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I want to show that:

If $G=\prod_{p\in P}\mathbb Z_p$, wherein $P$ is the set of all primes, then $\frac{G}{tG}$ is divisible.

I know that $tG$ is not a direct summand and if $x\in G$ wants to be divisible by every primes, it will be $0$. I wanted to know $tG$ for myself first, so I took an element in it: $$f=(a_1,a_2,...)\in tG\longrightarrow\exists n, nf=0 $$ Can I conclude here that for infinitely many $a_i\in \mathbb Z_{p_i}$, we necessarily have $a_i=0$? Is there any formal well-known description for $tG$? Thanks for any hint or references of where to start the main problem above.

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    What's $\,\Bbb Z_p\,$ for you: the finite cyclic group $\,\Bbb Z/p\Bbb Z\,$ or the $\,p-\,$ adics? And what is $\,t\,$?2012-10-11
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    @DonAntonio: $tG$ is torsion subgroup of abelian group $G$. And $\mathbb Z_p$ is integers mod $p$.2012-10-11
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    I think $tG$ consists of those elements for which all but finitely many components are the identity. As a hint, note that for $g \in G$ and a positive integer $n$, you can solve $nh_i = g_i$ for all but finitely many components of $g_i$ (i.e. for all primes not divising $n$).2012-10-11
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    I agree with @DerekHolt, and with you: if $na_i=0,$ then surely $p_i$ divides either $n$ or $a_i$, in particular divides $n$ if $a_i\neq 0,$ and $n$ is only divisible by finitely many $p$. Incidentally, Derek, that looks just about like an answer to me.2012-10-11
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    @Kevin: I thought I should let Babak fill in the details rather than write out a complete solution!2012-10-11
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    Naturally. It's my understanding that we're OK around here with listing not-entirely-complete solutions as answers when they do a good chunk of the job!2012-10-11
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    @KevinCarlson: OK. I am thinking.2012-10-11
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    @DerekHolt: The following way came to me cause of your nice hint: I want to show that if $p$ is a prime, for arbitrary $g+tG\in G/tG$ there is $x+tG\in G/tG$ that $g-px\in tG$. According to the hint, we can always find $h\in G$ with $h_i\in\mathbb Z_{p_i}$ that the i-th component in $g$, say $g_i$, is equal to $ph_i$. Of course if for any $i$, $p=p_i$ then $h_i$ will be zero. So for infinitely many $i$; $g-px$ wherein $x=(h_1,h_2,...)$ has many finitely zero components. This is what impose $g-px$ to be in $tG$. Did I do it right Prof.?2012-10-12
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    @KevinCarlson: Please have a look at the approach above. Thanks.2012-10-12
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    @Babak Sorouh: It would be better to say that you can find $h$ such that the $i$-th component $g_i$ of $g$ is equal to $ph_i$, except possibly for $p=p_i$. So then $g-px$ has at most a single nonzero component, namely the $p_i$ component.2012-10-12
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    @DerekHolt: Thanks for any steps in this problem.2012-10-12
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    @BabakSorouh, this'll work with Derek's adjustment for showing that $G/tG$ is divisible by primes-now to show that it's divisible by all $n$, you just have one more step in passing to the prime factorization of $n$. Then if you want $x\in G/tG$ such that $nx=g,$ you can find a solution in all but the (finitely many) $p_i$ which divide $n$, and set the others to 0. Feel free to answer your own question, now.2012-10-12

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