Is there a isomorphism between N and Q?

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Homework Help Overview

The discussion revolves around the existence of an isomorphism between the group of natural numbers (\mathbb{N}) and the group of rational numbers (\mathbb{Q}) or positive rationals (\mathbb{Q}^+). The original poster expresses confusion regarding the construction of such an isomorphism, particularly noting the difference in the number of generators between the two groups.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand the relationship between the groups and questions how a mapping could be constructed given the differing generators. Some participants clarify the nature of the groups involved and the operations typically associated with them.

Discussion Status

The discussion is exploring the conceptual differences between bijections and isomorphisms in the context of group theory. Participants are questioning assumptions about the nature of the groups and the implications of having a bijection between sets.

Contextual Notes

There is a noted confusion regarding the classification of \mathbb{N} as a group, with suggestions that \mathbb{Z} might be the intended group for discussion. The operations associated with \mathbb{Q} and \mathbb{Q}^+ are also clarified, indicating a need for precision in definitions.

twoflower
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Hi all,

I wonder if there is an isomorphism between the group of [itex]\mathbb{N}[/itex] and the group of [itex]\mathbb{Q}[/itex] (or [itex]\mathbb{Q}+[/itex]). I know there is a proof that there is a bijection between these sets, but I didn't find a way how to construct the isomorphism.

What confuses me a little is that (I think) the group of natural numbers has only one generator, while the group of (positive) rationals has more than one generator, so I can't see how the mapping would look like.

Thank you for any hints!
 
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Well you've answered your own question, since the image of the generator should generate the image of the entire group. Can you make this into a proof?
 
Last edited:
Er, N isn't a group. Did you mean Z, with addition as the operation?

(And, I assume you meant addition for Q and multiplication for Q+)


There's no reason to think that the existence of a bijection as sets implies that there is an isomorphism as groups.
 
Hurkyl said:
There's no reason to think that the existence of a bijection as sets implies that there is an isomorphism as groups.

It's a step in the right direction though.
 
Thank you, I've got it, I mixed the general set bijection and isomorphism between algebraic structures in my mind.
 

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