Order of Homomorphisms from Z to Z/2Z - Help Needed

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Discussion Overview

The discussion revolves around the number of group homomorphisms from the group of integers Z to the group Z/2Z. Participants explore the existence of these homomorphisms and the specific mappings involved.

Discussion Character

  • Exploratory, Technical explanation, Conceptual clarification

Main Points Raised

  • One participant notes the existence of two group homomorphisms from Z to Z/2Z, identifying the trivial homomorphism that maps all of Z to 0 in Z/2Z.
  • Another participant explains that since Z is generated by 1, there are two possible mappings for 1: to 0 or to 1 (mod 2).
  • A subsequent reply confirms the mapping of even integers to 0 and odd integers to 1, suggesting that if f(1)=1 (mod 2), then f(n)=n (mod 2) holds true.

Areas of Agreement / Disagreement

Participants appear to agree on the existence of two homomorphisms, but the discussion does not resolve whether there are additional mappings or clarify the implications of these mappings.

Contextual Notes

The discussion does not address potential limitations or assumptions regarding the definitions of homomorphisms or the properties of the groups involved.

Who May Find This Useful

This discussion may be useful for individuals studying group theory, particularly those interested in homomorphisms and the structure of cyclic groups.

jakelyon
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I have seen that there exists two group homomorphisms from Z to Z/2Z. However, I cannot seem to understand this. I mean I know that there exists a trivial grp hom. which sends all of Z to 0 in Z/2Z. But I cannot think of anymore. Any help?
 
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Well, Z is generated by 1. There are two places 1 can map to, 0 or 1 (mod 2). If it doesn't map to 0, it must map to 1. Do you see what function that really is?
 
You mean that I map all even to 0 and all odd to 1?
 
Yes :)
If f(1)=1 (mod 2), then f(n)=n (mod 2), which is 0 if n is even and 1 if n is odd.
 

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