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

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Homework Help Overview

The discussion revolves around the algebraic structure of the quotient group (Z cross Z)/<(1,2)>, with participants debating whether it is isomorphic to Z or Z cross Z_2. The context involves group theory and the properties of lattice points.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore different interpretations of the quotient group, with one suggesting a visual approach using lattice points, while others question the nature of the resulting group, particularly its finiteness and structure.

Discussion Status

The discussion is active, with various perspectives being shared. Some participants are seeking clarification on the reasoning behind the claims, while others are providing mathematical mappings to support their views. There is no explicit consensus yet, as differing opinions on the nature of the group remain.

Contextual Notes

Participants are navigating assumptions about the properties of the groups involved, including the implications of finite versus infinite structures. The original poster references a potential error in the textbook, which adds to the complexity of the discussion.

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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.
 

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