How to show this is a homomorphism?

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Homework Help Overview

The problem involves demonstrating that a function θ: Z6→Z2, defined as θ(x) being the remainder of x when divided by 2, is a homomorphism. The context is within group theory, specifically examining properties of homomorphisms between cyclic groups.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the representation of elements in Z6 and question how to express these elements in a general form. There is an attempt to connect the definition of homomorphism to the specific case of Z6 and Z2.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of elements in Z6 and how they relate to the concept of homomorphisms. Some guidance has been offered regarding the general form of integers and the application of the homomorphism definition, but no consensus or resolution has been reached.

Contextual Notes

There is a hint regarding the representation of odd integers and the need for a more detailed proof structure when applying the definition of homomorphism to the groups in question.

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How to show this is a homomorphism?

Homework Statement



Let θ: Z6→Z2 be given by θ(x)=the remainder of x when divided by 2 (as in the division algorithm)

Homework Equations





The Attempt at a Solution


I am stuck, this is all I have:

Let m,n be in Z6
θ(m +6 n)...

I'm not sure how to proceed. Any help is appreciated. Thanks
 
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If m is an element of Z6. What does it look like?
 


@kru- What do you mean? I imagine Z6 to look like a circle that starts and ends at 6...not sure if this is very accurate though.
 


I think kru_ is asking you how would you express an element of 6Z or 2Z or (any integer)Z in the general form (hint, odd integers are expressed as 2k + 1, where k is in Z).

From there, you should think what is the remainder of any element in 6Z divided by 2 and simply use the definition of homomorphism of groups.

If it is Z6 and Z2 (or Z(mod6) and Z(mod2) in other words) you are talking about, then you follow a very similar logic, just with a little more writing required in your proof.
 

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