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I am continuing to work on the problem of complementation of infinite groups in the sense of Zacher. The problem was motivated by an attempt to understand subquandles, about which very little can be said. I've realized as I go that the problem is much more vast than I anticipated. From what I can tell, the complementation of infinite groups is not well understood either. I cannot even manage to find a sufficient, but not unnecessarily restrictive, class of infinite group which admits complementation. There are examples of infinite abelian groups that do not even admit maximal subgroups, e.g., the Prüfer group. I considered Noetherian, which guarantees the existence of maximal subgroups. This is not a sufficient condition for complementation, however. Can you recommend good resources that discuss complementation, so that I can understand the problem as best as possible?

Edit: I'm adding a quandle/rack-theoretic argument for why all finite racks must be complemented. Highlighting where the proof doesn't work for finite groups might be helpful in narrowing down the extent to which the analogy follows. I'd think of quandles as a generalization of conjugacy classes of groups.

Definition. (Rack, Quandle) A rack is a set $R$ together with a binary operation $\triangleright$ such that for all $a,b,c\in R$ we have $a\triangleright(b\triangleright c)=(a\triangleright b)\triangleright(a\triangleright c)$, and such that there exists a unique $x\in R$ with $a\triangleright x=b$. A quandle is a rack $Q$ with the property that $x\triangleright x=x$ for all $x\in Q$.

Theorem. Let $R$ be a finite rack. The following statements hold:

(1) The intersection of all maximal subracks of $R$ is empty.

(2) For any two subracks $Q_1$ and $Q_2$ of $R$ with $T=\langle Q_1,Q_2\rangle$, the subrack $Q_1$ has a complement in $\mathcal{R}(T)$ which is a subset of $Q_2$.

The proof of this can be found here. Saki and Kiani also discuss $G$-racks and the equivalence between Boolean and unique complementation for $G$-racks.

Alex Byard
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    One "easy" class is elementary abelian groups, i.e., vector spaces over $\mathbf F_p$ for some $p$. But even this seems to be equivalent to the axiom of choice (https://math.stackexchange.com/a/1713573/23805). – Sean Eberhard Jun 14 '23 at 10:30
  • @SeanEberhard I've been looking into classes of groups such as residually finite, etc. I can prove residually finite groups admit maximal subgroups, this is true. The main issue I'm encountering right now arises when we try to go down infinite chains of subgroups. The position Artinian takes care of this, but I'm pretty sure it's equivalent to the axiom of choice as well, and I'm not sure if Noetherian, Artinian groups form an interesting class. – Alex Byard Jun 14 '23 at 14:14
  • @SeanEberhard Actually, no, Artinian doesn't even do it. The proof I'm trying to generalize won't work with this either. The idea is to pass from an order argument in the finite case to an inclusion argument in the infinite case. – Alex Byard Jun 14 '23 at 14:21
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    The problem with your approach is that each of the properties you are investigating (residually finite, Noetherian, Artinian) are finiteness properties, i.e., properties that assert that the group in question behaves like a finite group in certain ways. This would make sense if all finite groups were complemented, but that's far from true (e.g., $C_4$). – Sean Eberhard Jun 14 '23 at 14:35
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    What infinite nonabelian complemented groups do you know? Any? – Sean Eberhard Jun 14 '23 at 14:36
  • @SeanEberhard To investigate further, could you name some known properties/classes of infinite groups? However, perhaps the issue of complementation is where the intuition between quandles and groups diverges. I can prove that all finite quandles (and racks) are complemented, to which end these properties are perhaps interesting. I'll update the question with the known proof of this fact. I know some infinite nonabelian complemented quandles, e.g., P-racks, and more generally G-racks. Perhaps, the two issues are related but more disjoint than I previously thought. – Alex Byard Jun 14 '23 at 15:08
  • @SeanEberhard Also, I'm a bit confused. I'm under the intuition that all finite groups are $G$-racks, and all $G$-racks have their subobject lattice as Boolean, hence uniquely complemented. – Alex Byard Jun 14 '23 at 15:20
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    I have only the faintest idea what quandles and racks are, but $C_4$ has an obviously uncomplemented subgroup. Maybe work it out in that case, and, generally, find more concrete examples. – Sean Eberhard Jun 14 '23 at 15:22
  • @SeanEberhard Okay, I trust you. I'll take a look and see why it's wrong. I've updated the question with a paper from Saki and Kiani. See Theorem 3.1 and Corollary 3.2 for the proof I'm trying to generalize, in case you're interested. I was wrong about all $G$-racks being complemented. It is merely equivalent to being Boolean. I wonder if there is a lattice theoretic approach here. – Alex Byard Jun 14 '23 at 15:25
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    My guess is the issue is that a subrack of a group is not the same thing as a subgroup, so the property of a group being complemented as a group is not the same as it being complemented as a rack. So the analogy with groups is not very good – Sean Eberhard Jun 14 '23 at 15:31
  • @SeanEberhard Yes, that makes sense. It seems the group-rack analogy is very good in other places, though, so I still have hope. For example, you can embed free groups into a free group generated by two elements, while you can embed free quandles into a free quandle generated by three elements. A number of properties carry over, I just don't have an eye for why certain properties don't agree. – Alex Byard Jun 14 '23 at 15:34
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    For what it's worth, I think free groups might be complemented in the Zacher sense. – Sean Eberhard Jun 14 '23 at 15:45
  • @SeanEberhard I'll look at it today. My intuition is that finitely-generated quandles are complemented, and I think profinite quandles should be as well. Do you know if either of these holds for groups? If this holds for free groups, it should hold for finitely-generated groups, right? I'll see what I can figure out today and update the question, or make a new one tonight. – Alex Byard Jun 14 '23 at 16:05
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    Finitely generated: no of course not. E.g., $\mathbf Z$, or almost anything that comes to mind. Profinite: also no. Observation: the only abelian complemented groups are direct sums of elementary abelian groups. Again, it's really jumping ahead to try to guess some wide class of groups that works when you hardly know any examples. Instead try to find some examples. I know very few – Sean Eberhard Jun 15 '23 at 08:20
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    Also, I already told you why finiteness properties (including "finitely generated" and "profinite") are never going to be enough, since even "finite" is not enough. – Sean Eberhard Jun 15 '23 at 08:24
  • @SeanEberhard You're correct. The problem of complemented quandles is much different. – Alex Byard Jun 15 '23 at 15:34

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