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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).

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  • $\begingroup$ To be honest, I'm not 100% sure how to tag this question. $\endgroup$
    – Asaf Karagila
    Feb 7, 2020 at 9:40
  • $\begingroup$ Would the universal-algebra tag be appropriate? $\endgroup$
    – Shaun
    Feb 7, 2020 at 9:45
  • $\begingroup$ I'm not sure, maybe. Topological group theory is also related somehow, since we can see these filters as "open subgroups", especially if the filter is normal. $\endgroup$
    – Asaf Karagila
    Feb 7, 2020 at 9:47
  • $\begingroup$ How are you defining a free subgroup ultrafilter? $\endgroup$
    – user1729
    Feb 7, 2020 at 9:57
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    $\begingroup$ Okay. So finite index subgroups of residually finite groups might work. They always satisfy the intersection, supergroup and conjugation properties. The residual finiteness is there to ensure that there is no least element (without this assumption you can have least elements, e.g. consider $G=H\times F$ with $H$ infinite simple and $F$finite). The issue is the "ultra" bit... $\endgroup$
    – user1729
    Feb 7, 2020 at 11:08

1 Answer 1

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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.

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  • $\begingroup$ Ah, of course, working with Abelian groups! Thanks. (It's not a complete answer to my quandary, but it is helpful.) $\endgroup$
    – Asaf Karagila
    Feb 7, 2020 at 10:50
  • $\begingroup$ @AsafKaragila I am glad that you find it helpful. What would be a complete answer? Your question is interesting. $\endgroup$
    – Slup
    Feb 7, 2020 at 11:04
  • $\begingroup$ To be honest, I'm not 100% sure what would be a complete answer. But perhaps some general facts about ultrafilters of subgroups, rather than an example of a normal free ultrafilter... (The "usual" case sees $G$ as non-commutative, and with very few normal subgroups, although the Abelian cases do appear now and then.) $\endgroup$
    – Asaf Karagila
    Feb 7, 2020 at 11:06
  • $\begingroup$ @AsafKaragila Ah. I see. I agree it would be nice to have noncommutative example. $\endgroup$
    – Slup
    Feb 7, 2020 at 11:21

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