I was given a group and I have to prove
a) If $f\cdot g = f$ or $g\cdot f = f$ then $g = 1$.
Is it right to do it using just Identity axiom of group:
$f \cdot g = f = f \cdot1 \Longrightarrow g = 1 ?$
Thanks.
I was given a group and I have to prove
a) If $f\cdot g = f$ or $g\cdot f = f$ then $g = 1$.
Is it right to do it using just Identity axiom of group:
$f \cdot g = f = f \cdot1 \Longrightarrow g = 1 ?$
Thanks.
Two approaches:
Suppose e, e' are two identity elements. Then e = e \cdot e' = e' \cdot e = e'.
If e' \cdot f = f, then e' = e' \cdot f \cdot f^{-1} = f \cdot f^{-1} = e, and the same for right identities.
As $\color {blue}e_1$ is identity of $G$ (usage 1),
As $e_2$ is identity of $G$ (usage 1),
2a. ${\color {blue}e_{1}}\in G$
2b. ${\color {OliveGreen}e_{2}}\in G$
$e_2$ is identity of $G$ (usage 3),
As $e_1$ 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.,