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- #1

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?

- Thread starter oblixps
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- Thread starter
- #1

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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- #2

- Feb 7, 2012

- 2,702

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.

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?

To see that $G$ is isomorphic to \(\displaystyle \bigoplus_{n\in\mathbb{N}}\mathbb{Z}\), notice that each element $g\in G$ can be uniquely expressed in the form \(\displaystyle g = \prod_{p\in\mathcal{P}}p^{g_p}\), 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 \(\displaystyle \bigoplus_{n\in\mathbb{N}}\mathbb{Z}\).

It should be fairly obvious that if \(\displaystyle H = \bigoplus_{n\in\mathbb{N}}\mathbb{Z}\) then $H$ is isomorphic to $H\times H$ (because "twice infinity is infinity"), and it follows that $G$ is isomorphic to $G\times G.$

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- #4

- Feb 7, 2012

- 2,702

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

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- Feb 7, 2012

- 2,702

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.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.