# 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  $be_{b}=b$.  There are also such $x$ and $y$ that  $xb=e_{a}$  and  $ay=e_{b}$.  Thus we have

 $e_{a}=xb=x(be_{b})=(xb)e_{b}=e_{a}e_{b}=e_{a}(ay)=(e_{a}a)y=ay=e_{b},$

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  $a^{\prime}a=aa^{\prime\prime}=e$.  We then see that

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

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