How to prove the following: Groups of order less than 60, are solvable?
I tried to do this by showing that groups of order $p^n$, $p q$, $p^2 q$, $p q r$ for primes $p$, $q$, $r$ are solvable. In this way, almost every group of order less than 60 is eliminated. Precisely, it only remains to show that groups of order 24, 40, 48, 54, 56 are solvable. It seems to me that this is too complicated way of solving, so I would like to know is there any other more elegant (and shorter) solution to the problem.
Thanks in advance.
Burnsides Theorem state that if $G$ is a finite group of order $p^aq^b$ where $p$ and $q$ are primes, and $a$ and $b$ are non-negative integers are solvable. Also using sylow theorem one can show that any group of order $pqr$ where $p$, $q$ and $r$ are distinct primes are solvable. Hence any group of order less than 60 are solvable.
ANOTHER WAY
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.