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I feel a bit lazy, for I don't want to seek an answer myself in some textbook – because I fear it'd cost me too much time while someone here can easily just knock out the answer. I hope that's OK. (Do we have a policy on such questions?)

I'm just wondering: Why is it that (according to wikipedia) the signature of a group (or ring, field) contains a unary function “${}^{-1}$” for inversion? Does it make any difference to describing the theory of groups if someone leaves it out and replace its role by postulating the mere existence of invertible elements in the theory? (Like adding $\forall x \exists y\ (x\cdot y = 1 = y \cdot x)$ to the theory?)

(I have poor background in mathematical logic and model theory. So I apologise if the question was ill-put.)

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It makes no difference in terms of the strength of the theory. However, it may be more elegant to state the axioms using the inverse function, rather than having to "talk around" it with many additional quantifiers. In particular, when we have the inverse function symbol, the axioms of group theory can be stated with no existential quantifiers at all, and there are some techniques in model theory that work better with "universal" theories like that. –  Carl Mummert Nov 24 '12 at 19:19

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No, it doesn't make any difference. From the other group axioms you can prove that inverses, if they exist, are unique, so if you have $\forall x\exists y(xy=1=yx)$, you can use that to construct the inversion function uniquely whenever you need it.

Similarly, you can either consider the identity element to be part of the group ("a group is a quadruple $(A,\ast,(\cdot)^{-1},1)$ such that ..."), or just make it a group axiom that something that behaves like the identity exists. Again, since an identity is necessarily unique, this does not make a large difference.

You can find textbooks that take either approach. Sometimes the choice is informed by technical convenience, such as if one wants to fit groups into some general framework for algebraic structures that doesn't deal well with existentially quantified axioms. Then considering the identity and inversion to be group operations rather than axioms will be an advantage.

Note that things can go differently in algebraic structures other than groups. For example, a monoid is like a group except that it doesn't necessarily have inverses. A monoid must have an identity element, though, and just like for groups one easily proves that the identity element is unique. However, it turns out to be essential to insist that the identity in a monoid is a designated element (or in other words, an operation with no inputs), and not just one that is assumed to exist.

The problem is that monoid homomorphisms are supposed to take identities to identities, and this is not implied simply by requiring that the homomorphism preserves the binary operation. Otherwise a monoid homomorphism from the trivial monoid $\{0\}$ could take $0$ to any idempotent element of the target monoid, and we don't want that. (This is not a problem for groups, because groups have no idempotent elements other than the identity).

So if we want to define a "homomorphism" generally as "a map that commutes with all of the operations in our algebraic structure", then we had better consider the identity element of a monoid to be an operation.

(Arguably, this generic sense of "homomorphism" in an example of a "general framework for algebraic structures that doesn't deal well with existentially quantified axioms").

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Thanks. Is it formally possible to define new objects in a given theory? Say, naming the inverse of $x$, which is given by the group axioms, $-x$? Hm, I guess it is. Somehow this bugs me, though. –  k.stm Nov 24 '12 at 18:52
@K.Stm.: From a logical point of view (sorry, I overlooked your tags and ranted about algebra instead), the general property is: Whenever you have a theory the proves $\forall x\forall y\exists z.\phi(x,y,z)$, you can choose to extend the language of the theory with a new function symbol $f$ and add a new axiom $\forall x\forall y.\phi(x,y,f(x,y))$. The enhanced theory will then be a conservative extension of the old one -- that is, every theorem of the new theory that doesn't mention $f$ is necessarily a theorem of the original theory. (This is most easily shown by model theory). –  Henning Makholm Nov 24 '12 at 19:05
... The exception to this is if your original theory has axiom schemata of the "you can put any wff you like here" variety (such as the induction axiom in PA). Then the axiom schema in the enhanced theory becomes "you can put any wff _that doesn't mention $f$_ here", unless you invoke some stronger hypotheses -- such as $\forall x\forall y\exists!z.\phi(x,y,z)$. –  Henning Makholm Nov 24 '12 at 19:10

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