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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.

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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$2011-11-09
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    Just cancel the f using group axioms.2011-11-09
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    You could justify your approach using cancellation laws: http://www.proofwiki.org/wiki/Cancellation_Laws2011-11-09
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    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.2011-11-09
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    I figured this question must be a duplicate, but couldn't find one. How about that! :)2011-11-09
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    The title does not correspond to the body of the question.2011-11-09
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    The result in the title is proved [here](http://en.wikipedia.org/wiki/Elementary_group_theory#Identity_is_unique) (item *Basic theorems*).2011-11-09
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    Your approach does not work, you need inverses (or something which says that "if $gh=gk$ then $h=k$"). To see this, take the monoid (like a group, but without inverses) with three elements $\{e, a, b\}$ such that $ae=a=ea$, $be=b=eb$, $ab=a=ba$, $b^2=b$, $a^2=e$. This is simply the cyclic group of order $2$ ($\langle a, b\rangle\cong C_2$) with a sort of second identity adjoined ($e$ is your identity). However, $ab=ae=a$, but $b\neq e$.2011-11-10
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    @Didier Piau: Actually, the title does correspond the the body. If $gf=f\Rightarrow g=1$ then the identity element is unique...2011-11-10
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    @user1729: No. The identity element has to do with $gf=f$ for every $f$, the post is concerned with $gf=f$ for a given $f$. Of course, starting from this implication and adding some more steps, one can arrive at the conclusion you say, but the fact one can do so does not seem relevant in view of the level of the question asked.2011-11-10
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    @Didier Piau: The specific question asked is the key step in the proof. The other steps you would need to add are just noise...Also, I suspect that Egor was trying to prove the identity is unique and got stuck at this step...2011-11-10
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    @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