Does the language of group theory need a constant?

Question

I would like to verify the understanding of my claim.

In Model Theory: An Introduction by David Marker Example 1.2.5 defines the language of the theory of groups as follows to be $\mathcal{L}=\{., e\}$ where $.$ is a binary symbol and $e$ is a constant symbol called an identity element.

However, consider a reduced language $\mathcal{L_2}=\{.\}$. I claim that the identity element is definable, $\{e\}=\{e | e.e=e\}$ where $e.e=e$ is a defining formula.

Then we could just extend the theory of groups by the axiom:

$\exists e (e.e=e)$

Do we include a constant element in the language only for a convenience?

Answer 1

Yes. We only do it for the convenience.

Indeed one can write the axioms of the group in the following manner:

1. $\forall x\forall y\forall z(x\cdot(y\cdot z)=(x\cdot y)\cdot z)$ (Associativity).
2. $\forall x\exists y\forall z((x\cdot y)\cdot z=z)$ (Inverse).
3. $\exists e\forall x(e\cdot x=x\land x\cdot e=x)$ (Identity).

It's just easier to add this important constant to the language, rather than increasing the complexity of the formulas whenever we want to refer to that constant.

Note that we already do that with the $\cdot^{-1}$ operator, it's not in the language but it is definable. If one wants to simplify things even more, we can write the identity axiom in such way where $e$ is the unique such element.

Answer 2

You could make the argument, as Asaf did, that it is merely a convenience in the case of groups. But I'd like to argue that it's more than a convenience in the case of monoids.

It is true that for the theory of semigroups, given by a single associative binary function symbol $m$, we can adjoin the axiom $\exists_e \forall_x m(e, x) = m(x, e) = x$. The models of this new theory will of course be monoids. The trouble however is that homomorphisms of models in this new theory are not monoid homomorphisms; they are merely semigroup homomorphisms between monoids (nothing forces the identity to be preserved). By instead working with the signature $(m, e)$ which makes provision for an identity constant $e$, we ensure that we get the right notion of model homomorphism.

It so happens that semigroup homomorphisms between groups are in fact group homomorphisms, so in that case we can get away with the slight reduction to a single binary operation. But it's perhaps a good habit, generally speaking, to explicit include, within the signature, all the structure that is intended to be preserved by homomorphisms.