Singular $p$-subgroups of a finite group Let $G$ be a finite group and $p$ be a prime. Suppose that there is a $p$-subgroup $S$ of $G$ such that there is only one Sylow $p$-subgroup containing it. 
Let $N$ be a normal subgroup of $G$. Is it true that there is only one Sylow $p$-subgroup of $G/N$ containing $SN/N$?
I think the answer is yes, but I cannot prove it; if so, can you give me any hint to prove it? 
I know that the Sylow $p$-subgroup of $G/N$ are of the type $QN/N$ where $Q$ is a Sylow $p$-subgroup of $G$, but how can one use this?    
 A: Not sure if this is true in general, but it is when $\color{blue}{N \subseteq \bigcap_{P \in Syl_p(G)}N_G(P)}$, the intersection of all the normalizers of the Sylow $p$-subgroups of $G$. Call this intersection $D$. Note that $D \unlhd G$ and that it has a unique Sylow $p$-subgroup, namely $O_p(G)$, which is $\bigcap_{P \in Syl_p(G)}P$. So a "good" example for $N$ would be $N=O_p(G)$.
Anyhow, let $QN/N \in Syl_p(G/N)$, with $SN/N \subseteq QN/N$. Then $SN \subseteq QN$. But $N$ normalizes $Q$, so $Q \unlhd QN$, hence $Q$ is the unique Sylow $p$-subgroup of $QN$. But of course the $p$-subgroup $S \subseteq SN \subseteq QN$. Since by Sylow theory $S$ must be contained in some Sylow $p$-subgroup of $QN$, we get $S \subseteq Q$. And the choice of $S$ now gives $Q=P$, whence $PN/N=QN/N$.
Note (One day later ...) The general case is true, and I can only blame myself not having read the comments of @j.p.! We need the following.
Lemma Let $G$ be a a group, $H \leq G$ and $N \trianglelefteq G$. Then for any $g \in G$ we have $(H \cap N)^g=H^g \cap N$.
Proof Since $H \cap N \subseteq H$, $(H \cap N)^g \subseteq H^g$, and likewise $(H \cap N)^g \subseteq N^g=N$ (the latter equality follows from the fact that $N$ is normal. Hence $(H \cap N)^g \subseteq H^g \cap N$. Conversely, let $x \in H^g \cap N$, then $x \in N$ and $x=g^{-1}hg$ for some $h \in H$. Hence $h= gxg^{-1} \in N$, since $x \in N$ and $N$ is normal. So, $x=g^{-1}hg \in (H \cap N)^g$  and we are done.
Now assume that $SN/N \subseteq QN/N$, for some $Q \in Syl_p(G)$. Then $S \subseteq QN$. Now obviously, $Q \in Syl_p(QN)$ and hence the cardinality of Sylow subgroups in $QN$ is the same as in $G$! By Sylow and the fact that $S$ is a $p$-group, $S \subseteq Q^x$ for some $x \in QN$. But this $Q^x$ must actually be a Sylow subgroup of $G$ by the previous remarks. So, $Q^x=P$, and hence, $P \subseteq QN$. Finally we need to show that $PN=QN$ and that is where the Lemma comes in: $|QN|=\frac{|Q| \cdot |N|}{|Q \cap N|}=\frac{|Q^x| \cdot |N|}{|(Q \cap N)^x|}=\frac{|P| \cdot |N|}{|P \cap N|}=|PN|$. It follows that $PN=QN$, whence $PN/N=QN/N$ and we are done.

Additional remark It can be proved in general for a $p$-subgroup $S$, that $\#\{P \in Syl_p(G): S \subseteq P\} \equiv 1$ mod $p$. (Sketch of the proof: let $S$ act by conjugation on the set $\{P \in Syl_p(G): S \subseteq P\}$ and observe that $N_S(P)=P \cap S$.)
A: Stealing ideas from Nicky Hekster and j.p. you can prove your statement using these three easy facts about $\varphi : G \to G/N$:


*

*The image $\varphi(P)$ of a Sylow $p$-subgroup $P$ of $G$ is a Sylow $p$-subgroup of $G/N$. [Proof: It's a $p$-subgroup with index (in $G/N$) coprime to $p$.]

*The preimage $\varphi^{-1}(\bar{P})$ of a Sylow $p$-subgroup $\bar{P}$ of $G/N$ has index (in $G$) coprime to $p$, in particular its Sylow $p$-subgroups are also Sylow $p$-subgroups of $G$.

*The induced maps $\varphi : \mathcal{P}(G) \to \mathcal{P}(G/N), X \mapsto \varphi(X)$ and $\varphi^{-1} : \mathcal{P}(G/N) \to \mathcal{P}(G), X \mapsto \varphi^{-1}(X)$ are monotone with respect to $\subseteq$.
For a Sylow $p$-subgroup $\bar{P}$ of $G/N$ containing $SN/N$, we show that for the unique Sylow $p$-subgroup of $G$ containing $S$ we get $\bar{P} = PN/N$ (implying the uniqueness in $G/N$):
As $S\le SN = \varphi^{-1}(SN/N) \le \varphi^{-1}(\bar{P}) =: H$ holds by the last fact, any Sylow $p$-subgroup of $H$ containing $S$ (and such a Sylow $p$-subgroup exists) has to be already $P$ by the 2nd fact (and the uniqueness of $P$). As by the last fact $PN/N = \varphi(P)\le \varphi(H) = \bar{P}$, the equality $\bar{P} = PN/N$ follows as $PN/N$ is a Sylow $p$-subgroup of $G/N$ by the first fact.
