How many elements of each order are there in this $p$-group? [closed]

Question

Let $G$ be a countable Abelian $p$-group which equals a direct sum of at most countably many finite cyclic groups and at most countably many copies of the Prüfer $p$-group, where these finite cyclic groups all have order less than $p^k$ for some fixed $k\in \mathbb{N}$, and there are exactly $N$ copies of the Prüfer $p$-group. (Where $N$ can be $0$, finite, or infinite.)

Then my question is, how many elements of $G$ are there of order $p^{k+1}$, as a function of $N$?

Are there finitely many or infinitely many such elements, and does this function depend on any feature of the group other than $N$? One data point is that there are $p^{k+1}-p^k$ elements of order $p^{k+1}$ in the Prufer $p$-group.

Answer 1

We have $m=\phi(p^{k+1})=p^k(p-1)$. Since the orders of all cyclic factors $<p^k$, the number of Prufer factors should be at least 1, at least one coordinate of every element of order $p^{k+1}$ must have a Prufer coordinate of order $p^{k+1}$ and no coordinates of bigger orders.

So if the number $k_1$ of Prufer factors is $0$, the number $z$ of elements of order $p^{k+1}$ is 0. If $k_1>0$ and the number of cyclic non-trivial factors is infinite or $k_1=\infty$, then $z=\infty$. If $k_1>0$ is finite and the direct product of all cyclic factors has order $l$ then $z=l*m^{k_1}$.