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A group is a ordered pair $(G,*)$ which satisfies the following,

  1. $G$ is a set

  2. $*$ is a binary operation on $G$

  3. $*$ is associative

  4. $\exists e\in\mathbb{G} (\forall a\in\mathbb{G} a*e=e*a=a)$

  5. $\forall a\in\mathbb{G}\exists b\in\mathbb{G}(a*b=b*a=e)$

Note : $e$ is called THE identity element of this group $(G,*)$.

Note : $b$ is called THE inverse of a and denoted by $a^{-1}$.

Prove "THE", in other words, show identity is unique and inverse is unique.

I need a little bit help maybe and idea on how to start such problem.

Thank you

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closed as off-topic by T. Bongers, Cesareo, KReiser, user10354138, Brahadeesh Nov 27 '18 at 5:43

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  • $\begingroup$ It's a bit disappointing that this got onto the hot network questions list... after all, this is a "I have no idea post" that's answered by many elementary books in the field (e.g. Pinter, Dummit & Foote, and innumerable other textbooks...). In general, this is not a do-my-homework site and it's expected to show your own thoughts and the context where you encountered the problem. I'm voting to close as off-topic. But that really doesn't matter much, considering that five enthusiastic users have given you the exact same answers. $\endgroup$ – T. Bongers Nov 26 '18 at 22:30
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Let $e$ and $f$ be two elements where $ea = ae =a$ and $fa = af = a$ for all $a \in G$.

Then what is $ef$ equal to? It must be equal to something in $G$ as that is that is the definition of binary operation. And because $ea =ae =a$ for all $a\in G$ it's true for $f$ and $ef = fe = f$. But $fa = af =a$ for all $a \in G$ so $ef = fe = e$. Is it possible for $e$ and $f$ to be two different elements?

As for inverses: For $a$ let $a'$ be such that $a'a = aa' = e$ and let $a^{-1}$ be such that $a^{-1}a = a a^{-1} = e$.

What is $a^{-1}aa'$ equal to? Hint: use associativity. $(a^{-1}a)a' = ??$ and $a^{-1}(aa') = ???$.

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    $\begingroup$ Where you say $fa=ae=a$ in your first line, was that supposed to be $fa=af=a$? $\endgroup$ – David Z Nov 26 '18 at 19:57
  • $\begingroup$ Yes, it was. (Same thing though :) ) $\endgroup$ – fleablood Nov 26 '18 at 21:35
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Here's a little help; in both cases consider two such elements (two identity elements, or two inverses of the same element). Use the given relations to show that they are the same.

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Hint. Suppose $a$ and $b$ are both identities. What can you say about $ab$?

If $g$ seems to have two inverses can you find a similar kind of product that will let you conclude they are equal?

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  • $\begingroup$ Since a and b are both identities then their multiplication $ a \ast b$ should be equal to either a or b or ab ? $\endgroup$ – Scott Nov 26 '18 at 17:16
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    $\begingroup$ @Scott Almost. Of course $ab = ab$, so don't bother saying so. Then you know $ab$ is both $a$ and $b$ (better that than "either"). Now you have a proof that they are equal. $\endgroup$ – Ethan Bolker Nov 26 '18 at 17:20
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    $\begingroup$ @Scott No. All you've done is restate the meaning of "inverse". You are not trying to prove $e$ is the identity. You know that. You want to prove that an element $g$ (use a different letter so you don't mix this up with the first part) has two possibly different inverses $h$ and $k$, What can you conclude about $hgk$? $\endgroup$ – Ethan Bolker Nov 26 '18 at 17:41
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    $\begingroup$ @scott The equation following "also" makes no sense to me. And iou blew it at the end. STOP when you conclude that $h=k$! Going on to say "This implies $hg=e$" is nonsense. You knew that as one of the assumptions. Finally, it's not "$g$ should be equal to $h$", it's that $k=h$, which you proved two sentences ago. I've just been told by the system to avoid extended discussions in comments. There are several complete correct answers here. I've pretty much said everything I can. You're on your own now. $\endgroup$ – Ethan Bolker Nov 26 '18 at 18:52
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    $\begingroup$ 1) $e, g^{-1},$ and $\frac 1g$ are only labels. You can't assume they exist or have the properties they exist unless you declare them to have them. You don't know they exist and have names. In part 1 You need to prove if $\alpha$ and $\beta$ both act like an identity that $\alpha = \beta$. It won't make sense to say either is equal to $e$ because $e$ is just a symbol and BOTH $\alpha$ and $\beta$ deserve it. You need to prove that only one can have the label i.e $\alpha = \beta$. $\endgroup$ – fleablood Nov 26 '18 at 21:48
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Let $e,e'$ be identity elements of $G$. Then $e=e'e$, since $e'$ is an identity element, and $e'e=e'$, since $e$ is an identity element.

Let $u,v$ be inverses of $a$. Then $u= u e = u(av) = (ua)v = ev =v$, where $e$ is the identity element.

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Suppose there exist two identity elements $e_1$ and $e_2$ then:

  • $e_1*e_2=e_2*e_1=e_2$ (since $e_1$ is an identity);

  • $e_1*e_2=e_2*e_1=e_1$ (since $e_2$ is an identity);

  • So $e_1=e_2$.

If $y_1$ is an inverse of $x$ and $y_2$ is another one then:

$$ y_1=y_1*e=y_1*(x*y_2)=(y_1* x) * y_2=e* y_2=y_2 $$

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    $\begingroup$ Please use MathJax to format mathematical expressions. $\endgroup$ – Servaes Nov 26 '18 at 17:03
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    $\begingroup$ Welcome to stackexchange. This is a correct answer. That said, I wish you had waited a little longer to give the OP a chance to use hints to solve the problem, and so learn more. $\endgroup$ – Ethan Bolker Nov 26 '18 at 17:14
  • $\begingroup$ Thank you for the answer, i will consider this after i learn how can i conclude to this answer. $\endgroup$ – Scott Nov 26 '18 at 17:19
  • $\begingroup$ I get the first part and almost understand the second part. Thanks for the help. $\endgroup$ – Scott Nov 26 '18 at 17:46

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