Let $G$ be a finite group which has exactly eight Sylow 7 subgroups. Show that there exits a normal subgroup $N$ of $G$ such that the index $[G:N]$ is divisible by 56 but not by 49.
Now this is my first mode of thinking, if we have exactly one Sylow 7 subgroup and we know that there exits a subgroup of order 56, then I'm assuming that we would mean that we would have to have $56=2^3*7$ in order for this to be true. If it is to be a normal subgroup, then there must only exist one Sylow p-subgroup. I'm still working on how to show this but this is all I have so far.