A finite simple group with no more than 6 p-Sylow subgroups for any $p$ is cyclic [duplicate]

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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?).

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marked as duplicate by Chris Eagle, t.b., Kannappan Sampath, anon, robjohn♦Mar 31 '12 at 16:33

Suppose $G$ is a finite nonabelian simple group with $n_p$ $p$-Sylows for each prime $p$ dividing $|G|$. $G$ acts on its $p$-Sylows by conjugation. The kernel of this action is a normal subgroup of $G$. Since the action is nontrivial, the kernel must be trivial, so this action defines an embedding of $G$ into $S_{n_p}$.
Now suppose that $n_p\le 6$ for all $p$. As you observe, this forces $|G|$ to be divisible by $3$, with $4$ $3$-Sylows. Thus $G$ embeds in $S_4$. But $S_4$ has no nonabelian simple subgroups.