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

Let $G$ be a simple, finite group, so that for every prime $p$, it satisfies $k_p=|\operatorname{Syl}_p⁡ G| \le 6$. 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 $n = |G|$. 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?

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