$G$ is a reduced Group 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.

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 $p^{3\infty}(G) = 0$.
