A finite group which has exactly eight Sylow $7$-subgroups 
Let $G$ be a finite group which has exactly eight Sylow $7$-subgroups. Prove that there exist a normal subgroup $N$ of $G$ such that its index is divisible by $56$ but not by $49$. 

Give me some hints.
Thanks in advance.
 A: HINT: $G$ acts transitively by conjugation on the set of Sylow 7-subgroups, which gives us a homomorphism $G\to S_8$. What can bve said about its kernel? 
A: The proof is based on a basic fact about finite groups.
Theorem Let $H \subseteq G$ be a subgroup of index $n$. Then $G/core_G(H)$ is isomorphic to a subgroup of $S_n$.
Proof See I.M. Isaacs, Finite Group Theory, Theorem 1.1. Note, $core_G(H):=\bigcap_{g \in G}H^g$, which is a normal subgroup contained in $H$.
Now let us have a look at the question. Let $P \in Syl_7(G)$ and put $H=N_G(P)$ and $N=core_G(H)$. Then the Theorem tells us that $G/N$ is isomorphic to a subgroup of $S_8$. The order of the latter is $8 \cdot 7 \cdot6 \cdots 1$, hence $49$ cannot divide index$[G:N]$. We are done when we can show that $7$ divides index$[G:N]$. Assume the contrary, then the canonical image in $G/N$ of the Sylow $7$-subgroup $P$ would be trivial: $PN/N=\{\bar{1}\}$. This means $P \subseteq N$. Now apply the Frattini Argument - it follows that $G=NN_G(P)=NH=H$ (remember $N \subseteq H$). But this implies that $P \unlhd G$, and hence $\#Syl_7(G)=1$, a contradiction to $\#Syl_7(G)=8$.
