On sylow p-groups in finite groups 
Problem:
  Let $G$ be a finite group, $H\subset G$ a subgroup and $p$ a prime number. Show that:
(i) If $S\subset H$ is a Sylow $p$-subgroup of $H$, then there exists a Sylow $p$-subgroup $T\subset G$ of $G$ such that $S=T\cap H$.
(ii) If $G$ has a normal Sylow $p$-subgroup, then $H$ has a normal Sylow $p$-subgroup.
(iii) If $H\subset G$ is normal and $T\subset G$ a Sylow $p$-subgroup of $G$, then $T\cap H$ is a Sylow $p$-subgroup of $H$.

My ideas:
(i) Since $S$ is a sylow p-group of $H$ we know that $\operatorname{ord}(H)=p^l*m$ for $l\in\mathbb{N}$ and p,m coprime. And since $H$ is a subgroup of $G$ we know that $\operatorname{ord}(G)=p^k*m*q$ for $k\ge l$ and $q\in\mathbb{N}$. So $G$ has a sylow p-group $T$ such that $\operatorname{ord}(T)=p^k$. And since $k\ge l$ it follows that $S=T\cap H$.
(ii) Denote this normal sylow p-subgroup as $F$. We know that $\operatorname{ord}(F)=p^k$ for a $k\in\mathbb{N}$ and that $F$ is the only Sylow $p$-subgroup of $G$ (since $F$ is normal). Then $F\cap H$ is a Sylow $p$-subgroup of $H$ (using (i)) and it is still normal in $H$ since $H\subset G$.
(iii) Since $H$ is normal in $G$, $T\cap H$ is still normal in $H$. If we can show that $T\cap H$ is the only subgroup of $H$ of order $\operatorname{ord}(T\cap H)$ we would be done. This is where I'm stuck.
Any hints would be greatly appreciated.
 A: The last sentence in (i) is fishy. It does not follow that $S=T\cap H$. It is perfectly possible that $T\cap H$ is trivial, if you picked the wrong $T$. As an example consider $G=S_4$, $H\cong S_3=$ the subgroup of elements keeping $4$ fixed. If $p=3$ and $S=\langle(123)\rangle$ then you cannot select $T=\langle(124)\rangle$. You must use a stronger version/by-product of Sylow theorem's. Can you tell which? Hint: It deals with the existence of Sylow subgroups containing a given $p$-subgroup.
In (ii) you need to again explain more carefully, why $F\cap H$ is a Sylow subgroup of $H$. Part (i) says that any Sylow $p$-subgroup $S$ can be written in the form $S=T\cap H$ for some Sylow subgroup $T$ of $G$. Why are you allowed to assume that you are allowed to select $T=F$? Hint: You already said it earlier, but I wasn't entirely convinced that you fully appreciated it, because you had misunderstood what (i) says.
Zibadawa Timmy already pointed out the major flaw in your attempt (iii). Continuing with the above example of $G=S_4$. If we let $H=A_4$, then it's normal in $G$,
but if $T=\langle(123)\rangle$, then $T\le H$, but $T$ isn't normal in $H$.
Instead you could (should?) go via the route of the hint to part (i). First pick a Sylow subgroup $S$ of $H$ containing $T\cap H$. Then part (i) tells you that there exists a Sylow subgroup $T'$ of $G$ such that $T'\cap H=S$. But you should be able to show that $|T'\cap H|=|T\cap H|$. How? As 
$$(T\cap H)\subseteq S\subseteq (T'\cap H),$$
what conclusions can you draw?
