Abelian divisible p-groups and Prüfer group I've been asked to show tha if D is an abelian divisible p-group then is isomorphic to the sum of copies of $\mathbb{Z}(p^{\infty})$, which is to say that there exists a set $X$ such that $D\cong\sum_{x\in X}\mathbb{Z}(p^{\infty})$.
I took $X$ to be a basis of $D[p]=\lbrace a\in D\mid pa=0\rbrace$ over $\mathbb{Z}_{p}$.
For each $x\in X$ exists $x_{1}=x,x_{2},\dots,\in D$ such that $o(x_{1})=p$ and $px_{n+1}=x_{n}$ for each $n$ so I know that the subgroup generated by those $x_{1},x_{2},\dots$, say $H_{x}$, is isomorphic to $\mathbb{Z}(p^{\infty})$. So I want to prove $D\cong \sum_{x\in X}H_{x}$ as an external sum.
For that, I put $\mathcal{G}=\langle\bigcup_{x\in X}H_{x}\rangle$ (the internal sum) and I want to prove that $\mathcal{G}\cong \sum_{x\in X}H_{x}$. Thus, I'm trying to showing that $H_{x_{0}}\cap \langle\bigcup_{x\neq x_{0}}H_{x}\rangle=\lbrace 0\rbrace$ for each $x_{0}\in X$ where I got stuck.
Any hint to showing that $H_{x_{0}}\cap \langle\bigcup_{x\neq x_{0}}H_{x}\rangle=\lbrace 0\rbrace$ would be appreciate
 A: I'm writing here my answer for a homework. Hope it is correct
Let $M$ be a divisible abelian $p$-group. Take $x^1 \in M$ of order $p$ and let $x^{n+1}$ be the element in $M$ such that $px^{n+1} = x^n$. Note that $x^{n}$ has order $p^n$ and it is minimal, thus, they are different. Let us define $Z_x =  \langle x^i \rangle_{i=1}^n $. We claim $Z_x \to \mathbb{Z}/p^{\infty}\mathbb{Z} $ is a isomorphism
letting $\phi(x^n) = [1]_{p^n}$. We see that $\phi$ is homomorphism as for any $n$ and $k$ we have
\begin{equation*}
      \phi ( x^n + x^{n + k}) = \phi(p^kx^{n+k} + x^n)  = \phi((p^k + 1 )x^{n+k} ) = (p^k + 1 )[1]_{p^{n+k}} 
  \end{equation*}
and
\begin{equation*}
      \phi (x^n) + \phi(x^{n+k}) = [1]_{p^n} + [1]_p^{n+k} = [p^k]_p^{n+k} + [1]_p^{n+k} = (p^k + 1 )[1]_{p^{n+k}} 
  \end{equation*}
Then $ \phi ( x^n + x^{n + k}) =  \phi (x^n) + \phi(x^{n+k})$. Additionally $\phi$ bijective as it is defined from basis to basis.
Now, let us consider $D[p] = \{m \in M \mid pm = 0\}$. $D[p]$ is a $\mathbb{Z}/p\mathbb{Z}$-module. Then, $D[p]$ is a vector space over the field $\mathbb{Z}/p\mathbb{Z}$. Thus, it has a basis $Y$. Let us prove
\begin{equation*}
      M \cong \bigoplus_{y \in Y} Z_y \cong  \bigoplus_{y \in Y} \mathbb{Z}/p^{\infty}\mathbb{Z}
  \end{equation*}
For any $m \in M$ there is a minimum $k$ such that $p^k m = 0$. Then $p^{k-1} m \in D[p]$. Hence, there are $I_k$ set of indices, $b_i^k \in \mathbb{Z}$ and $y_i \in Y$ such that $p^{k-1} m = \sum\limits_{i \in I_k} b_i^k y_i$. We know that $y_i =  p^{k-1} y_i^{k}$. Hence, $p^{k-1} m = \sum\limits_{i \in I_k} b_i^k  p^{k-1} y_i^{k} =  p^{k-1}  \sum\limits_{i \in I_k} b_i^k y_i^k$ implying $p^{k-1}(m - \sum\limits_{i \in I_k} b_i^k y_i^k) = 0$ and $k-1$ is the minimal order of $m - \sum\limits_{i \in I_k} b_i^k y_i^k$. We can repeat the process with this element. Then, after $k-1$ steps we find
\begin{equation*}
m - \sum\limits_{i \in I_k} b_i^k y_i^k - \dotsb  - \sum\limits_{i \in I_1} b_i^1 y_i^1 = 0
\end{equation*}
Henceforth $m \in \sum_{y \in Y} Z_y$.
Let us prove by induction that for all $I$ with $|I| > 2$, $\sum_{i \in I} \sum_{j = 1}^{N_i} b_i^j y_i^j = 0$ implies that $b_i^j = 0$.
If $N_i = 1 $ we obtain a combination of $y_i$. Since they are in the basis $Y$ we got $b_i^1 = 0 $.
Suppose that if $N_i <k$ then the hypothesis is true and consider $N_i = k $ for all $i$. If we multiply the combination by $p^{k-1}$ we obtain that  $p^{k-1} \sum_{i \in I} b_i^k y_i^k = 0$. From what we said before $\sum_{i \in I} b_i^k y_i^k$ is a linear combination of elements of the form $y_j^{l}$ with $l < k$. Hence we obtain a new linear combination $\sum_{i \in I'}\sum_{j = 1}^{N_i} c_i^j y_i^j = 0$ such that $c_i^ j = 0$ and by extension $b_i^ j = 0$.
Finally, the last fact shows that for any $I$  and $i \in I$ if  $a_i \in Z_{y_i}$ and $\sum_{i \in I} a_i = 0$ it implies that $a_i = 0$
