If we have exactly 1 eight Sylow 7 subgroups, Show that there exits a normal subgroup $N$ of $G$ s.t. the index $[G:N]$ is divisible by 56 but not 49.

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.

• Consider the action by conjugation of $G$ on the $8$ Sylow $7$-subgroups, and let $N$ be the kernel of this action. – Derek Holt Apr 5 '15 at 21:07
• @DerekHolt, is there a way to solve it without using group actions? I'm not as familiar with that – cele Apr 6 '15 at 1:55
• I cannot think of any other way of solving this particular problem, and it looks to me as though this is the intended solution. You should learn about group actions which are fundamental to group theory. – Derek Holt Apr 6 '15 at 7:39
• Also, I don't think you can sensibly study Sylow's theorems and applications without using group actions. The standard proofs involve group actions, and textbooks generally cover groups actions before Sylow's theorems. – Derek Holt Apr 6 '15 at 10:36
• @DerekHolt Doesn't the simple group of order $\;168\;$ has exactly eight Sylow $\;7$-subgroups? – Timbuc Apr 10 '15 at 17:41

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 $\#Syl_7(G)=8$.
• But theorem $1$ is a result of group action :) – mesel Apr 12 '15 at 22:51