How do ultrafilters of subgroups look like?

Question

In my work related to the axiom of choice one of the central notions is a filter of subgroups. Namely, if $G$ is a group, a filter of subgroups is a non-empty collection of subgroups which is closed under supergroups and finite intersections.

But we are not interested in any filter, but specifically in normal filters which satisfy the clause that if $g\in G$ and $H$ is in the filter, then $g Hg^{-1}$ is also in the filter.

Are there also ultrafilters (maximal proper filters) of subgroups? Certainly there are principal ones, e.g. all the groups which contain a certain element of $G$ is the principal ultrafilter generated by the subgroup $\langle g\rangle$. But are there free ultrafilters?

If there are, are there any normal ultrafilters (ultrafilters which are also normal), or do the two conditions clash in some way?

I know that we can think about the filters of subgroups as just filters on $G$ in the standard set theoretic sense, and then considering all the subgroups generated by sets in the filter. But this doesn't help me understand how might a free and normal ultrafilter of subgroups can look like, especially since this is not a bijection between the two notions (look at the case where a subgroup can be generated by two disjoint subsets).

Answer 1

I think there are examples of "free ultrafilters" of subgroups which are normal, but I don't know, if this answers your question.

Consider the free commutative group $G = \mathbb{Z}^{\oplus \mathbb{N}}$ on $\mathbb{N}$ as the set of free generators. Then define a filter of subgroups

$$\mathcal{F} = \big\{H\subseteq G\,\big|\,\mbox{ there exists a cofinite subset }A\mbox{ of }\mathbb{N}\mbox{ such that }\mathbb{Z}^{\oplus A}\subseteq H\big\}$$

By Zorn's lemma there exists a maximal with respect to inclusion filter $\mathcal{U}$ of subgroups containing $\mathcal{F}$. It cannot be principal due to the fact that for every $g\in G$ there exists cofinite $A$ in $\mathbb{N}$ such that $g\not \in \mathbb{Z}^{\oplus A}$. On the other hand it is normal as $G$ is commutative.