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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.

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It might help you understand the group $G$ if you consider the group $\overline{G}:=G/\langle x_0 \rangle$. This has the presentation $\langle x_i\ (i \ge 1) \mid p^ix_i = 0 \rangle$, so it clearly just a direct sum of cyclic groups of order $p^i$. Note that the rank of $\overline{G}[p]$ is infinite and hence so is that of $G[p^2]/G[p]$, since $G[p] = \langle x_0 \rangle$. –  Derek Holt Nov 17 '12 at 11:22
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