Abstract Algebra/Group Theory/Group/Identity is Unique

Theorem

Each group only has one identity

Proof

0. Let G be any [[../Definition of a Group|group]]. Then G has an [[../Definition of a Group/Definition of Identity|identity]], say e₁.

1. Assume G has a different [[../Definition of a Group/Definition of Identity|identity]] e₂

As e1 is [[../Definition of a Group/Definition of Identity#Usage1|identity]] of G (usage 1),	As e2 is [[../Definition of a Group/Definition of Identity#Usage1|identity]] of G (usage 1),
2a. $e_1\in G$	2b. $e_2\in G$
e2 is [[../Definition of a Group/Definition of Identity#Usage3|identity]] of G (usage 3),	As e1 is [[../Definition of a Group/Definition of Identity#Usage3|identity]] of G (usage 3),
3a. $\mathrm{\forall }g\in G:g\ast e_2=g$	3b. $\mathrm{\forall }g\in G:e_1\ast g=g$
By 2a. and 3a.,	By 2b. and 3b.,
4a. $e_1\ast e_2=e_1$	4b. $e_1\ast e_2=e_2$

By 4a. and 4b.,

5. $e_1=e_2$, contradicting 1.

Since a right assumption can't lead to a wrong or contradicting conclusion, our assumption (1.) is false and identity of a group is unique.

Diagrams

1. Assume a group has two identities.

2. e1 * e2 = e1 as e2 is identity of G, and e1 is in G.

3. e1 * e2 = e2 as e1 is identity of G, and e2 is in G

4. The two identities are the same.

5. a group only has one identity