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Let $L$ be a language with a binary function symbol $\cdot$ and a constant symbol $e$. The axioms for groups are

$\forall x, e \cdot x=x\cdot e=x$

$\forall x \forall y \forall z, x \cdot (y\cdot z)=(x \cdot y) \cdot z$

$ \forall x \exists y, x \cdot y= y \cdot x =e$

But the axiom $\forall x \forall y \exists z, x \cdot y =z$ is not included in the axioms. Why is it the case? how do we guarantee that the product is also in the 'group' (the set we want to check whether is a group)

BetaY
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    Since $\cdot$ is a binary function symbol, by definition $x\cdot y$ is a term and there's no need to do further assumptions. In a model of the axioms, one should provide a function that will associate to each term an element. – egreg Dec 10 '17 at 10:58

2 Answers2

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As pointed out by egreg in the comments, you answer the question yourself by stating that $\cdot$ is a function symbol, and for function symbols your desired axiom $\forall x \forall y \exists z \ x \cdot y =z$ is just true by definition. This is because for a structure $M$ every $n$-ary function symbol will be interpreted as a function $f:M^n \to M$.

However, there might be another reason for your confusion. Note that we actually don't need function and constant symbols as we can express these two by relations: A constant is just a nullary relation, and an $n$-ary function may be expressed by a $n+1$-ary relation symbol $R_f$ by adding axioms for totality (i.e., what you asked for) and functionality. The interpretation of $R_f$ can then be $\{ (x_0,\dots,x_{n-1},y) \ | \ f(x_0,\dots,x_{n-1})=y \} $.

namsap
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For the same reason that there's no axiom reading $\exists x(x=e)$.

It is part of the concept of a "first-order language" (or rather, in the concept of an interpretation of the language, aka a structure over the language) that every function symbol must denote a total function whose values lie in the domain of the interpretation, and every constant symbol must denote a particular value in the domain of the interpretation.

Thus as soon as you have specified that $\cdot$ is a binary function in your language, it it implicit that the value of $x\cdot y$ always exists and always is in the carrier set of your group. It is not necessary to state this implicit fact again as an axiom formula.

Indeed, once you have a proof system for first-order logic, it should be able to prove $\forall x\forall y \exists z(f(x,y)=z)$ as a matter of logic, without appealing to any theory-specific axioms.

It is right that it is commonly stated as an explicit group axiom in introductory algebra texts that the product of two group elements must itself be an element of the group. That's necessary because those texts cannot assume that we're building upon a formalized logic that would allow this fact to be implicit.

hmakholm left over Monica
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