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.