sylow subgroup of $G/H$

I am in need of some help.

If $G$ is a finite group and $P$ is a $p$ sylow subgroup of $G$, and suppose $H$ is normal in $G$ then i have to prove $HP/H$ is a $p$- sylow subgroup of $G/H$.

I know that $HP/H \cong H/(P \cap H)$ but I don't know how to proceed further.

Pleae help,

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Firstly we have by the Correspondence theorem that $HP/H$ is a subgroup of $G/H$. Now do you want to show it is a $p$ - Sylow subgroup of $G/H$? – user38268 Jan 31 '13 at 10:30
@BenjaLim: Yes i need to do that – user60390 Jan 31 '13 at 10:31
Right the order of $HP/H$ equals $|P|/|P \cap H|$ and since $|P|$ is a power of a prime, the order of $HP/H$ is also a power of a prime. What are you having difficulty with? – user38268 Jan 31 '13 at 10:33

I think the important thing to recognize here is that $P$ is also a sylow $p$-subgroup of $HP$. So $|HP| = p^km$ where $p \nmid m$, and $|P| = p^k$. Since we also have:

$|HP| = \dfrac{|H||P|}{|H \cap P|}$

the highest power of $p$ that divides $|H|$ is the highest power of $p$ that divides $|H \cap P|$. Say $|H| = p^{k'}m'$ where $p \nmid m'$.

Then the highest power of $p$ that divides $|G/H|$ is clearly $p^{k-k'}$ which is also the highest power of $p$ that divdes $|P/(H \cap P)| = |HP/H|$ so that indeed $HP/H$ is a maximal $p$-subgroup of $G/H$.

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First note that $HP/H\cong P/H\cap P$ so $HP/H$ is a $p-$subgroup of $G/H$. Now suffices to show that $p\nmid |G/H:PH/H|$. Since $P\leq PH\leq G$ and $P$ is a sylow $p-$subgroup of $G$ this is verified.

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