# 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 |