Equivalence of Axiom Schemata for Groups/Warning
Jump to navigation
Jump to search
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 $3$