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

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