I'm reviewing for my final, and the following is a (non-homework) question that I'm stuck on:

Let $G$ be a finite and simple group, and suppose $|\mathrm{Syl}_p(G)| = p+1$.

(a) Prove that $G$ can be embedded into $S_{p+1}$

(b) Prove that $p^2 \nmid |G|$

(c) Prove that $C_G(P) = P$

(d) Prove that $|G|$ divides $p(p^2 - 1)$

I used (a) to prove (b), and (b) implies that $P\in \mathrm{Syl}_p(G)$ has order $p$. I used that to prove (c), but I'm not seeing how any of this implies (d). Any hints to show (d)?


Hint: Consider that we are given that $|G: N_G(P)| = p+1$ and we have shown in part (c) that $|C_G(P)| = p$. What can we say about $|N_G(P): C_G(P)|$?

  • $\begingroup$ So $[N_G(P) : C_G(P)] = [N_G(P):P]$, and $[G:P] = [G:N_G(P)][N_G(P):C_G(P)]$, so $|G| = p(p+1)[N_G(P):C_G(P)]$. I need to show then that $[N_G(P):C_G(P)]$ divides $p-1$. $\endgroup$ – Möbius Dickus Dec 7 '17 at 23:38
  • 1
    $\begingroup$ @MöbiusDickus that's what you need to do $\endgroup$ – Rolf Hoyer Dec 7 '17 at 23:45

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