# 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.

• I wonder what do you expect to happen if, for example, $\;G\;$ is non-abelian simple...then $\;G\;$ is trivially perfect, but it could be all its proper subgroups aren't (for example, take $\;A_4\;$) ... – DonAntonio Oct 18 '13 at 6:05
• When a condition is not automatically satisfied by the group itself (like perfect in this case), being maximal with respect to that property usually does not require being a proper subgroup. – Tobias Kildetoft Oct 18 '13 at 7:33
• Just take the intersection of the terms in the derived series. (For infinite groups, these terms are indexed by ordinals.) – Derek Holt Oct 18 '13 at 7:54

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.