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From a universal algebraic perspective, let's say we have two isomorphic groups. Then can I speak of their isomorphic nature by saying the binary operations of multiplication of the two groups are "isomorphic" in that they encode the same structure?

jf9rj4
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  • How do you intend to give rigorous meaning to "encode the same structure" if not by "isomorphism"? – Zhen Lin Dec 08 '12 at 14:53
  • If for example I had some algebraic objects that had more than one $n$-ary operation. It may be the case that we had two algebraic objects that agreed completely with respect to one of the $n$-ary operations but not to the others. Then I want a notion of "isomorphism" with respect to the $n$-ary operations. – jf9rj4 Dec 08 '12 at 14:56
  • That's easy enough: given a signature $\Sigma$ and a subset $\Sigma' \subseteq \Sigma$, any $\Sigma$-structure gives rise to a $\Sigma'$-structure in a natural way, called a reduct; so what you are talking about is isomorphisms of reducts. – Zhen Lin Dec 08 '12 at 14:59
  • Can I ask- in model theory, is there a definition for what a morphism is? – jf9rj4 Dec 08 '12 at 15:22
  • There are various possibilities. Since you say you are doing universal algebra, the most relevant one is the usual notion of "homomorphism", i.e. a map that preserves the interpretation of function symbols and relation symbols in the signature. – Zhen Lin Dec 08 '12 at 15:44
  • Informally it means that that the operation tables are identical after some renaming of the elements (i.e. a set-theoretic bijection of the undelying sets). Algebraically we don't care about the "names" or any internal structure that the elements may have, only how they are related to each other under the basic operations of the algebraic structure. – Bill Dubuque Dec 09 '12 at 00:54
  • Bill, wouldn't that be an isomorphism? – jf9rj4 Dec 09 '12 at 03:54

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