What to call a group $G$ such that $\mathrm{Gp}\vdash A\iff G\models A$ for sentence $A$ of first-order logic, where $\mathrm{Gp}$ is the group axioms For an instance,
I want to know how to call a group $G$ such that
$$ \mathrm{Gp} \vdash A \iff G \models A $$
for any sentence $A$ of the first-order logic,
where $\mathrm{Gp}$ is the group axioms.
That is, I am wondering what the name of a model of a given axioms is,
which satisfies only the theory of the axioms in a specific formal system.
 A: No such group exists.
For every structure $M$ and every sentence $\varphi$ in the appropriate language, either $M\models\varphi$ or $M\models\neg\varphi$. This means that, since the theory of groups is not complete, every group $G$ will satisfy some sentence which isn't in the theory of groups. There are lots of examples:

*

*Either $G\models\forall x,y(x=y)$ or $G\models\exists x,y(x\not=y)$, even though neither of those sentences is a consequence of the group axioms.


*Either $G\models\forall x,y(x*y=y*x)$ or $G\models\exists x,y(x*y\not=y*x)$, even though neither of those sentences is a consequence of the group axioms.
And so on. Indeed, the theory of a structure is always complete - so we only have $$\forall\varphi(M\models\varphi\iff T\vdash\varphi)$$ in case $M\models T$ and $T$ is complete.

Completeness can go away once we consider classes of structures as opposed to individual structures. For a class of structures $\mathbb{K}$, let $$Th(\mathbb{K})=\{\varphi: \forall M\in\mathbb{K}(M\models\varphi)\}=\bigcap_{M\in\mathbb{K}}Th(M).$$ In general $Th(\mathbb{K})$ is not complete, and $Th(\mathbb{Groups})$ is as expected exactly the set of sentences which are consequences of the group axioms.
But this is a very different situation.
A: While Noah and Gerald have completely answered the question as written, I just want to point out that the question becomes more interesting if we restrict attention from all sentences of first-order logic to some interesting subclass of sentences.
For example, an equational sentence is one of the form $\forall \overline{x}\, t(\overline{x}) = t'(\overline{x})$, where $t$ and $t'$ are terms. Now it's quite possible to have a group $G$ such that for any equational sentence $\varphi$, $G\models \varphi \iff \text{Grp}\vdash \varphi$. Such a group $G$ would have to be non-abelian (since $\text{Grp}$ does not prove $\forall x\forall y\, xy = yx$), have infinite exponent (since $\text{Grp}$ does not prove $\forall x\,x^n = e$ for any $n>0$), etc. An example of such a group $G$ is the free group on two generators.
Equational theories (those axiomatized by equational sentences) are the main object of study in the field of universal algebra. For an equational theory $T$, a model $M\models T$ satisfies $M\models \varphi \iff T\vdash \varphi$ for all equational sentences $\varphi$ if and only if $M$ generates the variety of algebras axiomatized by $T$. This means that every model of $T$ is a quotient of a substructure of a cartesian power of $M$. Universal algebraists must have a name for such a model $M$ - unfortunately, I don't know what it is. At the risk of overloading a word that has been too many times already, I might call such a model "generic".
Now you could do the same thing with other classes of sentences, like the existential ($\Sigma_1$/$\exists_1$) or universal $(\Pi_1$/$\forall_1$) sentences. The point is that if your class of sentences is closed under negation (like the class of all first-order sentences), then the argument in the other answers shows that you can only have a "generic" model of $T$ with respect to that class if $T$ is complete (for that class of sentences). But incomplete theories can have "generic" models with respect to classes of sentences which are not closed under negation.
A: So you want a group $G$ such that the sentences satisfied by $G$ are exactly those sentences satisfied by all groups?
I think there is no such group $G$.  Consider the two sentences:
$$
\forall x, \; x^2 = e,\\
\exists x, \; x^2 \ne e.
$$
Then your group $G$ satisfies one of these, but the theory $\mathrm{Gp}$ proves neither of them.
