Equivalence of Axiom Schemata for Groups/Warning

Theorem

Suppose we build an algebraic structure with the following axioms:

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

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 $3$