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Let $P$ be a Sylow p-subgroup of a finite group $G$ and let $H$ be a normal subgroup of $G$. Show that if $p$ does not divide $[G:H]$, then $P \subseteq H$

I do not understand how to work this problem. If P is a Sylow p subgroup, then the order is prime or a power of a prime. We have to show that $P \subseteq H$, so does that mean that P is cyclic? But we must also show that $[G:H]$, so how do you prove $P \subseteq H$ but disprove $p$ does not divide $[G:H]$?? Any help in understanding would be appreciated.

After looking at the comments and taking some time to look at the problem, i am still somewhat confused. Here is a little of what I've understood.

If I let $|G|=p^an$ where $p \not| n$, then what I am trying to say is that I want to prove $p^an=|G|=|H|$ where $p$ in in $|H|$ and $p$ is not in $[G:H]$ which essentially means, that $|H|=p^an, p\not| n$ and $p \not | [G:H]$

Which I believe will show that $S$ will be a conjugate with $P$. $S$ is also a Sylow p-subgroup of $H$ Here is where I have stopped.

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Observe that under the canonical homomorphism, and using the second (or third or something) isomorphism therem:

$$PH/H\cong P/(P\cap H)$$

It's clear the rightmost expression has order a power of $\;p\;$, yet the leftmost expression is clearly a subgroup pf $\;G/H\;$ , which by assumption is not divisible by $\;p\;$ .

From here that it must be that

$$\;PH/H=1\implies PH=H\implies P\le H$$

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    $\begingroup$ Or the same argument elementwise: If $g\in G$, then the order of its image $gH$ in $G/H$ divides $[G:H]$ and also divides the order of $g$. Hence for $g\in P$, $g$ has orde ra power of $p$ while $p\nmid [G:H]$, so the order of $gH$ is $1$, i.e., $g\in H$. $\endgroup$ – Hagen von Eitzen Mar 23 '15 at 6:33
  • $\begingroup$ Ok, I am still trying to understand the set up of this problem. I will add some comments to my original post $\endgroup$ – cele Mar 24 '15 at 16:03

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