Is H Isomorphic to Z if H is an Infinite Cyclic Subgroup?

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Homework Help Overview

The discussion revolves around the isomorphism between an infinite cyclic subgroup H of the integers Z. Participants are tasked with demonstrating that H and Z are isomorphic through the properties of group homomorphisms.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss defining a function f from Z to H and explore its properties, including injectivity, surjectivity, and operation preservation. There is a specific focus on demonstrating surjectivity, with some participants expressing uncertainty about the necessity of justification for certain steps.

Discussion Status

Some guidance has been provided regarding the surjectivity of the function, with one participant suggesting that it is evident based on the definition of the subgroup. However, there remains a conversation about the importance of justifying each step in the argument.

Contextual Notes

Participants are navigating the definitions and properties of infinite cyclic groups and their relation to the integers, while also addressing the expectations for rigor in mathematical proofs.

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Homework Statement



Suppose H is an infinite cyclic subgroup of Z. Show that H and Z are isomorphic.

Homework Equations



We know that any infinite cyclic group H isomorphic to Z.
H = <a> ≠ <0>
|a| = ∞

The Attempt at a Solution



Define f : Z → H | f(k) = ak for all k in Z. We want to show that f is injective, surjective and operation preserving.

( Note, I think that map is correct. Since Z is an additive group in this case i think ak is the same as saying ka )

Case : Injectivity. Suppose that f(k) = f(q) for integers k and q. We want to show k = q. So :

f(k) = f(q)
ka = qa
(k-q)a = 0

We know that |a| = ∞ and (k-q) is in Z. Hence our equation reduces to k-q = 0 and thus k=q. Therefore f is injective.

Case : Surjectivity. Pick r in H. Then f(k) = ka = r for some integer k. ( Having a bit of trouble with this one ). So :

Case : Operation preserving ( Homomorphic ). To show f is a homomorphism, we must show that f(k+q) = f(k) + f(q) for integers k and q. So :

f(k+q) = ka + kq = f(k) +f(q)

Hence f is a homomorphism.

Now if I could get a bit of help showing that f is surjective, that will mean f is a bijective homomorphism aka an isomorphism and thus Z≈H and H≈Z.
 
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Zondrina said:

Homework Statement



Suppose H is an infinite cyclic subgroup of Z. Show that H and Z are isomorphic.

Homework Equations



We know that any infinite cyclic group H isomorphic to Z.
H = <a> ≠ <0>
|a| = ∞

The Attempt at a Solution



Define f : Z → H | f(k) = ak for all k in Z. We want to show that f is injective, surjective and operation preserving.

( Note, I think that map is correct. Since Z is an additive group in this case i think ak is the same as saying ka )

Case : Injectivity. Suppose that f(k) = f(q) for integers k and q. We want to show k = q. So :

f(k) = f(q)
ka = qa
(k-q)a = 0

We know that |a| = ∞ and (k-q) is in Z. Hence our equation reduces to k-q = 0 and thus k=q. Therefore f is injective.

Case : Surjectivity. Pick r in H. Then f(k) = ka = r for some integer k. ( Having a bit of trouble with this one ). So :

Case : Operation preserving ( Homomorphic ). To show f is a homomorphism, we must show that f(k+q) = f(k) + f(q) for integers k and q. So :

f(k+q) = ka + kq = f(k) +f(q)

Hence f is a homomorphism.

Now if I could get a bit of help showing that f is surjective, that will mean f is a bijective homomorphism aka an isomorphism and thus Z≈H and H≈Z.

That is pretty good. The DEFINITION of <a> is Za. So if r is in <a> then r=ka for some k in Z. So f(k)=r. It's kind of obviously surjective.
 
Dick said:
That is pretty good. The DEFINITION of <a> is Za. So if r is in <a> then r=ka for some k in Z. So f(k)=r. It's kind of obviously surjective.

Yeah I didnt know whether I had to actually justify it or to say it was trivial. I guess I got my answer then.

Thanks
 
Zondrina said:
Yeah I didnt know whether I had to actually justify it or to say it was trivial. I guess I got my answer then.

Thanks

You ALWAYS have to justify it. Quoting a definition and showing why it works is a fine way to do that. Just saying 'it's trivial' is criminally lame. Don't do that! You didn't think it's trivial, why should someone else? Please forget my 'kind of obvious' comment.
 
Last edited:

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