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September 12th, 2017, 10:07 AM  #1 
Senior Member Joined: Jan 2015 From: usa Posts: 103 Thanks: 0  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 07:52 AM. 
September 12th, 2017, 10:18 AM  #2 
Senior Member Joined: Aug 2012 Posts: 2,043 Thanks: 583 
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 10:23 AM. 
September 12th, 2017, 10:32 AM  #3 
Senior Member Joined: Jan 2015 From: usa Posts: 103 Thanks: 0 
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 10:48 AM. 
September 12th, 2017, 11:35 AM  #4 
Senior Member Joined: Aug 2012 Posts: 2,043 Thanks: 583 
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 11:43 AM. 
September 29th, 2017, 05:08 AM  #5 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 
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 07:50 AM. 

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