Show that G is isomorphic to G x G

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Discussion Overview

The discussion revolves around the isomorphism between the group of positive rational numbers under multiplication and the direct product of itself. Participants explore methods to demonstrate this isomorphism, particularly through the lens of infinite direct sums and products of integers.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses uncertainty about how to construct an explicit isomorphism for the group of positive rational numbers, G.
  • Another participant suggests showing that G is isomorphic to the direct sum of infinitely many copies of the integers, providing a method involving prime factorization.
  • A concrete example is given to illustrate the mapping from G to the direct sum of integers.
  • A question is raised about whether the result would hold if an infinite direct product of integers were used instead of a direct sum.
  • It is argued that the direct product would not work because it results in an uncountable group, while the group of positive rationals is countable.
  • A later reply considers the case of the direct product of countably many copies of integers, suggesting that it would also lead to an isomorphism with itself, similar to the direct sum argument.
  • Another participant agrees with this assertion, noting that countably many copies remain countable regardless of how they are combined.

Areas of Agreement / Disagreement

Participants generally agree on the isomorphism of G with the direct sum of integers and the implications of using countably infinite structures. However, there is disagreement regarding the applicability of the direct product in this context.

Contextual Notes

The discussion highlights the distinction between direct sums and direct products in the context of countability, but does not resolve the broader implications of these structures on isomorphism.

oblixps
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Let G be the group of positive rational numbers under multiplication. Show G is isomorphic to G x G.

i'm not sure how to start this. i am trying to come up with an explicit isomorphim but i can't seem to find one.

can someone give me some hints on how to approach this?
 
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oblixps said:
Let G be the group of positive rational numbers under multiplication. Show G is isomorphic to G x G.

i'm not sure how to start this. i am trying to come up with an explicit isomorphim but i can't seem to find one.

can someone give me some hints on how to approach this?
I think the easiest method would be to show that each of these groups is isomorphic to the direct sum of infinitely many copies of the integers.

To see that $G$ is isomorphic to $$\bigoplus_{n\in\mathbb{N}}\mathbb{Z}$$, notice that each element $g\in G$ can be uniquely expressed in the form $$g = \prod_{p\in\mathcal{P}}p^{g_p}$$, where $\mathcal{P}$ denotes the set of primes. Each $p\in\mathcal{P}$ is raised to an integer power, and all but finitely many of these powers are $0$. The map $g\mapsto (g_p)_{p\in\mathcal{P}}$ gives the required isomorphism.

If that seems a bit abstract, here's a concrete example. Suppose $g = \frac{63}{50}$. By factorising the numerator and denominator as products of primes, you see that $g = 2^{-1}3^25^{-2}7^111^013^0\ldots$. The map taking $g$ to $(-1,2,-2,1,0,0,\ldots)$ associates $g$ with an element of $$\bigoplus_{n\in\mathbb{N}}\mathbb{Z}$$.

It should be fairly obvious that if $$H = \bigoplus_{n\in\mathbb{N}}\mathbb{Z}$$ then $H$ is isomorphic to $H\times H$ (because "twice infinity is infinity"), and it follows that $G$ is isomorphic to $G\times G.$
 
i wouldn't have thought of taking the direct sum of infinitely many copies of Z. seems like an interesting example.

would the result also be true if we took an infinite direct product instead of direct sum?

thanks for your reply!
 
oblixps said:
would the result also be true if we took an infinite direct product instead of direct sum?
Direct product would not work here. The direct product of infinitely many copies of the integers is uncountable. So it could not be isomorphic to the group of positive rationals, which is countable.
 
ah i see. now forgetting about the group of rationals, if we just let G be the direct product of countably many copies of Z, we would also have G isomorphic to G x G right? it seems like it would be the same argument as in the case for a direct sum.
 
oblixps said:
ah i see. now forgetting about the group of rationals, if we just let G be the direct product of countably many copies of Z, we would also have G isomorphic to G x G right? it seems like it would be the same argument as in the case for a direct sum.
Yes, that's correct. In both cases, if you take countably many copies and then countably many more copies, you still have countably many copies.
 

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