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
- No prime $p_1$ satisfies $k_{p_1}=2$.
- The other options are that $p_1=2,3,5$ and $k_{p_1}=3,4,6$ respectively.
How do I continue from here?
On the contrary, let's assume that $G$ is nonabelian simple group, which implies that $60=|A_5|\le |G|$. As you mentioned, for any prime $p\in \pi(G)$, $k_p \equiv 1 \pmod{p}$. Since $G$ is nonabelian simple group $k_p> 1$, for every $p\in \pi(G)$, hence $p<k_p\le 6$, which implies $\pi(G)\subseteq\{2,3,5\}$. On the other hand, by Burnside's $p^\alpha q^\beta$-theorem, we get that $\pi(G)=\{2,3,5\}$. Now we see that $k_3=4$, which is a contradiction, because that way $G\le S_4$ (actually $G\le A_4$, since $G$ is assumed to be a simple group).