
Abstract Algebra Abstract Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
September 12th, 2017, 11:07 AM  #1 
Senior Member Joined: Jan 2015 From: usa Posts: 103 Thanks: 1  Abilization of a group
Let $G$ be a group, $N$ a normal subgroup and $(A,\rho)$ an abilization of $G$. Let $B=\rho(N)$ and $\lambda:G/N\to A/B$ be given by $\lambda(gN)=\rho(g)B$. I want to show that $(A/B,\lambda)$ is an abilization of $G/N$ (in this problem, abilization is meant in the sense of universal property). Thanks in advance for your help. Last edited by skipjack; September 29th, 2017 at 08:52 AM. 
September 12th, 2017, 11:18 AM  #2 
Senior Member Joined: Aug 2012 Posts: 2,135 Thanks: 621 
What progress have you made on the problem? Any thoughts? Anything you've tried? You will especially get these comments on Stackexchange. https://math.stackexchange.com/quest...ionofagroup Over there you have to show that you've made some effort of your own. Last edited by Maschke; September 12th, 2017 at 11:23 AM. 
September 12th, 2017, 11:32 AM  #3 
Senior Member Joined: Jan 2015 From: usa Posts: 103 Thanks: 1 
I am blocked from the beginning and I don't know what to do, since tomorrow I have a quiz. please help me if you can. Last edited by skipjack; September 12th, 2017 at 11:48 AM. 
September 12th, 2017, 12:35 PM  #4 
Senior Member Joined: Aug 2012 Posts: 2,135 Thanks: 621 
General steps for getting started on a proof: * Write down the exact definition of every technical term, especially the ones that are a little fuzzy. For example I'm sure you know what a group is, but perhaps you should write down exactly what's an Abelianization, what's a normal subgroup, what's a universal property. * Write down exactly what you need to show in order to complete the proof. In other words write down the thing that if you show it, you're done. More often than not, in classroom situations the above two steps are enough to make the proof pretty much write itself. Last edited by Maschke; September 12th, 2017 at 12:43 PM. 
September 29th, 2017, 06:08 AM  #5 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 895 
By the way, I believe the word you want is "abelianization", making an arbitrary group into an "Abelian" group, named for Niels Abel, also known as a "commutative" group, not "abilization".
Last edited by skipjack; September 29th, 2017 at 08:50 AM. 

Tags 
abilization, group 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Last nontrivial group in the derived series of a solvable group  fromage  Abstract Algebra  0  December 4th, 2016 03:39 AM 
Group Theory Proofs, least common multiples, and group operations HELP!  msv  Abstract Algebra  1  February 19th, 2015 12:19 PM 
Show that group Z2 x Z2 is not isomorphic to the group Z4  Vasily  Abstract Algebra  6  June 5th, 2012 03:58 PM 
Group of units of Z/pZ is a cyclic group  sunflower  Abstract Algebra  0  October 15th, 2010 02:20 PM 
fundamental group, free group  mingcai6172  Real Analysis  0  March 21st, 2009 03:35 PM 