There is a consensus, that "isomorphy" (the "is isomorphic to"-relation) is the right kind of sameness between universal algebras (say groups, (single-sorted) vector spaces, lattices, ...), because isomorphic objects share all their "algebraic properties". I wonder though:

Is there a (meta)theorem, that tells us exactly which properties, I can possibly come up with in the language of said algebraic structure, are indeed shared by isomorphic algebras?

If not, is there at least a general (meta)theorem stating, that a big class of (which?) properties are shared by isomorphic algebras?

This question might be related to model theory, although I have to say I know nothing about that.

  • $\begingroup$ There's an easy answer here: all properties. Any counterexample you might be thinking of will have to involve something not expressible in the language of the algebraic structure. $\endgroup$ – Zhen Lin Sep 27 '15 at 9:44
  • $\begingroup$ @ZhenLin and the proof of your claim... is it so trivial, that nobody ever bothered to write it down or can I read up on that somewhere? --- No, I didn't think of any counterexamples, just want to know it for sure. $\endgroup$ – Stefan Perko Sep 27 '15 at 9:54
  • $\begingroup$ @StefanPerko I think you should try writing down the proof yourself. Take a property $P$ that you can express in the language of the algebraic structure (e.g. for a group, it's a property that you can express using multiplication, inversion, the neutral element, and logical operations like quantifiers etc), and then try to show that if it's satisfied by some algebra, then it's satisfied by every isomorphic algebra. You'll see that everything just works. $\endgroup$ – Najib Idrissi Sep 27 '15 at 9:57
  • $\begingroup$ @NajibIdrissi I'll try that then, thanks. $\endgroup$ – Stefan Perko Sep 27 '15 at 9:59
  • $\begingroup$ @StefanPerko It is trivial because it is built into the very definition of isomorphism. $\endgroup$ – Zhen Lin Sep 27 '15 at 10:00

Two isomorphic algebraic structures are elementarily equivalent, that is, both satisfy the same first-order sentences.


Initially when I answered this question I was under the impression that somehow isomorphic structures could disagree on second-order properties. Now, seeing the answers to this question and the comments to the present one more closely, I can see the absurd of isomorphic structures disagreeing on properties build up on their language no matter higher the underlying logic.

Anyway, I'll keep this answer for reference even if it's not a complete one.

  • 2
    $\begingroup$ I upvoted, since I think this is a useful contribution, but I'd like to point out that it doesn't really answer the question, since there are may properties preserved under isomorphism that are not expressible by first-order sentences. Maybe it would be better to say that two isomorphic structures satisfy the same sentences of any logic which takes only the symbols in the language of the structure as primitive. $\endgroup$ – Alex Kruckman Sep 29 '15 at 15:54

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