# 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).

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Exercise $7.12$ - 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $2$: The Definition of Group Structure: $\S 26 \gamma$ - 1974: Thomas W. Hungerford:
*Algebra*... (previous) ... (next): $\text{I}$: Groups: $\S 1$ Semigroups, Monoids and Groups: Proposition $1.3$ - 1982: P.M. Cohn:
*Algebra Volume 1*(2nd ed.) ... (previous) ... (next): $\S 3.2$: Groups; the axioms: Theorem $1$