Simple groups with order a power of two times fifteen ($|G|=2^e\cdot 15⟹G\cong A_5$)

For a finite simple group of this order, $e=0$ gives a normal Sylow $5$-subgroup, and $e=1$ gives a nontrivial sign homomorphism from the regular action, because an involution swaps fifteen pairs. Thus $e\ge 2$. The Sylow $2$-count divides $15$; it cannot be one or three. If it is five, a Sylow subgroup's normalizer has index five. If it is fifteen, the strengthened Sylow congruence from intersections produces two Sylow subgroups meeting in index two. Their common subgroup is normal in both, so its normalizer contains both and has index three or five. Simplicity excludes index three. The resulting faithful degree-five action embeds $G$ in $A_5$ and forces order $60$, proving the conclusion.