Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

up vote 3 down vote accepted

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.

share|improve this answer
    
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

As you yourself and Mariano mention, you can create this largest perfect subgroup from below, by taking the subgroup K generated by all perfect subgroups. Any particular element in K is a product of finitely many elements in Hi, and your finitely-many proof shows that product is a product of commutators from the perfect subgroups Hi.

However, you can also find this subgroup from above. I like this method better because the first method requires having a large supply of perfect subgroups (and at face value requires having them all, even K itself). The second method just requires being able to find derived subgroups (and in the infinite case, intersections).

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.