Error in My Book: Z cross Z/<(1,2)> = Z_2

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[SOLVED] error in my book

Homework Statement


My book says that (Z cross Z)/<(1,2)> = Z. I say it equals Z cross Z_2. This is easy to see if you draw it out on a lattice plane. Right?


Homework Equations





The Attempt at a Solution

 
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I think the books right. Can you spell out your reasoning a little more?
 
You can choose any lattice point on the line y=0 or the line y=1 and get a unique line with slope 2 that goes through that point.
 
Hmm. And I think it's just Z_2. I mean, it has to be a finite group doesn't it?
 
Construct a map \phi:\mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Z} by \phi(a,b)=2a-b. It's easy to check that this is a surjective homomorphism, and its kernel is <(1,2)>, so by the isomorphism theorem:

\mathbb{Z} \times \mathbb{Z}/&lt;(1,2)&gt; \cong\mathbb{Z}
 
Bing! Sure. It's not (Z/Z)x(Z/(2*Z)). Thanks, StatusX.
 
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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