Equivalence of Axiom Schemata for Groups/Warning

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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