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Given a variety (in the sense of universal algebra) $\mathscr{V}$ axiomatized by a finite set of equations $E$, say that $\mathscr{V}$ is consistently gappy iff it is consistent with $\mathsf{ZF}$ that there is an algebra $A\in\mathscr{V}$ which is uncountable but has only countably many subalgebras (since $E$ is finite this makes sense). For example, the variety of groups is consistently gappy via the "dyadic circle" $\mathbb{Z}[{1\over 2^\infty}]/\mathbb{Z}$ (see this old answer of mine for details), and a similar argument works for rings. On the other hand, lattices - or any other variety in which all operations are idempotent - are not consistently gappy since they always have at least as many substructures as elements.

My question is:

Which (finitely-based) varieties are consistently gappy?

I'm tempted to suggest that the following is a sufficient condition for consistent gappiness of a nontrivial variety $\mathscr{V}$:

  • There is some binary term operation $\star$ such that $\star$ is associative in $\mathscr{V}$ and for some $a\in A\in\mathscr{V}$ we have $\vert\{a^{\star n}: n\in\mathbb{N}\}\vert=\infty$.

However, I have no actual evidence for this.

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  • $\begingroup$ Didn't we prove a theorem towards that sort of result in our paper? $\endgroup$ Commented Aug 12 at 22:07
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    $\begingroup$ @AsafKaragila You mean Theorem 3.2? I do think that (or something of the same flavor) will be a useful tool, but we're still left with the problem of understanding which $\mathscr{V}$s are guaranteed to have elements with appropriately few "large-automorphism-group-fixed" substructures. $\endgroup$ Commented Aug 12 at 23:17
  • $\begingroup$ I am not sure what you mean by "via" and I don't understand the details of the construction in your link - the group you've named is called the Prufer $2$-group and it is, of course, countable, even in ZF. $\endgroup$ Commented Aug 13 at 2:12
  • $\begingroup$ @QiaochuYuan This has to do with external vs. internal isomorphisms. Forcing over the universe for simplicity, we form a symmetric extension $V\subset M\subset V[G]$; that is, while $V[G]\models\mathsf{AC}$ (since $V$ does), we merely have $M\models\mathsf{ZF}$. Now in $M$ there is a group $X$ such that $M\models$ "$X$ is uncountable and has only countably many subgroups." In $V[G]$, we have that $X$ is countable and in fact isomorphic to a group $Y\in V$ such that $V\models$ "$Y$ is countable," but this data isn't in $M$ alone. In this particular case, $Y$ is the Prufer 2-group. $\endgroup$ Commented Aug 13 at 2:28
  • $\begingroup$ (The linked answer is phrased in terms of models of ZFA (= ZF with atoms), which is easier to work with for elementary constructions like this. Again, the point is internal vs. external; if $V(A)$ is the whole universe with countable set of atoms $A$, we form a subsystem $W(A)$ with the property that $W(A)$ thinks $A$ is uncountable but $V(A)$ sees that $A$ is countable. See Jech's book The axiom of choice for this sort of approach. Crucially, (via a blackbox) when we "de-atomize" this approach we get exactly a symmetric extension argument a la my previous comment.) $\endgroup$ Commented Aug 13 at 2:31

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