Equivalence of Axiom Schemata for Finite Group

Theorem

The following axiom schemata for the definition of a finite group are logically equivalent:

Finite Group Axioms

An algebraic structure which fulfils the finite group axioms:

A finite group is an algebraic structure $\struct {G, \circ}$ which satisfies the following four conditions:

\((\text {FG} 0)\)	$:$		Closure	\(\ds \forall a, b \in G:\)	\(\ds a \circ b \in G \)
\((\text {FG} 1)\)	$:$		Associativity	\(\ds \forall a, b, c \in G:\)	\(\ds a \circ \paren {b \circ c} = \paren {a \circ b} \circ c \)
\((\text {FG} 2)\)	$:$		Finiteness	\(\ds \exists n \in \N:\)	\(\ds \order G = n \)
\((\text {FG} 3)\)	$:$		Cancellability	\(\ds \forall a, b, c \in G:\)	\(\ds c \circ a = c \circ b \implies a = b \)
					\(\ds a \circ c = b \circ c \implies a = b \)

These four stipulations are called the finite group axioms.

Group Axioms

An algebraic structure of finite order which fulfils the group axioms:

A group is an algebraic structure $\struct {G, \circ}$ which satisfies the following four conditions:

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

These four stipulations are called the group axioms.

Proof

Let $S$ be an algebraic structure of finite order which fulfils the group axioms.

It is to be shown that $S$ also fulfils all the finite group axioms.

We have that $\text {FG} 0$ and $\text {FG} 1$ are the same as $\text G 0$ and $\text G 1$ and so a priori $\text {FG} 0$ and $\text {FG} 1$ are fulfilled.

By hypothesis $S$ is of finite order.

That is, there exists some (strictly) positive integer $n$ such that $s$ has exactly $n$ elements.

That is, $\text {FG} 2$ is fulfilled.

As $S$ fulfils the group axioms, it is by definition a group.

Therefore the Cancellation Laws hold:

$b a = c a \implies b = c$

$a b = a c \implies b = c$

That is, $\text {FG} 3$ is fulfilled.

Thus, $S$ fulfils all the finite group axioms.

$\Box$

Let $S$ be an algebraic structure which fulfils the finite group axioms.

It is to be shown that $S$ also fulfils all the group axioms.

We have that $\text G 0$ and $\text G 1$ are the same as $\text {FG} 0$ and $\text {FG} 1$ and so a priori $\text G 0$ and $\text G 1$ are fulfilled.

By definition, an algebraic structure which fulfils group axioms $\text G 0$ and $\text G 1$ is a semigroup.

From Cancellable Finite Semigroup is Group, it then follows that $S$ is a group.

That is, $S$ fulfils all the group axioms $\text G 0$, $\text G 1$, $\text G 2$ and $\text G 3$.

Hence the result.

$\blacksquare$