# A simple group with $|\operatorname{Syl}_p⁡ G| \le 6$ is cyclic

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?

• $4$ 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 $20$. – Derek Holt May 23 at 19:08
• @DerekHolt changed, thank you. Why does $k_2=5$ implies that $G\le A_5$? – Roy Sht May 23 at 20:43
• 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 $A_5$. – Derek Holt May 23 at 21:25
• 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. – Jyrki Lahtonen Jun 12 at 17:01