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 June 27th, 2017, 03:41 AM #1 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 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 04:15 PM.
 July 5th, 2017, 02:51 PM #2 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 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?
 July 5th, 2017, 03:08 PM #3 Senior Member     Joined: Sep 2015 From: USA Posts: 1,696 Thanks: 861 $g^{-1} \otimes g = e$ $g \otimes g^{-1} \otimes g = g \otimes e$ $e \otimes g = g \otimes e$ $g = g \otimes e$
 July 5th, 2017, 03:23 PM #4 Senior Member   Joined: Aug 2012 Posts: 1,681 Thanks: 437 This seems on point but I didn't read it. https://math.stackexchange.com/quest...mplies-a-group
 July 5th, 2017, 07:17 PM #5 Global Moderator   Joined: Dec 2006 Posts: 18,442 Thanks: 1462 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 $g^{-1}g = g^{-1}(ge)\implies g = ge$.
 July 5th, 2017, 10:26 PM #6 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 @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'
July 5th, 2017, 10:31 PM   #7
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Quote:
 Originally Posted by agentredlum @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 .

 July 5th, 2017, 10:45 PM #8 Global Moderator   Joined: Dec 2006 Posts: 18,442 Thanks: 1462 Using a different browser (if available) might help. Alternatively, click on image below. Identity.PNG Thanks from agentredlum
July 5th, 2017, 10:50 PM   #9
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Quote:
 Originally Posted by agentredlum @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?

July 5th, 2017, 10:53 PM   #10
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Quote:
 Originally Posted by skipjack 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 10:56 PM.

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