A question about groups of order $504=2^3\cdot 3^2 \cdot 7$ I was looking at past qualifying exams in algebra and came across the following problem which I can't solve.

The problem asks to show that if a group $G$ of order $504=2^3\cdot 3^2 \cdot 7$ has a normal subgroup of order $2^3$, then it has at most $8$ Sylow $7$-subgroups.

Applying the Sylow theorems gives that the number of Sylow $7$-subgroups is either $1$, $8$, or $36$. So we only need to exclude the last possibility.
If $P$ is a  Sylow $7$-subgroup of $G$ and if $G$ has $36$ Sylow $7$-subgroups, then $N_G(P)$ has order $14$. I don't see what to do with that information though.
Any help would be appreciated.
 A: Following Derek Holt's hint, consider the canonical epimorphism:
$$ G\to G/N $$
where $N$ is the unique Sylow $2$-subgroup of $G$.
By correspondence theorem, we know that there is a bijection of subgroups in $G/N$ and subgroups in $G$ containing $N$. 
Furthermore, Sylow theorems tell us that there is only one Sylow $7$-subgroup in $G/N$.
However, note that every Sylow $7$-subgroup $H$ in $G$ can be made into a subgroup of order $56$, namely $NH\supset N$ since $N$ is normal. Hence we conclude that $NH$ is a subgroup of order $56$ of $G$ containing $N$ for any Sylow $7$-subgroup $H$. Moreover, it is the only subgroup of order $56$ containing $N$.
Now we have $$NH_1=NH_2=NH_3=\cdots=NH_k$$
 where $H_i$ s are distinct Sylow $7$-subgroups, and we must have $k<10$, since $H_i\cap H_j=\{e\}$, the unit, if $i\ne j$ and there are at most $$ \left\lfloor\frac{56}{7-1} \right\rfloor=9 $$ possibilities. 
As you have claimed that the only possibilities of numbers of subgroups of order $7 $ are $1,8$ and $36$ and we are through.
