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:

Group Axioms (Left)
A group is an algebraic structure $$\left({G, \cdot}\right)$$ which satisfies the following four conditions:

Group Axioms (Right)
A group is an algebraic structure $$\left({G, \cdot}\right)$$ which satisfies the following four conditions:

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 Inverse for All is Right Inverse 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.