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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:

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

Thanks.

  • 0
    @user1729, How do you know? Guesses, presumptions, surmises... Must we decipher the OP's intentions here? [I would prefer not to](http://en.wikipedia.org/wiki/Bartleby,_the_Scrivener#Plot_summary).2011-11-10

2 Answers 2

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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?2014-02-20