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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 $\,t\,$? – DonAntonio Oct 11 '12 at 11:23
@DonAntonio: $tG$ is torsion subgroup of abelian group $G$. And $\mathbb Z_p$ is integers mod $p$. – Babak S. Oct 11 '12 at 11:27
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 $n$). – Derek Holt Oct 11 '12 at 11:45
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. – Kevin Carlson Oct 11 '12 at 11:58
@Kevin: I thought I should let Babak fill in the details rather than write out a complete solution! – Derek Holt Oct 11 '12 at 12:33

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