Clarification on proof that all groups of order $< 60$ are solvable I've manged to prove that all groups of order $< 60$ are solvable, using Burnside's theorem. However, I found an alternate proof here Question about solvable groups It states that:
"Note that $A_5$ is the smallest non-abelian simple group and its order is $60$. Therefore in any subnormal series of any group of order less than $60$, $A_5$ is not a composition factor. Hence all group of order less than $60$ are all solvable."
Could anyone explain why the fact that in any subnormal series of any group of order $< 60$, $A_5$ is not a composition factor implies that all groups of order less that $60$ are solvable? It's not clear to me why that follows immediately.

Let $N$ be a maximal normal subgroup. Then $G/N$ is simple and abelian (because it is smaller than $A_5$), hence cyclic. $N$ is strictly smaller than $G$, hence we may assume by induction that itis solvable, hence $G$ is solvable.
