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September 30th, 2017, 11:15 PM  #11  
Senior Member Joined: Aug 2012 Posts: 1,572 Thanks: 379  Quote:
No I still don't have the right to left inclusion [I deleted an erroneous proof if you saw it before you replied] and my brain has had enough of this one. At least for tonight. I think right to left might be the hard part of this exercise. Your question about $I_n$ shows that it's naive to hope that an arbitrary $X$ can be expressed as a commutator. Rather in the general case it's only a product of commutators. I am not sure how to get started on this. I'm sure the right to left direction would be prerequisite to working on the $I_n$ problem. In other words we know that $(0,0,I_n)\in \{(0,0,X)\}$ but we don't know that $(0,0,I_n)\in G'$ till we have the right to left inclusion. Last edited by Maschke; October 1st, 2017 at 12:10 AM.  
October 2nd, 2017, 01:25 PM  #12 
Member Joined: Jan 2015 From: usa Posts: 67 Thanks: 0 
@Maschke, There is no news? you didn't manage to prove the other inclusion? 
October 2nd, 2017, 01:43 PM  #13  
Senior Member Joined: Aug 2012 Posts: 1,572 Thanks: 379  Quote:
Questions like this would be better off posted on math.stackexchange.com, where you'll find professional mathematicians who can explain how to do a problem like this. On that site you will be asked to show what you've done so far and where you're stuck. It's worth the trouble since that's by far the best place to ask questions like this. Unless ABVictor shows up to help us out!  
October 3rd, 2017, 07:13 AM  #14 
Global Moderator Joined: Dec 2006 Posts: 18,053 Thanks: 1395  
October 3rd, 2017, 08:31 AM  #15  
Senior Member Joined: Aug 2012 Posts: 1,572 Thanks: 379  The angle brackets indicate the subgroup generated by the commutators. Oh I see. I have a curly brace that shouldn't be there. Should be $\langle [x,y] : x, y \in G \rangle$. Had a bit of a hard time finding a reference. Quote:
Last edited by Maschke; October 3rd, 2017 at 08:39 AM.  

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