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June 27th, 2017, 02:41 AM   #1
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Simple question about a group

1) We have a finite non-empty set of well-defined elements $\ \ \text{G}$

2) There is a well-defined associative binary operation on this set $\ \ \otimes$

3) The set contains a left identity element $\ \ \text{e} \in \text{G } \ $ such that for every element $\ \ \text{g} \in \text{G} \ $ ,

$ \text{e} \otimes \text{g} = \text{g} $

4) For every element $\ \ \text{g} \in \text{G} \ \ $ there exists a left inverse $\ \ \text{g}^{-1} \ $ such that

$\text{g}^{-1} \otimes \text{g} = \text{e}$

Using only the above information, is it possible to prove that the left identity $\ \ \text{e} \ \ $ must also be a right identity? If so, how do we prove

$\text{g} \otimes \text{e} = \text{g} \ \ $ ?


Last edited by skipjack; July 5th, 2017 at 03:15 PM.
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July 5th, 2017, 01:51 PM   #2
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I tried a few approaches without success. Starting to think it is impossible to prove the identity must commute with every element given the above information. Note: We cannot assume this is a commutative binary operation.

Any ideas?

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July 5th, 2017, 02:08 PM   #3
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$g^{-1} \otimes g = e$

$g \otimes g^{-1} \otimes g = g \otimes e$

$e \otimes g = g \otimes e$

$g = g \otimes e$
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July 5th, 2017, 02:23 PM   #4
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This seems on point but I didn't read it. https://math.stackexchange.com/quest...mplies-a-group
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July 5th, 2017, 06:17 PM   #5
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The set $\text{G}$ doesn't need to be finite.

As romsek seemed to assume that $g \otimes g^{-1} = e$, the proof below doesn't use that.

I let juxtaposition imply the binary operation $\otimes$.

$g^{-1}g = e = ee = (g^{-1}g)e = g^{-1}(ge)$,
but $ca = cb \implies c^{-1}(ca) = c^{-1}(cb) \implies ea = eb \implies a = b$,
so $g^{-1}g = g^{-1}(ge)\implies g = ge$.
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July 5th, 2017, 09:26 PM   #6
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@Romsek , we can't assume $ \ \ \otimes \ \ $ is commutative

@Masche , the LaTeX in the link won't render in my browser

@skipjack , last 2 lines of your post are rendering as 'Undefined control sequence \implies'

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July 5th, 2017, 09:31 PM   #7
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Math Focus: Yet to find out.
Quote:
Originally Posted by agentredlum View Post
@Masche , the LaTeX in the link won't render in my browser

@skipjack , last 2 lines of your post are rendering as 'Undefined control sequence \implies'

Maybe something wrong with your browser, both work for me .
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July 5th, 2017, 09:45 PM   #8
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Using a different browser (if available) might help. Alternatively, click on image below.
Identity.PNG
Thanks from agentredlum
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July 5th, 2017, 09:50 PM   #9
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Quote:
Originally Posted by agentredlum View Post
@Masche , the LaTeX in the link won't render in my browser
It's stackexchange, it's got to be working. Did something change at your end?
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July 5th, 2017, 09:53 PM   #10
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Quote:
Originally Posted by skipjack View Post
Using a different browser (if available) might help. Alternatively, click on image below.
Attachment 8987
This is brilliant in its simplicity and I'm kicking myself for not making the connection. Specifically the part about $ \ \ c^{-1}a = c^{-1}b \ \ $ --> $ \ \ a = b $

Many thanks


Last edited by skipjack; July 5th, 2017 at 09:56 PM.
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