# uniqueness of inverse (for groups)

Lemma
Suppose $(G,\ast )$ is a group. Then every element in $G$ has a
unique inverse^{}.

*Proof.* Suppose $g\in G$. By the group axioms we know that there
is an $h\in G$ such that

$$g\ast h=h\ast g=e,$$ |

where $e$ is the identity element^{} in $G$. If there is also a ${h}^{\prime}\in G$
satisfying

$$g\ast {h}^{\prime}={h}^{\prime}\ast g=e,$$ |

then

$$h=h\ast e=h\ast (g\ast {h}^{\prime})=(h\ast g)\ast {h}^{\prime}=e\ast {h}^{\prime}={h}^{\prime},$$ |

so $h={h}^{\prime}$, and $g$ has a unique inverse. $\mathrm{\square}$

Title | uniqueness of inverse (for groups) |
---|---|

Canonical name | UniquenessOfInverseforGroups |

Date of creation | 2013-03-22 14:14:33 |

Last modified on | 2013-03-22 14:14:33 |

Owner | waj (4416) |

Last modified by | waj (4416) |

Numerical id | 5 |

Author | waj (4416) |

Entry type | Result |

Classification | msc 20-00 |

Classification | msc 20A05 |

Related topic | UniquenessOfAdditiveIdentityInARing |

Related topic | IdentityElementIsUnique |