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These are concluding remarks from Dummit and Foote Abstract Algebra, Ch 3, sec 3:

We have, in the course of the proof of the isomorphism theorems, encountered situations where a homomorphism $\varphi$ on the quotient group $G/N$ is specified by giving the value of $\varphi$ on the coset $gN$ in terms of the representative $g$ alone. In each instance we then had to prove $\varphi$ was well defined, i.e., was independent of the choice of $g$. In effect we are defining a homomorphism, $\Phi$, on $G$ itself by specifying the value of $\varphi$ at $g$. Then independence of $g$ is equivalent to requiring that $\Phi$ be trivial on $N$, so that

  • $\varphi$ is well defined on $G/N$ if and only if $N\le \ker \Phi$.

And then after some more commentary, they include this image: enter image description here

Although I don't quite understand what is going on in this part of the section, I would like to see a proof of bullet assertion. Here is my attempt:

$\implies$: Suppose $\varphi$ well defined then, and let $n\in N$, then $\Phi(n)=\varphi\circ\pi(n)=\varphi(nN)=\varphi(N)=e_H$.

But, I can't show reverse implication. I think that since we have not shown yet that $\varphi$ is indeed well defined, using $\Phi=\varphi\circ\pi$ will be illegal.

Also, if you can attach other reading or lecture note link, that discusses aforementioned topic in more detail, I will be obliged.

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The idea is that if $N \leqslant \ker \Phi$ and we have two representatives $x, x'$ of the same coset $xN$ then $x' = xn$ for some $n \in N$ and $\Phi(n) = 1_H$ so

$$\Phi(x) = \Phi(x)1_H = \Phi(x)\Phi(n) = \Phi(xn) = \Phi(x')$$

Therefore the definition $\varphi(xN) = \Phi(x)$ is well defined since the value $\Phi(x)$ we get does not depend on the representative we choose.

This result is known as the "Universal Property of Quotient Groups" and is part of a larger discussion of Noether's three "isomoprhism theorems." You can find more about this in essentially any introduction to the subject. If you still looking for more information, I'd recommend looking at multiple sources to get different viewpoints. Online sources will, of course, be the easiest to get a hold of.

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  • $\begingroup$ Thank you very much for this elaborate reply. And also for telling the name of this property. $\endgroup$ – Silent Oct 22 '18 at 5:24

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