# Equivalence of Axiom Schemata for Groups/Warning

## Theorem

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).

## Proof

Let $\struct {S, \circ}$ be the algebraic structure defined as:

- $\forall x, y \in S: x \circ y = x$

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

From Element under Left Operation is Right Identity, 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 More than one Right Identity then no Left Identity, there is no left identity and therefore no identity element.

Hence $\struct {S, \circ}$ is not a group.

$\blacksquare$

## Sources

- 1968: Ian D. Macdonald:
*The Theory of Groups*... (previous) ... (next): $\S 1$: Some examples of groups - 1974: Thomas W. Hungerford:
*Algebra*... (previous) ... (next): $\text{I}$: Groups: $\S 1$: Semigroups, Monoids and Groups: Exercise $3$