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# alternative definition of group

The below theorem gives three conditions that form alternative group 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 $a^{{\prime}}$.

## Mathematics Subject Classification

20A05*no label found*20-00

*no label found*08A99

*no label found*

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