Suppose that $G$ is a finite group such that $(G, 15)=1$. Why $G$ is solvable group?

We note that the Suzuki groups are only nonAbelian simple groups of order prime to $3$ and $5$ is a prime divisor of the Suzuki groups. Let $G$ be unsolvable group, then $G$ has the following normal series: $1\unlhd K\lhd M\unlhd G$ such that $M/K$ is a nonAbelian simple group (and or $M/K\cong S\times $ $ S\cdot \cdot \cdot \times S$ where $S$ is nonAbelian simple group). As $ 3\nmid G$, then $M/K$ is a Suzuki group ( and or $M/K\cong S\times $ $ S\cdot \cdot \cdot \times S$ where $S$ is a Suzuki group). On the other hand $5\nmid G$, then $M/K$ is not isomorphic to a Suzuki group (and or $ M/K\ncong S\times $ $S\cdot \cdot \cdot \times S$ where $S$ is a Suzuki group), a contradiction. 


Because as you go through the list of finite simple noncyclic groups, you can observe that their orders are divisible by either 3 or 5. In fact, only a few of them, for instance, $\,^2B_2(8)$, will have orders not divisible by 3... Off course, this argument depends on the classification... Maybe, there is a trick to show it directly... 

