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.