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(I want to link to similar question with a very good answer: Question about Algebraic structure?)

An algebraic structure is an ordered tuple of sets. One of these is called the underlying set, and the others are operations of various arity. Since operations are functions, which are sets of ordered pairs, this is why we can interpret the components of an algebraic structure to be sets.

For example, a group is a quadruple $(G,0,-,+)$ where

  • $G$ is the underlying set,
  • $0 \subseteq G^0 \times G$ is a nullary operation,
  • $- \subseteq G \times G$ is a unary operation, and
  • $+ \subseteq G^2 \times G$ is a binary operation.

My question is why we choose an ordered tuple to describe the algebraic structure. For instance, does it make a difference if I define a group to be $(G,+,-,0)$, where I list the operations in order of descending, rather than ascending, arity? If the order doesn't matter, why don't we just define a group to be $\{G,0,-,+\}$, rather than an ordered tuple?

Thanks!

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  • $\begingroup$ For all intents and purposes, it does not matter, and the use of ordered tuples is just the convention. Perhaps it is motivated by the fact that the ordering of the items are usually standard (i.e. for groups, it is some flavor of "set, operation, operation" and for measure spaces, it is "set, sigma-algebra, measure"). $\endgroup$ – angryavian Oct 8 '14 at 19:55
  • $\begingroup$ Your definition of group doesn't capture everything. It doesn't capture the fact that $+$ is associative, that $0$ is the identity, or that $-$ provides inverses under $+$. $\endgroup$ – Hayden Oct 8 '14 at 19:57
  • $\begingroup$ At the risk of being pedantic, won't many operations require representation as ordered triples or even $n$-tuples? $\endgroup$ – Todd Wilcox Jan 15 '15 at 18:34
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The reason is really silly in some sense. Since all the objects ($G,+$ and so on) are sets, how can you know which one is the group, and which one is the addition?

You can say, well, $+\subseteq G^2\times G$. But there are sets $X$ such that $X^2\times X\subseteq X$. So it's really not so obvious as much as we might want it to be.

On the other hand, in an ordered set, you can say that the first element is the group, the second is the operation $+$, and so on.

If you want to be fully formal, a structure is really just an ordered pair $(M,\Sigma)$ where $\Sigma$ is the interpretation function which maps the function, relation and constant symbols of the language to their interpretation in $M$. Just when the language we work in is simple enough, we might skip that and write directly the tuple (knowing full well that the reader will not be confused if we wrote $(G,0,+,-)$ or $(G,+,0,-)$).

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  • $\begingroup$ Thanks Asaf! Where can one find out more about interpretation functions? $\endgroup$ – Mathemanic Oct 8 '14 at 20:10
  • $\begingroup$ Most books about predicate logic, I guess. $\endgroup$ – Asaf Karagila Oct 8 '14 at 20:11
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Suppose that $(G,+,-,0)$ is a group. Then $(G,-,+,0)$ is not a group. Hence, the order is, in a sense, essential. On the other hand, which order you choose is pretty much arbitrary, as long as you stick with one.

Less obvious are algebraic structures like lattices and quandles, where there are two binary operations on a set that satisfy identical equational axioms. In this case, changing the order of the two binary operations does result in a structure of the same sort, but the structures need not be identical, or even isomorphic.

(If one is in the mood to be really pedantic, one could notice that the underlying set in $(G,+,-,0)$ (or just $(G,+)$) is actually redundant, as it is implicitly specified by the binary operation $+$. In other words, a group "is" just a binary operation.)

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  • $\begingroup$ Thanks for your answer! So it seems that listing the operations in descending arity is best. However, is it only convention, or is there another reason to prefer descending arity instead of ascending arity? For some reason, $(G,0,-,+)$ seems more natural to me, since you're starting with a rudimentary pointed set $(G,0)$, and then tacking on more "higher level" operations to construct a more and more sophisticated structure. $\endgroup$ – Mathemanic Oct 8 '14 at 20:26
  • $\begingroup$ @Mathemanic. I don't see any formal reason to prefer either order. I suppose to the extent there is a preference, it is just a matter of custom. $\endgroup$ – James Oct 8 '14 at 20:38

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