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.


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?

  • $\begingroup$ Can you give me more details? $\endgroup$ – user111636 Jan 5 '14 at 15:33
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    $\begingroup$ @user111636 that's about as explicit a hint as Hagen could give without writing out the full solution. Now time for you to think. $\endgroup$ – Igor Rivin Jan 5 '14 at 17:14
  • $\begingroup$ @Hagen I see that $|G/N|$ divides $|S_8|$. I see that $8||G/N|$. I see that 49 does not divide $|G/N|$, but i do not see why 56 does not divide $G/N$. Clearly i am missing something obvious, sorry. $\endgroup$ – user114539 Jan 6 '14 at 6:25
  • $\begingroup$ i mean why 56 divides. $\endgroup$ – user114539 Jan 6 '14 at 12:46
  • $\begingroup$ I can prove that there is a subgroup $H$ of $S_8$ such that $H$ also has exactly eight Sylow $7$-subgroups. What should I do then? $\endgroup$ – user111636 Jan 7 '14 at 2:12

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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