# 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 $\struct {G, \circ}$ which satisfies the following four conditions:

 $(G 0)$ $:$ Closure Axiom $\displaystyle \forall a, b \in G:$ $\displaystyle a \circ b \in G$ $(G 1)$ $:$ Associativity Axiom $\displaystyle \forall a, b, c \in G:$ $\displaystyle a \circ \paren {b \circ c} = \paren {a \circ b} \circ c$ $(G_L 2)$ $:$ Left Identity Axiom $\displaystyle \exists e \in G: \forall a \in G:$ $\displaystyle e \circ a = a$ $(G_L 3)$ $:$ Left Inverse Axiom $\displaystyle \forall a \in G: \exists b \in G:$ $\displaystyle b \circ a = e$

### Group Axioms (Right)

A group is an algebraic structure $\struct {G, \circ}$ which satisfies the following four conditions:

 $(G 0)$ $:$ Closure Axiom $\displaystyle \forall a, b \in G:$ $\displaystyle a \circ b \in G$ $(G 1)$ $:$ Associativity Axiom $\displaystyle \forall a, b, c \in G:$ $\displaystyle a \circ \paren {b \circ c} = \paren {a \circ b} \circ c$ $(G_R 2)$ $:$ Right Identity Axiom $\displaystyle \exists e \in G: \forall a \in G:$ $\displaystyle a \circ e = a$ $(G_R 3)$ $:$ Right Inverse Axiom $\displaystyle \forall a \in G: \exists b \in G:$ $\displaystyle a \circ b = e$

## 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.

$\blacksquare$

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:

 $(0)$ $:$ Closure Axiom $\displaystyle \forall a, b \in G:$ $\displaystyle a \circ b \in G$ $(1)$ $:$ Associativity Axiom $\displaystyle \forall a, b, c \in G:$ $\displaystyle a \circ \paren {b \circ c} = \paren {a \circ b} \circ c$ $(2)$ $:$ Right Identity Axiom $\displaystyle \exists e \in G: \forall a \in G:$ $\displaystyle a \circ e = a$ $(3)$ $:$ Left Inverse Axiom $\displaystyle \forall x \in G: \exists b \in G:$ $\displaystyle b \circ a = e$

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