I am backing to read my previous question and learn more facts which the Masters left for me within comments. One of them appeared here Verifing some properties about $G$.
There; I was verifying that the group
$G=\langle x_0,x_1,x_2,\ldots|px_0=0,x_0=p^nx_n,\;\; n\geq1\rangle$ which is an infinite $p$-primary group has $\bigcap_{n=1}^\infty p^nG\neq0$ ( In fact, a very nice comment given by @Jack Schmidt paved the way of proving later). When I was working on the problem above and reading other approaches, suddenly, I felt the presentation of $G$ is a bit similar to the presentation of $\mathbb Z({p^\infty})$:
$\langle x_0,x_1,x_2,\ldots|px_0=0,x_0=px_1,,x_1=px_2,\ldots,,x_{n-1}=px_n,\ldots, \;\; n\geq1\rangle$
So I asked him if these two structures are the same and he said patiently:
it is a different group. It's "ulm invariant", rank(G[p^2]/G[p]) = infinity is different than for the Prüfer group, with rank 1. Here G[n] = { g : ng = 0 }. This is a group where an element x0 has infinite height but is not divisible...
My question is circling around rank(G[p^2]/G[p]) = infinity. I know that $G[p^2]=\{g\in G|p^2g=0\}$ and the other and know that the Ulm invariant is defined as: $\dim_{\mathbb Z_p}\frac{p^nG\cap G[p]}{p^{n+1}G\cap G[p]}$ when the group is $p$-primary. Honestly, I didn't catch his equality well and I am thinking how he could get $G$'s Ulm inv. by another formula. Thanks for your time.