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Let $G$ be a group and $P\in Syl_p(G)$, H is normal in G. I want to show that $P\cap H\in Syl_p(H)$.

So I let $P_0\in Syl_p(H)$. $P\cap H$ is a p subgroup of $H$, so by Sylow 2nd Theorem, $P\cap H \leq P_0$.

And by Sylow's 2nd and 3rd theorem, I get that there exists $g\in G$ such that $P_0 \leq gPg^{-1}$.

I think I want to prove that $P_0 \leq P\cap H$ next in order to conclude that $P_0=P\cap H$ but got stuck at this part.

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2 Answers 2

up vote 1 down vote accepted

You're almost there.

So, you have chosen a $P_0\in Syl_p(H)$ such that $P\cap H\le P_0$. Then, we have $P_0\le gPg^{-1}$. Also, $P_0\le H$, so, as $gHg^{-1}=H$, it means $$g^{-1}P_0\,g\le P\cap H\,.$$ Assuming everything is finite, by calculating sizes, we are ready, as $|g^{-1}P_0\,g|=|P_0|$ and both are included in $H$, so $|P\cap H|=|P_0|$ also follows.

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and this implies that $P_0=P \cap H$, right? –  Akaichan Apr 3 '13 at 15:00

Hint: $PH$ is a subgroup when $H$ is a normal subgroup. Apply the formula

$$|PH| = \frac{|P||H|}{|P \cap H|}$$

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Is the subgroup generated by $P_{0}$ and $P$ a p-subgroup of G? I couldn't see your hint. Thanks! –  Ergin Suer Jan 11 at 21:04

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