June 27th, 2017, 02:41 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 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: 3,372 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, 02:08 PM  #3 
Senior Member Joined: Sep 2015 From: USA Posts: 1,939 Thanks: 1006 
$g^{1} \otimes g = e$ $g \otimes g^{1} \otimes g = g \otimes e$ $e \otimes g = g \otimes e$ $g = g \otimes e$ 
July 5th, 2017, 02:23 PM  #4 
Senior Member Joined: Aug 2012 Posts: 1,888 Thanks: 525 
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: 18,957 Thanks: 1604 
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: 3,372 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, 09:31 PM  #7 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,591 Thanks: 546 Math Focus: Yet to find out.  
July 5th, 2017, 09:45 PM  #8 
Global Moderator Joined: Dec 2006 Posts: 18,957 Thanks: 1604 
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: 1,888 Thanks: 525  
July 5th, 2017, 09:53 PM  #10  
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  Quote:
Many thanks Last edited by skipjack; July 5th, 2017 at 09:56 PM.  

Tags 
group, question, simple 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
kind of silly/simple question with probably a simple answer.  GumDrop  Math  4  October 4th, 2016 04:34 PM 
Simple group  limes5  Abstract Algebra  2  January 5th, 2014 02:47 AM 
nonabelian simple finite group  moont14263  Abstract Algebra  4  November 24th, 2012 12:59 PM 
Simple, not so simple question about areas of triangles  jkh1919  Algebra  1  November 20th, 2011 08:14 AM 
Is group with certain order simple?  zolden  Abstract Algebra  3  January 25th, 2009 03:32 PM 