Show ZXZ/<1,1> is an infinite cyclic group.

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Homework Help Overview

The discussion revolves around demonstrating that the group ZXZ/<1,1> is an infinite cyclic group. Participants explore the properties of infinite cyclic groups and the implications of the factor group structure.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the characterization of infinite cyclic groups, questioning the nature of the elements in the factor group and their relationships. There are attempts to express elements in canonical forms and to clarify the implications of specific elements being in left cosets.

Discussion Status

The discussion includes various interpretations of the group structure and the properties of the elements involved. Some participants suggest that ZXZ/<1,1> is isomorphic to Z, while others question the implications of certain elements and their roles as generators. There is a recognition of the infinite cyclic nature of the group, but no explicit consensus is reached.

Contextual Notes

Participants are navigating the definitions and properties of groups, particularly focusing on the implications of modding out by the subgroup <1,1>. The discussion reflects uncertainty about the relationships between elements and their representations within the group structure.

Daveyboy
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Homework Statement



Show ZXZ/<1,1> is an infinite cyclic group.

Homework Equations





The Attempt at a Solution


<1,1> = {...(-1,-1), (0,0), (1,1),...}

implies ZXZ/<1,1> = {<1,0>+<1,1>, <0,1>+<1,1>} which is isomorphic to ZXZ.

But ZXZ is not cyclic, is my description of the factor group wrong , is it not isomorphic to ZXZ, or am I missing other points?
 
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what characterize an infinite cyclic group?

1. generated by a single element
2. infinite order

can you put an element canonically into a certain form?
so say you have (m, n), what can I write it as?
 
if (m,n) is an element of one of the left cosets of the factor group then,
if m=n (m,n) is in <1,1>. if m/=n then (m,n) = (m,n) + <1,1>. (because (0,0) is
generated by <1,1>). I'm not sure what this implies though.
 
so actually <1,0> + <1,1> generated all elements of ZXZ. Is it acceptable to say ZXZ/<1,1> is isomorphic to Z+0Z which is isomorphic to Z? (Which is generated by <1> or <-1>)
 
<1,1> is just the identity. So, yeah.
 
... this is the additive group, (0,0) is the identity, and I'm using the notation <1,1> to imply that this is the cyclic generator. I hope that clears up any confusion.

I'm not sure if the conclusion I've come to is correct.

ZXZ/<1,1> = {Z+<1,1>}
Clearly Z is infinite and cyclic generated by <1> or <-1>.
So, ZXZ/<1,1> is a cyclic infinite group.
 
well, (0,0)=(1,1) since you are modding out the diagonal set.
(1,0) is the generator (or using coset notations (1,0)+<(1,1)>). Your conclusion is definitely correct (as in post 4)
 
awesome thanks.
 

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