If we have exactly $8$ 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.

Question

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.

Answer 1

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 $\#Syl_7(G)=8$.

Answer 2

In your case, $G$ acts transitively by conjugation on the set $\operatorname{Syl}_7(G)$, where $|\operatorname{Syl}_7(G)|=8$. Therefore, there's a homomorphism $\varphi\colon G \rightarrow \operatorname{Sym}(\operatorname{Syl}_7(G))$, whence (First Homomorphism Theorem):

$$G/\operatorname{ker}\varphi \cong \operatorname{im}\varphi \le \operatorname{Sym}(\operatorname{Syl}_7(G)) \tag 1$$

and thence (Lagrange Theorem):

$$[G:\operatorname{ker}\varphi]=|G/\operatorname{ker}\varphi| \mid 8! \tag 2$$

Now, by contrapositive, let's suppose $49 \mid [G:\operatorname{ker}\varphi]$; thence, by $(2)$, $49\mid8!$: contradiction. Therefore, $49 \nmid [G:\operatorname{ker}\varphi]$. So, if we prove that $56 \mid [G:\operatorname{ker}\varphi]$, then take $N=\operatorname{ker}\varphi$, and we are done. To this aim, let's remind that ($\operatorname{Stab}(P)=N_G(P)$):

$$\ker\varphi=\bigcap_{P\in {\rm{Syl}_7(G)}}N_G(P) \tag 3$$

Therefore:

\begin{alignat}{1} 56 \mid [G:\operatorname{ker}\varphi] &\iff 56 \mid \frac{|G|}{|\bigcap_{P\in {\rm{Syl}_7(G)}}N_G(P)|} \\ \tag 4 \end{alignat}

Now, by the Orbit-Stabilizer Theorem:

$$8\cdot|N_G(Q)|=|G|, \forall Q \in \operatorname{Syl}_7(G) \tag 5$$

and thence:

\begin{alignat}{1} 56 &\mid \frac{|G|}{|\bigcap_{P\in {\rm{Syl}_7(G)}}N_G(P)|} \iff \\ 56 &\mid \frac{|G|}{|N_G(Q)|}\cdot\frac{|N_G(Q)|}{|\bigcap_{P\in {\rm{Syl}_7(G)}}N_G(P)|} \iff \\ 56 &\mid 8\cdot\frac{|N_G(Q)|}{|\bigcap_{P\in {\rm{Syl}_7(G)}}N_G(P)|} \iff \\ 7 &\mid \frac{|N_G(Q)|}{|\bigcap_{P\in {\rm{Syl}_7(G)}}N_G(P)|} \\ \tag 6 \end{alignat}

But this latter holds as a corollary ($p=7$) of this general:

Lemma. Let $G$ be a finite group, $p$ a prime divisor of $|G|$ and $\operatorname{Syl}_p(G)$ the set of the Sylow $p$-subgroups of $G$. Assume further that $|\operatorname{Syl}_p(G)|>1$. Then:

$$p \mid \frac{|N_G(Q)|}{|\bigcap_{P\in {\rm{Syl}}_p(G)}N_G(P)|}, \space\forall Q \in \operatorname{Syl}_p(G) \tag 7$$

Proof. See here and the therein accepted answer. $\space\space\space\Box$

So $N:=\operatorname{ker}\varphi$ is the sought normal subgroup of $G$.