Algebraic Structures: Does Order Matter? (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!

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.)
