Prove that every group $G$ has a unique maximal perfect subgroup $R$ and $R$ is fully-invarinat in $G$

Solution 1:

As you yourself and Mariano mention, you can create this largest perfect subgroup from below, by taking the subgroup K generated by all perfect subgroups. Any particular element in K is a product of finitely many elements in H_i, and your finitely-many proof shows that product is a product of commutators from the perfect subgroups H_i.

However, you can also find this subgroup from above. I like this method better because the first method requires having a large supply of perfect subgroups (and at face value requires having them all, even K itself). The second method just requires being able to find derived subgroups (and in the infinite case, intersections).

A perfect subgroup of G is contained in [G,G], and so also in $G^{(2)}=[[G,G],[G,G]]$, and so on in $G^{(n+1)} = [ G^{(n)}, G^{(n)} ]$. It is then of course in the intersection $G^{(\infty)} = \cap_{n=2}^\infty G^{(n)}$. In a finite group, $G^{(\infty)}$ is the unique largest perfect subgroup and the unique smallest normal subgroup with solvable quotient. In infinite groups, one may need to go further, defining $G^{(\infty+1)}=[G^{(\infty)},G^{(\infty)}]$, and so on transfinitely (taking intersections as limit ordinals). Such a series always stabilizes (a group only has a specific cardinal worth of subgroups at all), and by definition it stabilizes at a perfect subgroup containing all other perfect subgroups.

Solution 2:

The subgroup generated by the union of all the perfect subgroups of a group is perfect—and it is therefore the unique maximal perfect subgroup.