# identity element is unique

Theorem. The identity element^{} of a monoid is unique.

Proof. Let $e$ and ${e}^{\prime}$ be identity elements of a monoid $(G,\cdot )$. Since $e$ is an identity element, one has $e\cdot {e}^{\prime}={e}^{\prime}$. Since ${e}^{\prime}$ is an identity element, one has also $e\cdot {e}^{\prime}=e$. Thus

$${e}^{\prime}=e\cdot {e}^{\prime}=e,$$ |

i.e. both identity elements are the same (in inferring this result from the two first equations, one has used the symmetry^{} (http://planetmath.org/Symmetric) and transitivity of the equality relation).

Note. The theorem also proves the uniqueness of e.g. the identity element of a group, the additive identity (http://planetmath.org/Ring) 0 of a ring or a field, and the multiplicative identity^{} (http://planetmath.org/Ring) 1 of a field.

Title | identity element is unique |

Canonical name | IdentityElementIsUnique |

Date of creation | 2013-03-22 18:01:20 |

Last modified on | 2013-03-22 18:01:20 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 11 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 20M99 |

Synonym | neutral element is unique |

Synonym | uniqueness of identity element |

Related topic | Group |

Related topic | UniquenessOfInverseForGroups |

Related topic | ZeroVectorInAVectorSpaceIsUnique |

Related topic | AbsorbingElement |