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.
Asked
Active
Viewed 1,339 times
4
-
See this post for the genuine question. $A_5$ is the first simple non-abelian group. – Dietrich Burde Apr 05 '21 at 15:22
2 Answers
4
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
- 31,015
3
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.
