# identity element 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