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As it is defined; dG is a group generated by all the divisible subgroups of G. So, if the group G is reduced then dG=0. Does it mean that if G is a reduced group then G has no any non zero divisible subgroups? Thanks for help.

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    Yes; if $dG=0$ then $G$ has no nonzero divisible subgroups, because $dG$ contains every divisible subgroup of $G$ by definition.2011-05-03
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    @Arturo: Thanks for help.2011-05-03
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    @Arturo: Can any reduced group G be taken to a non reduced one by an injection?2011-05-03
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    @Basil: I'm not sure what you mean. Certainly, you can take any group $G$ and embed it into, say, $G\oplus\mathbb{Q}$, which is not reduced. Far more interesting is the fact that every abelian group can be decomposed as a direct sum of a divisible group and a reduced group, $G = dG\oplus R$, with $R$ reduced.2011-05-03
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    @Arturo: That is right. What I am looking for is that, if we have an exact sequence 0->K->G->H->0 such dK=0 and dH=0 then necessarily dG=0. It seems to me; dH=0 is enough for showing dG=0. But dK=0 is useless here as I see.2011-05-03
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    @Basil: The image of a divisible group is divisible; so if you write $G=dG\oplus R$, then $\phi(dG)$ is a divisible subgroup of $H$. If $dH=0$, then $dG$ lies in the kernel of the map $G\to H$, hence it lies in the image of $K\to G$. But $K\to G$ is an embedding, so that means that $dG\subseteq K$; since $K\leq G$ implies $dK\leq dG$, then $dG=d(dG)\subseteq dK\subseteq dG$, so $dK=dG$, hence from $dK=0$ you get $dG=0$. Not "useless" at all. (Consider $0\to\mathbb{Q}\to\mathbb{Z}\oplus\mathbb{Q}\to \mathbb{Z}\to 0$; then $dH=0$, but this is *not* enough to show $dG=0$).2011-05-03
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    @Basil: Could you add the exact sequence question to the body of your post to give context to the answer I've given?2011-05-03

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Okay, this has gotten too extended in the comments, so let me summarize it here:

Since $dG$ is, by definition, the subgroup of $G$ generated by all divisible subgroups of $G$. A group $G$ is reduced if and only if $dG=0$, which occurs if and only if the only divisible subgroup of $G$ is trivial.

When $G$ is abelian, $dG$ is a direct summand of $G$ (because divisible groups are injective in the category of abelian groups), so every group $G$ can be written in the form $G=dG\oplus R$, where $R$ is reduced; this follows by considering the quotient map $G\to G/dG$, and showing that $G/dG$ has no nontrivial divisible subgroups.

It is easy to verify that the homomorphic image of a divisible group is divisible; that if $M\leq N$ are groups, then $dM\leq dN$; and that $d(dG) = dG$.

So, if we have an exact sequence $$0\to K \stackrel{\iota}{\longrightarrow}G\stackrel{\varphi}{\longrightarrow} H \to 0$$ then $\iota(dK)\subseteq dG$ and $\varphi(dG)\subseteq dH$. If $dH=0$, then it follows that $\varphi(dG)=0$, so $dG\in\mathrm{ker}(\varphi) = \mathrm{Im}(\iota)$, so $dG\subseteq \iota(K)$. Hence, $dG=d(dG)\leq \iota(dK)\leq dG$, and we can conclude that $dG=\iota(dK)$. If in addition you know that $dK=0$ as well, then it follows that $dG=0$. That is, an extension of reduced groups is necessarily reduced.

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    @Aurturo: The point that I didn't notice was: if M≤N are groups, then dM≤dN ; and that d(dG)=dG . Thanks for your answer.2011-05-04