I came across a proof where it is assumed that a simple group is solvable, but I can't really understand how.
Is this true? If yes, help to prove it would be very helpful.
PS : Question was to prove that all groups of order < 50 are solvable.
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See this post for the genuine question. $A_5$ is the first simple non-abelian group. – Dietrich Burde Apr 05 '21 at 15:22
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If $G$ is a solvable group then there exists a finite chain of subgroups $$G\geq G^1\geq G^2\geq...\geq G^n\geq1$$ such that $G^{i+1}=[G^i, G^i]$.
Now, $[G, G]$ is a normal subgroup of $G$ (why?), and so if $G$ is non-trivial and simple then either $[G, G]=G$ or $[G, G]=1$. In the first case, $G$ is not solvable (as $G^i=G$ for all $i$). In the second case, $G$ is abelian (why?). Therefore,
A simple group is solvable if and only if it is abelian.
user1729
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@ZiruiWang For $A, B$ subsets of a group $H$, we write $[A, B]$ to mean the subgroup generated by the set of commutators of the form $[a, b]$, $a\in A, b\in B$, i.e. $[A, B] = \langle [a, b]\mid a\in A, b\in B\rangle$. So $[G, G]$ is the derived (commutator) subgroup of $G$. – user1729 Apr 22 '24 at 09:31
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No, if a simple group is solvable it must be abelian. Because since it has no normal groups it won't be possible to find a factor group.
The smallest non-abelian simple group is $A_5$ and it has order $60$, so every simple group of order less than $50$ is in fact abelian.
