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.

• You mean less than $60$, as the alternating group on $5$ letters is a simple group of order $60$. Jun 6, 2013 at 14:39
• I thought there was a non-solvable group of order 60? Edit: Andreas beat me to it by 4 seconds. Jun 6, 2013 at 14:39
• What nonabelian simple groups do you know, and what size are they? Alternatively, do you know the Sylow Theorems? Jun 6, 2013 at 14:41
• My question is edited. Yes, I know Sylow theorems. Jun 6, 2013 at 14:45
• math.stackexchange.com/questions/353552/… would handle 24, 40, 56. 54 is just silly: 4k+2 are always solvable and 2p^n is obvious. 48 falls to math.stackexchange.com/questions/398307/… Jun 6, 2013 at 15:23

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.
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.
• Why does it follow that all groups of order less than 60 are solvable from the fact that $A_5$ is not a composition factor in any subnormal series of order less than 60? It seems like a bit of a leap in logic. Oct 22, 2014 at 0:08