Let $G$ be a simple, finite group, s.t. for every prime $p$, it satisfies $k_p=\left|\operatorname{Syl}_p⁡ G\right| \le 6$. Show that $G$ is cyclic.

My attempt: Let $n=p_1^{e_1}p_2^{e_2}\ldots p_r^{e_r}$ be the (distinct) prime factorization of $n = \left|G\right|$. If $n$ is prime, $G$ is cyclic. So we assume $e_1\ge 1$ and $e_2\ge1$. From Sylow's theorems, we have $k_{p_1}\mid\prod_{i\ne1} p_i^{e_i}$ and $k_{p_1}\equiv1\ (\operatorname{mod} p_1)$, and the same applies for $p_2$. Sicne $G$ is simple, $k_{p_{1,2}}\ne1$, and it is given that $k_{p_{1,2}}\le6$. The options for $p_1$ are

  1. No prime $p_1$ satisfies $k_{p_1}=2$.
  2. The other options are that $p_1=2,3,5$ and $k_{p_1}=3,4,6$ respectively.

How do I continue from here?

  • 1
    $\begingroup$ $4$ is not prime, so you cannot have $p_1=4$. Also $k_p = 1,2,3,4$ are impossible in a nonabelian simple group, so you are left with $k_2=5$ and $k_5=6$ to think about. Then $k_2=5 \Rightarrow G \le A_5$, so $|G|$ divides $20$. $\endgroup$ – Derek Holt May 23 at 19:08
  • 1
    $\begingroup$ @DerekHolt changed, thank you. Why does $k_2=5$ implies that $G\le A_5$? $\endgroup$ – Roy Sht May 23 at 20:43
  • 4
    $\begingroup$ The conjugation action of $G$ on the set of Sylow $5$-subgroups induces and embedding $G \to S_5$ but then $G$ simple implies that it embeds into $A_5$. $\endgroup$ – Derek Holt May 23 at 21:25
  • $\begingroup$ Roy, if you completed this exercise following Derek Holt's hint, feel free to post your solution as an answer. That way you get more feedback on the details. $\endgroup$ – Jyrki Lahtonen Jun 12 at 17:01

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