Normal Subgroups: Why Every Kernel is a Homomorphism

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Homework Statement


Expain why every normal subgroup is the kernel of some homomorphism.


The Attempt at a Solution


Every kernel is a normal subgroup but the reverse I can't show rigorously. It seems possible how to show?
 
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What's the obvious map from G to G/N? What's its kernel?
 
I would have thought that would be easy- it's the direction emphasised in Algebra texts! Of course, this says "some" homomorphism- you have to pick the homomorphism carefully.

If H is a normal subgroup G, then we can define the "quotient group", G/H. There is an obvious homomorphism from G to H. What is the kernel of that homomorphism?

Darn, I had to stop in the middle to take a telephone call and morphism got in in front of me!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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