Show the map is isomorphic

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In summary: It's just that once you've written out 'f(k)=ka' then it's an easy step to 'f(k)=r'. In summary, to show that an infinite cyclic subgroup H of Z is isomorphic to Z, we can define a bijective homomorphism f : Z → H | f(k) = ak for all k in Z. By showing that f is injective, surjective and operation preserving, we can conclude that f is an isomorphism and therefore, H ≈ Z and Z ≈ H.
  • #1
STEMucator
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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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  • #2
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.
 
  • #3
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
 
  • #4
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:

1. How do you determine if two maps are isomorphic?

Two maps are considered isomorphic if there exists a one-to-one and onto mapping between the vertices of the two maps, such that the edges between the vertices are preserved. This means that for every vertex in the first map, there is a corresponding vertex in the second map, and the edges between these vertices are the same in both maps.

2. Can two maps with different number of vertices be isomorphic?

No, two maps with different number of vertices cannot be isomorphic. In order for two maps to be isomorphic, they must have the same number of vertices and edges, and the same connectivity between these vertices.

3. What is the significance of determining if two maps are isomorphic?

Determining if two maps are isomorphic is important in various fields such as graph theory, topology, and computer science. It helps in understanding the structure and relationships between different maps and can also be used in data analysis and pattern recognition.

4. How do you prove that two maps are isomorphic?

In order to prove that two maps are isomorphic, you must first establish the one-to-one and onto mapping between the vertices of the two maps. Then, you must show that the edges between the corresponding vertices are the same in both maps. This can be done by either drawing the maps and visually comparing them, or by using mathematical techniques such as graph isomorphism algorithms.

5. Can two maps with different edge lengths be isomorphic?

Yes, two maps with different edge lengths can still be isomorphic as long as the connectivity between the vertices is the same. The length of the edges does not affect the isomorphism of the maps, only the structure and relationships between the vertices and edges matter.

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