alternative definition of group

The below theorem gives three conditions that form alternative postulates.  It is not hard to show that they hold in the group defined ordinarily.

Theorem.

Let the non-empty set $G$ satisfy the following three conditions:
I.     For every two elements $a$, $b$ of $G$ there is a unique element $ab$ of $G$.
II.   For every three elements $a$, $b$, $c$ of $G$ the equation  $(ab)c=a(bc)$  holds.
III. For every two elements $a$ and $b$ of $G$ there exists at least one such element $x$ and at least one such element $y$ of $G$ that  $xa=ay=b$.
Then the set $G$ forms a group.

Proof.  If $a$ and $b$ are arbitrary elements, then there are at least one such $e_a$ and such $e_b$ that  $e_{a}a=a$  and  $b{e}_b=b$.  There are also such $x$ and $y$ that  $xb=e_a$  and  $ay=e_b$.  Thus we have

$$e_a=xb=x(b{e}_b)=(xb)e_b=e_a{e}_b=e_a(ay)=(e_{a}a)y=ay=e_b,$$

i.e. there is a unique neutral element $e$ in $G$.  Moreover, for any element $a$ there is at least one couple $a^{\prime }$, $a^{\prime \prime }$ such that  $a^{\prime }a=a{a}^{\prime \prime }=e$.  We then see that

$$a^{\prime }=a^{\prime }e=a^{\prime }(a{a}^{\prime \prime })=(a^{\prime }a)a^{\prime \prime }=e{a}^{\prime \prime }=a^{\prime \prime },$$

i.e. $a$ has a unique neutralizing element $a^{\prime }$.

Title	alternative definition of group
Canonical name	AlternativeDefinitionOfGroup
Date of creation	2013-03-22 15:07:58
Last modified on	2013-03-22 15:07:58
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	13
Author	pahio (2872)
Entry type	Theorem
Classification	msc 20A05
Classification	msc 20-00
Classification	msc 08A99
Related topic	Characterization
Related topic	ACharacterizationOfGroups
Related topic	DivisionInGroup
Related topic	MoreOnDivisionInGroups
Related topic	LoopAndQuasigroup