Possible Duplicate:
Proving a theorem about a finite simple group
My homework: Let $G$ be a finite simple group such that for any prime $p$, the group $G$ has at most 6 $p$-Sylow subgroups. Prove that $G$ is cyclic.
My attempt: If $G$ is finite, simple and cyclic then it's actually $\mathbb{Z}_p$ for some prime $p$, and for any prime $p$, $\mathbb{Z}_p$ does have the required property. So the question is equivalent to proving the $G$ is of prime order.
First, if $G$ is of prime-power order then we're done because, as a $p$-group, its center is nontrivial and then as a simple group, we must have $G=Z(G)$, but again, since $G$ is simple (and by Cauchy's theorem) we have that $G$ is of prime order.
Now, if $G$ is not of prime power order, by simplicity it has more than one $p$-Sylow subgroups for any prime $p$, but then by Sylow's theorem it has at least $p+1$ $p$-Sylow subgroups so $p+1 \leq 6$, that is, $p \leq 5$.
So the possible prime factors of $|G|$ are $2, 3$ and $5$. Actually, by Sylow's theorem, $|G|$ is divisible by $2$ and $3$ and possibly by $5$ as well.
How do I continue (and is there a simpler argument?).