Let $G$ be an infinite non-abelian group. Is there any statement like this in group theory : If its commutator subgroup is simple then $G$ is simple?
Normally if $G$ is non-abelian, simple, its commutator subgroup is equal to $G$. But I wonder its inverse.
This is a question about Thompson's Group $F$. This group has the simple commutator subgroup. I want to prove that $F$ is simple.
Best Answer
The Thompson group $F$ has the $2$-generator presentation $$\langle A , B\mid [ A B^{-1} , A^{-1}B A ],\, [ A B^{-1}, A^{-2} B A^2 ]\rangle,$$ as described in this Wikipedia page.
You can see immediately from this that $F/[F,F] \cong {\mathbb Z}^2$.