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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.

RobPratt
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alans
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  • You mean less than $60$, as the alternating group on $5$ letters is a simple group of order $60$. – Andreas Caranti Jun 06 '13 at 14:39
  • I thought there was a non-solvable group of order 60? Edit: Andreas beat me to it by 4 seconds. – Charles Jun 06 '13 at 14:39
  • What nonabelian simple groups do you know, and what size are they? Alternatively, do you know the Sylow Theorems? – Mark Bennet Jun 06 '13 at 14:41
  • My question is edited. Yes, I know Sylow theorems. – alans Jun 06 '13 at 14:45
  • http://math.stackexchange.com/questions/353552/group-of-order-8p-is-solvable-for-any-prime-p would handle 24, 40, 56. 54 is just silly: 4k+2 are always solvable and 2p^n is obvious. 48 falls to http://math.stackexchange.com/questions/398307/let-g-be-a-simple-group-of-order-n-let-h-be-a-subgroup-of-g-of-index-k-show-th – Jack Schmidt Jun 06 '13 at 15:23
  • A more efficient is Burnside's $p^a q^b$ which handles all but 30 and 42, which fall to “square-free”. If we add obviously normal sylows and 4k+2 to the batch, then we can handle up to and including 250, only counting elements for 132. – Jack Schmidt Jun 06 '13 at 15:32
  • The paper Solvable Groups - A Numerical Approach shows that any group of order up to 100 and not 60 is solvable. – hengxin Apr 27 '14 at 07:09
  • For the orders $24, 48, 54$ you can use Sylow's theorems, and the fact that if a group has a subgroup of index $k$, then it has a normal subgroup of index a multiple of $k$ and a divisor of $k!$. For $40$, How many $5$-Sylow subgroup may you have? For $56$, if the number of $7$-Sylow subgroups is $8$, how many elements of order $7$ can you count, and how many elements are left? – Andreas Caranti Feb 22 '22 at 13:17

1 Answers1

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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.

dei
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    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. – user141592 Oct 22 '14 at 00:08