What we know about sylow subgroups of $G/P$? Let $G$ be a finite group and $P$ be it's normal $p$-subgroup. What we can say about the sylow subgroup of $G/P$?
For example if we know that a $q$-sylow $(q\neq p)$ subgroup of $G/P$ is normal and non abelian what we can say about $q$-sylow subgroup of $G$?
 A: Note: my answer does not assume that $P$ is a normal $p$-group (I didn't see this when I first read the question), but merely any normal subgroup.  
If $q: G \rightarrow G/P$ is the quotient map, then for every Sylow $p$-subgroup $S$ of $G$, the image $q(S)$ is a Sylow $p$-subgroup of $G/P$.  This is easy to see: one just has to check that it has the right order. 
From this it follows that if $G$ has a normal Sylow $p$-subgroup $S$, then the Sylow $p$-subgroups of $G/P$ are normal (so in fact there is exactly one).  Also, if $S$ is commutative, so is $q(S)$.  The converse clearly does not hold: e.g. suppose $G = P$ is a nonabelian $p$-group.  (However if $P$ is a $p$-group and $S$ is a Sylow $q$-subgroup for a prime $q \neq p$, then $q(S) = SP/P \cong S/(S \cap P) \cong S$, as in Boris Novikov' answer.)
A: Let $Q$ be a $q$-Sylow subgroup in $G$, $Q'$ be its image in $G/P$. Since $P\cap Q=1$, then $Q\cong Q'$ (so if $Q'$ is Abelian, $Q$ is also). 
If $Q'$ is normal, $Q$ can be n0n-n0rmal. An example: for $G=S_3$, $P$ its Sylow 3-group, $Q'=G/P$.
A: The key point is that the Sylow $q$-subgroups of $G/P$ (for any prime $q$) are precisely the homomorphic images of the Sylow $q$-subgroups of $G$ under the canonical homomorphism $G\to G/P$. Also, homomorphic images of abelian groups are always abelian. E.g., this is elaborated in both Pete's and Boris' excellent answers. I've provided details in my answer below. 
Firstly, let's consider the case of primes $q\neq p$: any Sylow $q$-subgroup of $G/P$ (in fact, any subgroup of $G/P$) is of the form $H/P$ for some subgroup $H$ of $G$ containing $P$. Now, $H=QP$ where $Q$ is a Sylow $q$-subgroup of $H$. (Why?) Since $Q$ is also a Sylow $q$-subgroup of $G$ (why?), it follows that $H/P$ equals the image of $Q$ under the canonical homomorphism $G\to G/P$. 
Now let's consider the case $q=p$. If a $p$-subgroup $P\subseteq G$ is normal in $G$, then $P$ must be contained in all Sylow $p$-subgroups of $G$. (Exercise; use the Sylow theorems.) The intersection of all Sylow $p$-subgroups of $G$ is referred to as the $p$-core of $G$ and is the unique maximal normal $p$-subgroup of $G$. 
Therefore, the Sylow $p$-subgroups of $G/P$ are precisely the Sylow $p$-subgroups of $G$ which contain $P$. 
I hope this helps!
