Unique "maximal" perfect subgroup. There is a problem that I have come across that I have yet to find a sufficient answer to: Prove that every group G has a unique maximal perfect subgroup R. 
The usual "proof" that has been given is to take all perfect subgroups of G and the the group generated by all of these groups, call it M, is perfect (True. No problem there.) and M is maximal in G. How are we so sure that the group M does not equal G? I am fine with having the unique perfect subgroup be all of G, but G is not maximal in itself. At least as to my understanding, a maximal subgroup needs to be proper. (Much like a maximal ideal is proper by definition.) So yes, in some group like A5 (in general just a non abelian simple group), the whole group is perfect, but the group is NOT a maximal subgroup.
See the discussion at this question for further details
This could all come down our definition of maximal. It is not hard to use Zorn's lemma to get a true (i.e. proper) maximal perfect subgroup in G, but getting uniqueness is the difficulty then. So, if we can get a true maximal perfect subgroup that is unique, that would be great. 
 A: It is possible that $M = G$, and allowing this is the natural way to define a "maximal perfect subgroup". 
Generally, when we say that $M$ is a "maximal $\mathscr{P}$-subgroup" of $G$, we mean that $M$ is maximal among the set of $\mathscr{P}$-subgroups of $G$. That is, if $L$ is a $\mathscr{P}$-subgroup and $M \leq L$, then $M = L$. This definition allows $M = G$.
When we talk about "maximal subgroups", we talk about subgroups that are maximal among the proper subgroups of $G$. So the terminology might be a bit confusing.
A: It might seem weird to say that $G$ can be a maximal perfect subgroup because it is not proper. This definition is problematic when we are working with regular maximal subgroups because it means that there is no maximal subgroup other than $G$. However, when we are talking about perfect subgroups, $G$ might not be perfect. For example, if $G$ is a nontrivial abelian group, then $G'=\{1\}\neq G$. When working in this context, knowing that $G$ is a perfect group can bring us a good deal of information, so we should not exclude this from our consideration when defining what it means to be a maximal perfect subgroup.
