Proving a theorem about a finite simple group I need to prove this theorem that I had in an exam and I am stuck.

Let $G$ be a finite simple group and we assume that for every prime $p$ the number of $p$-sylow sub-groups is $\leq 6$. Prove that $G$ is cyclic.

my lead: I proved that $|G| = 2^a3^b5^c$ (no other primes divides the size of $G$).
and now I've thought that maybe I can prove that $a=1$ or $b=1$ or $c=1$ and continue from there (I've proved before that 
 A: We can assume that $G$ is not trivial.
If $G$ has a normal $p$-Sylow subgroup, then $G$ must have order $p^k$ since $G$ is simple. Groups of prime power order have a nontrivial center, so $G$ has order $p$ and is thus cyclic.
Suppose then that no Sylow subgroup of $G$ is normal. We will show that this leads into a contradiction. Now for each prime divisor $p$ of $G$ there are $2$, $3$, $4$, $5$ or $6$ $p$-Sylow subgroups. The amount of $p$-Sylow subgroups is $\equiv 1 \mod p$, so the only prime divisors of $G$ can be $2$, $3$ and $5$. 
If $G$ has order divisible by $3$, then there are four $3$-Sylow subgroups. Therefore there exists a homomorphism $\phi: G \rightarrow S_4$ with $\operatorname{Ker}(\phi)$ contained in the normalizer of a Sylow $3$-subgroup. Since $G$ is simple, the kernel is trivial and thus $G$ is isomorphic to a subgroup of $S_4$. Now $S_4$ has order $2^3 \cdot 3$, so $G$ has three $2$-Sylow subgroups. Thus we can embed $G$ into $S_3$, but this isn't actually possible when $G$ has four $3$-Sylow subgroups.
Therefore the order of $G$ is $2^a5^b$ for some integers $a$ and $b$. If $b > 0$, then $G$ would have six $5$-Sylow subgroups. This is not possible, because $6$ does not divide $2^a5^b$. Thus $G$ has order $2^a$ and must be cyclic, contradicting the assumption that $G$ has no normal Sylow subgroups.
