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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?

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How do you intend to give rigorous meaning to "encode the same structure" if not by "isomorphism"? –  Zhen Lin Dec 8 '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 8 '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 8 '12 at 14:59
    
Can I ask- in model theory, is there a definition for what a morphism is? –  jf9rj4 Dec 8 '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 8 '12 at 15:44

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