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Let $G$ be a finite group which has exactly eight Sylow 7 subgroups. Show that there exists a normal subgroup $N$ of $G$ such that the index $[G:N]$ is divisible by 56 but not by 49.

I will start by saying that I previously asked this question before here 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. I had to take some time to think about this problem and here is what I have gotten so far: Please help fill in any gaps or make corrections.

Now,$P$ is a Sylow p-subgroup. We also know that $8=[G:N_G(P)]$ The Third Sylow Theorem states that If $G$ is a finite group and $p$ a prime, then the number of Sylow p-subgroups of $G$ divides $|G|$ and is of the form $kp+1$ for some $k \ge 0$
So, then $[G: N_G(P)] = |{gPg{-1}/g \in G}|$ So, if P is a Sylow 7-subgroup, then $[G:N_G(P)]=8$ If we consider $N_G(P)$, $G \to A(S),$ where $S={gNg(p)/g \in G}$ and $|S|=8$

$h \to \tau: S \to S : \tau_h(gNg(P))=hg Ng(P)$ Now, I'm not sure if in order to say that, must I check that it is one to one and onto, therefore making it a homormorphism?? I mean, I would say that it is a homomorphism, which tells us $N=ker \varphi$ is a normal subgroup of $G$ which then means that $N \subseteq N_G(P)$ and then $|G/N|/8!$. This then means that $[G:N]/8!$

So, $[G:N=[G:N_G(P)][N_G(P):N]$ where $[G:N_G(P)]=8$ But I'm still a little confused on how to finish the problem. I know that I must somehow show that $7/[G:N]$ or find a contradiction

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    $\begingroup$ The claim, as redacted in the title, is false if we require $\;N\neq G\;$: there exists a simple group of order $\;168\;$ with exactly $\;8\;$ sylow $\;7$-subgroups. $\endgroup$ – Timbuc Apr 10 '15 at 17:40
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    $\begingroup$ @Timbuc But in that case the claim is true with $N=1$. I don't think that there was any intention to exclude $N=1$. $\endgroup$ – Derek Holt Apr 10 '15 at 18:21
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    $\begingroup$ @DerekHolt Perhaps so, Derek. It is only that I think that in many of these cases the non-proper subgroups are put aside, but of course: it could be they wanted to take that possibility into account. $\endgroup$ – Timbuc Apr 10 '15 at 18:23
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    $\begingroup$ To show that $7$ divides $|G:N|$, you should show that the image of $P$ in $A(S)$ is nontrivial. To do that, show that if $Q$ is a Sylow $7$-subgroup with $Q \ne P$, then $P \not\le N_G(Q)$. $\endgroup$ – Derek Holt Apr 10 '15 at 18:24
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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 $\#Syl_7(G)=8$.

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