June 27th, 2017, 02:41 AM  #1 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,275 Thanks: 203  Simple question about a group
1) We have a finite nonempty set of welldefined elements $\ \ \text{G}$ 2) There is a welldefined 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. 
July 5th, 2017, 01:51 PM  #2 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,275 Thanks: 203 
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, 02:08 PM  #3 
Senior Member Joined: Sep 2015 From: CA Posts: 1,264 Thanks: 650 
$g^{1} \otimes g = e$ $g \otimes g^{1} \otimes g = g \otimes e$ $e \otimes g = g \otimes e$ $g = g \otimes e$ 
July 5th, 2017, 02:23 PM  #4 
Senior Member Joined: Aug 2012 Posts: 1,414 Thanks: 342 
This seems on point but I didn't read it. https://math.stackexchange.com/quest...mpliesagroup 
July 5th, 2017, 06:17 PM  #5 
Global Moderator Joined: Dec 2006 Posts: 17,466 Thanks: 1312 
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, 09:26 PM  #6 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,275 Thanks: 203 
@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, 09:31 PM  #7 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,225 Thanks: 420 Math Focus: Yet to find out.  
July 5th, 2017, 09:45 PM  #8 
Global Moderator Joined: Dec 2006 Posts: 17,466 Thanks: 1312 
Using a different browser (if available) might help. Alternatively, click on image below. Identity.PNG 
July 5th, 2017, 09:50 PM  #9 
Senior Member Joined: Aug 2012 Posts: 1,414 Thanks: 342  
July 5th, 2017, 09:53 PM  #10  
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,275 Thanks: 203  Quote:
Many thanks Last edited by skipjack; July 5th, 2017 at 09:56 PM.  

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