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

A group is called perfect if it equals its derived subgroup.

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

If this unique maximal perfect subgroup does exist, it is obviously fully-invariant. Now I have to prove the existence.

Let $S$ denote the set of perfect subgroups of $G$. As $\{ 1 \} \in S$, $S \neq \phi$. So, I think of proving it by showing that, for any two elements $H_1$ and $H_2$ of $S$, the subgroup $\langle H_1, H_2 \rangle$ of $G$ generated by $H_1$ and $H_2$ is still in $S$. But what will happen if $G$ doesn't satisfy the maximal condition (every set of subgroups has a maximal element) and $S$ doesn't have a maximal element?

Thanks very much.

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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.

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How do we know that our group generated by the union is not the entire group itself? It then is not truly maximal if it is the entire group. –  Vladhagen Oct 17 '13 at 19:13
@Vladhagen, in that case, the group $G$ itself is perfect, so it is its own maximal perfect subgroup! –  Mariano Suárez-Alvarez Oct 17 '13 at 19:51
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.