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

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


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

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 and closure under taking left inverses).

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

Then there is a left inverse of $$g$$, which we will call $$a$$, such that $$ag=e$$.

Also, there is a left inverse of $$ga$$, which we will call $$b$$, such that $$b(ga)=e$$.

So:

$$ $$ $$ $$ $$ $$ $$

Thus, $$a$$ is also a right inverse of $$g$$, making it a two-sided inverse.

Also:

$$ $$ $$

Thus, $$e$$ is also a right identity, making it a two-sided 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 and is therefore omitted.