Let $p$ be a prime number and $G$ an abelian group. The group $G$ is said to be $\textbf{primary}$ if every element of $G$ has order power of $p$. For every natural number $n$, we define $$\ker(p^n)=\{x\in G\colon p^nx=0\}.$$ Then $\ker(p)$ and $G/pG$ are vector spaces over the field $\mathbb{Z}/p\mathbb{Z}$. I am interested in the relationship between the dimensions of these two vector spaces. For example, if $G$ is a direct sum of cyclic groups, then we have \begin{equation} \quad\dim\ker(p)=\dim G/pG.\quad (*) \end{equation} If $G=\langle x_0,x_1,x_2,\cdots\mid px_0=0,x_0=px_1=p^2x_2=\cdots\rangle$, then $$\ker(p)=\langle x_0,x_1-px_2,px_2-p^2x_3,\cdots\rangle,\quad G/pG=\langle\overline{x_1},\overline{x_2},\overline{x_3},\cdots\rangle.$$ In this case, $(*)$ holds. But $(*)$ is not ture for the Prüfer group $\mathbb{Z}(p^{\infty})$. From these examples, I guess \begin{equation} \dim\ker(p)\geqslant\dim G/pG.\quad (**) \end{equation} Actually, $(**)$ is always true.
$\textbf{Question.}$ If $G$ is reduced, i.e., $G$ has no (nonzero) divisible subgroups, is $(*)$ always true?
$\textbf{Proof of $(**)$.}$ Since $G=\bigcup_{n\geqslant 1}\ker(p^n)$, we have $$G/pG=\bigcup_{n\geqslant 1}\big(\ker(p^n)+pG\big)/pG.$$ For every $n\geqslant 1$, since $\big(\ker(p^n)+pG\big)/pG$ is a vector space, we can write $$\big(\ker(p^n)+pG\big)/pG=\big(\ker(p^{n-1})+pG\big)/pG \oplus\bigoplus_{i\in I_n}\langle x_i+pG\rangle.$$ It follows that $$G/pG=\bigoplus_{n\geqslant 1}\bigoplus_{i\in I_n}\langle x_i+pG\rangle.$$ Now we consider the set $B=\{p^{n-1}x_i\colon i\in I_n,n\geqslant 1\}\subseteq\ker(p)$. It is not hard to show that $B$ is linearly independent. Therefore, we obtain that $$\dim\ker(p)\geqslant|B|=\dim G/pG.$$ We complete the proof.
I notice that the question is related to the basic subgroups of $G$. But I have no idea to prove it. Any comments and suggestions are welcome. Thank you very much for your kind help.
