# 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 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