Abstract Algebra/Group Theory/Group/Identity is Unique

Theorem

Each group only has one identity

Proof

0. Let G be any group. Then G has an identity, say e₁.

1. Assume G has a different identity e₂

As e1 is identity of G (usage 1),	As e2 is identity of G (usage 1),
2a. ${\color {blue}e_{1}}\in G$	2b. ${\color {OliveGreen}e_{2}}\in G$
e2 is identity of G (usage 3),	As e1 is identity of G (usage 3),
3a. $\forall \;g\in G:g\ast {\color {OliveGreen}e_{2}}=g$	3b. $\forall \;g\in G:{\color {Blue}e_{1}}\ast g=g$
By 2a. and 3a.,	By 2b. and 3b.,
4a. ${\color {blue}e_{1}}\ast {\color {OliveGreen}e_{2}}={\color {blue}e_{1}}$	4b. ${\color {blue}e_{1}}\ast {\color {OliveGreen}e_{2}}={\color {OliveGreen}e_{2}}$

By 4a. and 4b.,

5. ${\color {blue}e_{1}}={\color {OliveGreen}e_{2}}$, 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