Finite solvable group Suppose that $G$ is a finite group such that $(|G|, 15)=1$. Why $G$ is solvable group?
 A: We note that the Suzuki groups are only non-Abelian 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 non-Abelian simple group (and or $M/K\cong S\times $ $
S\cdot \cdot \cdot \times S$ where $S$ is non-Abelian 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.
A: 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...
