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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:

$f \cdot g = f = f \cdot1 \Longrightarrow g = 1 ?$


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If you have $f \cdot g = f$, then multiply by the inverse element of $f$. Then you get $f^{-1} \cdot f \cdot g = f^{-1} \cdot f = 1$ – user9413 Nov 9 '11 at 18:44
Just cancel the f using group axioms. – simplicity Nov 9 '11 at 18:45
You could justify your approach using cancellation laws: – Martin Sleziak Nov 9 '11 at 19:00
The identity axiom of groups does not say that if $xy = xz$ then $y=z$. The identity axiom of groups says that there exists $1$ such that for all $x$, $x1=1x = x$. You are trying to use the identity axiom to justify something it does not directly say. – Arturo Magidin Nov 9 '11 at 19:10
I figured this question must be a duplicate, but couldn't find one. How about that! :) – Bill Cook Nov 9 '11 at 19:59

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.

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  1. Let $G$ be any group. Then $G$ has an identity, say $e_1$.
  2. Assume $G$ has a different identity $e_2$

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.,

  1. ${\color {blue}e_{1}}={\color {OliveGreen}e_{2}}$, contradicting 1.
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Did you copy and paste this from somewhere? Care to indicate the source? – Memming Feb 20 '14 at 6:12

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