It is known that upon not accepting the Axiom of Choice (AC), there exist models of ZF in which there are projective spaces (over a division ring) with a trivial automorphism group. (This is a truly remarkable result, if I may say.)

Now let $G$, more generally, be any given group. Do there exist models of ZF (without AC) in which there are projective spaces $\mathbb{P}$ over some division ring, for which the automorphism group of $\mathbb{P}$ is isomorphic to $G$ ?

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    $\begingroup$ Can you link to the paper with the result you mention? $\endgroup$ Dec 4, 2022 at 17:30

1 Answer 1


On abstract metamathematical grounds (having nothing to do with projective spaces), I claim that it is relatively consistent with ZFC and indeed with ZFC+Con(ZF) and much more that the answer to your question is negative. This doesn't mean the answer is negative, since it could also be consistent that the answer is positive, making the property independent of ZFC+Con(ZF). It could also be that a positive answer to your question is a consequence of a stronger theory, such as ZFC plus large cardinals.

The fact that this answer has nothing at all to do with projective spaces shows that perhaps the question that was asked isn't precisely the question that one might have wanted to ask.

In particular, the role of Con(ZF) is important for your question as it was asked. Your initial remark is not quite stated correctly, since what we need for the initial result is not to reject AC, but (presumably) to assume Con(ZF), in order to get the model of ZF as you mentioned.

But I shall argue that neither Con(ZF) nor indeed any iteration of consistency statements Con${}^n$(ZF) is sufficient in general to get your more general statement. The main reason is that it is consistent with these theories that some groups $G$ simply cannot appear in a model of ZF at all.

To see this, observe first that Con(ZF) and the stronger iterated statements are strictly weaker than the assertion that there is an $\omega$-standard model of ZF. One can see that it is weaker because all $\omega$-standard models fulfill the same consistency statements as the ambient model, and so if Con${}^n$(ZF) implied the existence of an $\omega$-standard model of ZF, then it would imply its own consistency, contrary to the incompleteness theorem.

So if ZF+Con${}^n$(ZF) is consistent, then there is a model of this theory that thinks there are no $\omega$-standard models of ZF. In such a set-theoretic world, there are many models of ZF, but no $\omega$-standard models of ZF. But in this case, no model of ZF could have a group that is isomorphic to the standard integers $\mathbb{Z}$, because this can happen only in an $\omega$-standard model.

In summary, if ZF+Con(ZF) is consistent, then there are models of this theory in which the answer to your question is no.

  • $\begingroup$ In the 1st paragraph you repeatedly use "ZFC" but this seems confusing to me because the question is about ZF, right? Is the repeated use of "ZFC" just a typo? $\endgroup$ Dec 4, 2022 at 18:28
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    $\begingroup$ The question is about whether there are certain kinds of models of ZF, yes, but we may ask this question with ZFC as our background theory. My consistency result provides a model of ZFC in which there are no models of ZF realizing G. This is a stronger result than merely providing a ZF model like that. $\endgroup$ Dec 4, 2022 at 19:05
  • $\begingroup$ This answers the question as stated, although I wonder if the OP intended to ask a slightly different question. Maybe something like, in ZF, what nontrivial facts can be proved about the automorphism group of a projective space over a division ring? $\endgroup$ Dec 4, 2022 at 20:33
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    $\begingroup$ I find this remark to be one of the reasons people often roll their eyes at mathematicians, and why mathematicians roll their eyes at logicians. There is a certain leeway that is assumed when discussing amongst people. Very clearly the question is to be read "Given a group $G$, can we extend the universe to fail AC and have a projective space whose automorphism group is $G$?". I agree that make this sort of remark, as you did, is important, but it is a glorified comment, not really an answer. $\endgroup$
    – Asaf Karagila
    Dec 5, 2022 at 1:05
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    $\begingroup$ Well, depending on the construction, one can simply apply this to the framework from my paper "Iterating failures of choice". So the question is reduced into is there a symmetric extension that takes care of a single group? Not to mention that the question did very explicitly used $\forall G\exists M$ quantification. I agree your question, $\exists M\forall G$ is natural, but it is going a bit far to presume that mathematicians don't understand scope of quantifiers. $\endgroup$
    – Asaf Karagila
    Dec 5, 2022 at 13:35

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