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Let $G$ is an abelian group and $tG$ is its torsion subgroup. If $p$ is a prime, how to show that:

$$t\bigg(\prod_{n=1}^{\infty}\mathbb Z_{p^n}\bigg)\ncong\sum_{n=1}^{\infty}\mathbb Z_{p^n}$$

I just know that $\sum_{n=1}^{\infty}\mathbb Z_{p^n}< t\bigg(\prod_{n=1}^{\infty}\mathbb Z_{p^n}\bigg)$. Thanks

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I thinjk it is to denote "torsion", @Rasmus , but: what has G to do with the whole question, anyway? –  DonAntonio Oct 27 '12 at 10:59

2 Answers 2

up vote 7 down vote accepted

They cannot be isomorphic, because they are of different cardinality.

To see this, we will find uncoutably many elements of $t\bigg(\prod_{n=1}^{\infty}\mathbb Z_{p^n}\bigg)$. To do this, for an arbitrary sequence $\tau:\mathbb N\to\{0,1\}, \tau(i)=\tau_i$, define an element $$g_\tau=(\tau_11,\tau_2p,\tau_3p^2,\tau_4p^3,\tau_5p^4,\ldots)$$ of the torsion group. Every such element has order $p$ and for distinct sequences $\sigma,\tau:\mathbb N\to\{0,1\}$, we get distinct elements $g_\tau$, $g_\sigma.$ The set $2^{\mathbb N}$ of all such sequences is uncountable and $\tau\mapsto g_\tau$ is an injection $2^{\mathbb N}\to t\bigg(\prod_{n=1}^{\infty}\mathbb Z_{p^n}\bigg)$. Therefore this torsion group is uncountable.

On the other hand, $\sum_{n=1}^{\infty}\mathbb Z_{p^n}$ only has countably many elements. (Since it can be written as a countable union of countable sets $\sum_{n=1}^N\mathbb Z_{p^n}$.)

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Thanks for the complete answer. You and @Rasmus showed me other aspects of this structure, $\prod_{n=1}^{\infty}\mathbb Z_{p^n}$ with details. –  B. S. Oct 27 '12 at 12:20

One element that belongs to $t\big(\prod_{n=1}^{\infty}\mathbb Z_{p^n}\big)$ but not to $\sum_{n=1}^{\infty}\mathbb Z_{p^n}$ is $$ (1,p,p^2,p^3,\ldots). $$ That shows that the obvious inclusion is not an isomorphism. To see that there cannot be an abstract isomorphism think about cardinality.

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Thanks Rasmus. I think, I should spend more time to know $\prod_{n=1}^{\infty}\mathbb Z_{p^n}$ better. Thanks for your time. –  B. S. Oct 27 '12 at 12:16

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