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If I have structure $(S, \cdot)$, where $\cdot$ has type $(2)$, i.e., $\cdot : S \times S \rightarrow S$ and $(S', \circ)$, where $\circ$ has type $(3)$, i.e., $\cdot : S' \times S' \times S' \rightarrow S'$ then is it possible that $(S, \cdot)$ and $(S', \circ)$ are isomorphic or $(S', \circ)$ is embed in $(S, \cdot)$?

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What do you mean by isomorphism in this case? Do you mean a bijective map $\phi: S\times S \rightarrow S^{'3}$ such that $\phi(x\cdot y) = \phi(x)\circ\phi(y)$? –  Vishal Jul 21 '13 at 13:35
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Your question is ill posed. From a logical point of view it doesn't make any sense to compare two structures which have different signature: as in this case where you have a structure with a binary operation and another one with a ternary operation. That's because there's no easy way to see how to define a morphisms between structures which have different signature.

What could make sense is a comparison of the clones which can be obtained from the structure above. The clones associated to two structure are themselves some sort of structure which can be considered as a sort of closure or completion of the given structures, and for such structures become meaningful make a comparison.

If there's an isomorphism between the clones of two structures then the two structures are two presentation of essentially the same algebraic object: the two presentation are different because one in one case the clone is generated via a binary operation in the other case with a ternary operation. So basically isomorphisms of clones generated by different structures means that the theories (in the logical sense) of the two structures are essentially the same.

An embedding amount to prove that on structure have enough information to codify the other structure inside itself. This means that the theory of one structure is in some sense richer or stronger of the second one.

I hope this help.

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The relation of generating isomorphic seems to be a special case of what is sometimes called cryptomorphism. –  Andreas Blass Jul 21 '13 at 15:57
    
Let me give example. If it possible that ternary semigroup isomorphic to semigroup? –  nameless Jul 23 '13 at 18:11
    
@Aolf if you consider simply your two structure is not so clear what an isomorphism should be. What more important is not clear what a semigroup with a ternary operation should be. –  Giorgio Mossa Jul 23 '13 at 19:42
    
Anyway if you consider a monoid $(M,\cdot,e)$ where $\cdot \colon M \times M \to M$ is the ordinary operation and $e$ is the identity then the clones induced by $\cdot$ and the one induced by the structure having carrier $M$ and ternary operation the $c \colon M \times M \times M \to M$ with $c(a,b,c)=a\cdot b \cdot c$ (which is well defined since a product in a monoid is associative) these two clones are isomorphic: because $a \cdot b = a \cdot b \cdot e= c(a,b,e)$ and so the generator of each clone can be rewritten via the generator of the other clone. –  Giorgio Mossa Jul 23 '13 at 19:44
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@Aolf no it cannot. –  Giorgio Mossa Jul 24 '13 at 15:47
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