Example of a group such that the perfect maximal subgroup is $1$.

Question

I an looking for an example of a group say $G$ such that the perfect maximal subgroup is $1$.

---

I cannot find any references or come up with an example of my own. Could someone please cite a reference?

Answer 1

If you start with a finite group $G$ and construct the derived sequence $G_0=0$, $G_1=[G_0,G_0]$, $G_2=[G_1,G_1]$, $\dots$, then the intersection $\bigcap_{i\geq0}G_i$ is the unique maximal perfect subgroup of $G$ (in the sense that it is a perfect subgroup and contains all other perfect subgruops of $G$) If that intersection is trivial, then clearly one of the $G_i$ is trivial already, and your your is solvable.

If the group is infinite, you need to continue the derived sequence transfinitely, and intersect the whole thing to find the maximal perfect subgroup. Its triviality will not imply that the group is solvable, in general, because it may be the case that none of the initial, "finite" steps was trivial. Groups for which the transfinite sequence has trivial intersection are called hypoabelian.

Answer 2

Nonabelian free groups have the desired property, and they are not solvable.