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Here is the problem:

Let $G=\langle x_0,x_1,x_2,\ldots\ |px_0=0,x_0=p^nx_n, \text{all } n\geq1\rangle$. Prove that $G/\langle x_0\rangle$ is a direct sum of cyclic groups and is reduced.

The first part is easy because if we put the relation $x_0=0$ to other relations in $G$; $G$ would be a direct sum of cyclic groups. For another part, I feel that the first part is usefull, but I don't know how to link these together. The following ideas just came to me:

  • if I assume it is not reduced; $dG$ would be a proper subgroup and so $G/dG$ is reduced.

or

  • To show that $\{0\}$ is the only divisible subgroup of $G$.

Thanks for your time.

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    What does "reduced" mean here?2012-10-12
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    $G/\langle x_0 \rangle$ is not a direct sum of cyclic groups because it is not abelian.2012-10-12
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    @ChrisEagle: The abelian group G is said to be reduced if dG=0.2012-10-12
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    @BabakSorouh: "direct sum of cyclic" groups, known as DSC groups are pretty awesome to study. over the integers, DSC is closed under submodules, but not quotients (everything is a quotient of a DSC). No nonzero DSC group is divisible, so DSC groups are reduced. DSC groups have some properties that make them easily handled by "finite means", as in DonAntonio's answer. In this sense they are called "pure-projective". The dual notion of "pure injective" is also called "algebraically compact" and in more interesting in the torsion-free case.2012-10-12
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    @JackSchmidt: Dear Jack, I made an awful mistake here and don't know what to do? :( I really wanted to be hint about $G$ not $G/$. I made a really bad typo here. How can I say that to Don. Sorry.2012-10-12
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    You mean you want to show $G/\langle x_0 \rangle$ is DSC, and $G$ is reduced?2012-10-12
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    Exactly. I wanted to do that.2012-10-12

2 Answers 2

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Consider $p^\infty G := \bigcap_{n=1}^\infty p^n G$. For each $n$, $p^n (A/B) = (p^nA+B)/B$, but for $A=G$, $B=\langle x_0 \rangle$, $p^n(A/B) \leq \langle \bar x_n, \dots \rangle$, so $p^n A \leq \langle x_0, x_n,x_{n+1}, \ldots\rangle$, so $p^\infty A = \langle x_0 \rangle$. Hence every divisible subgroup of $G$ is contained within $p^\infty G = \langle x_0 \rangle$ of order $p$, so $dG=0$.

When working with abelian $p$-groups, you'll mostly be concerned with these $p^n A$ and $A[p^n]$ subgroups, and their transfinite counter parts.

$p^{\infty+1}G = p(p^\infty(G)) = 0$, but $dG \leq p^{\infty+\infty}(G) = p^\infty(p^\infty(G)) \leq p^{\infty+1}(G)$.

A fun exercise is to find an abelian $p$-group with $p^{2\infty}(G) \neq 0$, but $p^{3\infty}(G) = 0$.

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    That is very interesting working with that infinity powers. I am working on one of the Rotman’s book and have not seen these magic powers. Thank you for sharing me your knowledge. You pointed that, since every divisible subgroup is contained in $p^{\infty}G\cong \mathbb Z_p$ then $dG=0$. Is this fact arisen from the point that **finite groups are not divisible** ? However, I am thinking to find out why did you say "**Hence every...of order $p$**".2012-10-13
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Suppose $\,A\leq G/\langle\,x_0\,\rangle\,$ is a divisible subgroup and let $\,a\in A\,$ . Then

$$\forall\,n\in\Bbb N\,\,\exists\,a_n\in A\,\,s.t.\, a= na_n$$

Now, we can write $\,a=w(x_i)+\langle\,x_0\,\rangle\,$ ,where $\,w(x_i)\,$ is a word in a finite number of generators $\,x_i\,$ of $\,G\,$.

Well, let now $\,m:=\max\,\{i\;;\;x_i\,\,\text{appears in the word}\,\,w\}$ , and take now $\,n:=p^m\,$ above...can you take it from here?