Equivalence of Axiom Schemata for Groups

Theorem
In the definition of a group, the axioms for the existence of an identity element and for closure under taking inverses can be replaced by the following two axioms:


 * Given a group $G$, there exists at least one element $e \in G$ such that $e$ is a left identity;
 * For any element $g$ in a group $G$, there exists at least one left inverse of $g$.

Alternatively, we can also replace the aforementioned axioms with the following two:


 * Given a group $G$, there exists at least one element $e \in G$ such that $e$ is a right identity;
 * For any element $g$ in a group $G$, there exists at least one right inverse of $g$.

Thus we can formulate the group axioms as either of the following:

Proof
Suppose we define a group $G$ in the usual way, but make the first pair of axiom replacements listed above:
 * the existence of a left identity
 * every element has a left inverse.

Let $e \in G$ be a left identity and $g \in G$.

Then, from Left Inverse for All is Right Inverse, each left inverse is also a right inverse with respect to the left identity.

Also from Left Identity while exists Left Inverse for All is Identity we have that the left identity is also a right identity.

Also we have that such an Identity is Unique, so this element can rightly be called the identity.

So we have that:
 * $G$ has an identity;
 * each element of $G$ has an element that is both a left inverse and a right inverse with respect to this identity.

Therefore, the validity of the two axiom replacements is proved.

The proof of the alternate pair of replacements (existence of a right identity and closure under taking right inverses) is similar.

Warning
Suppose we build an algebraic structure with the following axioms:

Then this does not (necessarily) define a group (although clearly a group fulfils those axioms).

Let $\left({S, \circ}\right)$ be the algebraic structure defined as:
 * $\forall x, y \in S: x \circ y = x$

That is, $\circ$ is the left operation.

From Left Operation All Elements Right Identities, every element serves as a right identity.

Then given any $a \in S$, we have that $x \circ a = x$ and as $x$ is an identity, axiom $(3)$ is fulfilled as well.

But from the complementary result of More than One Left Identity then No Right Identity, there is no right identity and therefore no identity element, so $\left({S, \circ}\right)$ is not a group.