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:

Group Axioms
An algebraic structure of finite order which fulfils 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 $\text {FG} 0$ 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.

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 $\text G 0$ 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.