# A finite group which has exactly eight Sylow $7$-subgroups

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.

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## 1 Answer

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?

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Can you give me more details? –  user111636 Jan 5 '14 at 15:33
@user111636 that's about as explicit a hint as Hagen could give without writing out the full solution. Now time for you to think. –  Igor Rivin Jan 5 '14 at 17:14
@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. –  user114539 Jan 6 '14 at 6:25
i mean why 56 divides. –  user114539 Jan 6 '14 at 12:46
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? –  user111636 Jan 7 '14 at 2:12