup of $H$ (as $P$ is the unique maximal $p$ subgroup of $G$) and $P \cap H$ is also a subgroup of P and subgroups of $p$ groups are $p$ groups.

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  • $\begingroup$ Your answer seems to suggest that this all works even without the normality assumption on $P$ which is wrong. For instance suppose $H$ was another Sylow $p$-subgroup of $G$, then clearly $H\cap P$ would not be a maximal $p$-subgroup of $H$. One should suppose that $P'$ is a maximal $p$-subgroup of $H$ containing $P\cap H$ and then note that since $P$ is the unique maximal $p$-subgroup of $G$, then $P'$ is contained in $P$ and hence $P'$ is contained in $P\cap H$. $\endgroup$
    – Nex
    Commented Nov 7, 2017 at 3:54
  • $\begingroup$ Alternatively you could use the hint of Bungo above to prove maximality. (Just note that the left hand side is clearly not divisible by $p$). $\endgroup$
    – Nex
    Commented Nov 7, 2017 at 4:00
  • $\begingroup$ I think I could've been clearer, but the argument I am using is the same as yours. $\endgroup$ Commented Nov 7, 2017 at 4:33

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