Divisibility in an abelian group Let $G$ a abelian group. I recall that:

*

*Let $g\in G$, $n\in\mathbb{N}^+$; i write $n \mid g$ iff exists $x\in G$ such that $g=nx$.

*$G$ is said divisible iff $n \mid g$ $\forall g\in G, n\in\mathbb{N}^+$.

*$G$ has a largest divisible subgroup: the divisible part ${\rm Div}(G)$.

*$\forall p\in\mathbb{P}, \forall g\in G$ the $p$-height is defined as $h_p(g)=\max\{n\in\mathbb{N}: g\in p^nG\}$. I pose $h_p(g)=\infty$ iff $ g\in p^nG$ $ \forall n\in\mathbb{N}$.

I know that: $G$ is divisible iff $h_p(g)=\infty$ $\forall g\in G, p\in\mathbb{P}$.
I am asking if:

*

*${\rm Div}(G)=\{g\in G: n \mid g$ $\forall n\in\mathbb{N}^+\}$.

*${\rm Div}(G)=\{g\in G: h_p(g)=\infty$ $\forall p\in\mathbb{P}\}$.

In 1) and 2) the inclusion $\subseteq$ is obvious; I'm not sure that $\supseteq$ it's true, and I ask for help.
Thanks
 A: $\DeclareMathOperator{\Div}{Div}\newcommand{\QED}{\qquad \blacksquare}$The equalities do not hold. I give a counterexample of a group $G$ and an element $g \in G$ which belongs to both the $X$s but not in $\Div(G)$.
Let us set up some terminology and notation first.
Notation. $\mathbb{N}$ denotes the set of positive integers, i.e., $\mathbb{N} = \{1, 2, 3, \ldots\}$.
Definition. An element $g \in G$ is called divisible if $n \mid g$ for all $n \in \mathbb{N}$.

Let $F = \mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z} \oplus \cdots = \mathbb{Z}^{\mathbb{N}}$ be the direct sum of countably many copies of $\mathbb{Z}$.
We shall use $e_{n}$ to denote the element with $1$ in the $n$-th coordinate and $0$ in all other coordinates.
Let $H$ be the subgroup of $F$ generated by elements of the form $e_{1} - n e_{n}$ for $n \in \mathbb{N}$, i.e.,
\begin{equation*} 
 H := \langle e_{1} - 2 e_{2}, e_{1} - 3 e_{3}, e_{1} - 4 e_{4}, \ldots\rangle.
\end{equation*}
Lastly, set $G = F/H$. For an element $x \in F$, we shall denote its image in $G$ by $\overline{x}$.
Let $X$ be as in the first part, i.e., $X$ is the set of all divisible elements of $G$.
Claim 1. $\overline{e_{1}} \in X$.
Proof. Let $n \ge 1$ be given. Then, consider $e_{n} \in F$. By definition of $H$, we have $e_1 - n e_{n} \in H$ or $\overline{e_{1}} = n \overline{e_{n}}$ in $G$. $\QED$
Now, we show that $\overline{e_{1}} \notin \Div(G)$. In particular, this shows that $X \neq \Div(G)$. We do this by proving the following claim.
Claim 2. Let $x \in F$ be any element such that $\overline{e_{1}} = 2 \overline{x}$. Then, there exists $K \in \mathbb{N}$ such that $K \nmid \overline{x}$ in $G$.
Note that $\overline{e_{1}} \notin \Div(G)$ follows immediately from Claim 2. Indeed, if $\overline{e_{1}}$ were in $\Div(G)$, then there must be $\overline{x} \in \Div(G)$ such that $2\overline{x} = \overline{e_{1}}$. But the above claim shows that no such $\overline{x}$ is divisible.
Proof of Claim 2. This proof is a bit ugly but fairly straightforward.
Let $x$ be as given. Then, write
\begin{equation*} 
 x = (n_{1}, \ldots, n_{N}, 0, 0, \ldots).
\end{equation*}
(Each $n_{i}$ is an integer. Moreover, as will be clear from the next equations, we must have $N \geqslant 2$.)
Then, we have
\begin{equation} \tag{$1$}
 e_{1} - 2x = (1 - 2n_{1}, -2n_{2}, \ldots, -2n_{N}, 0, 0, \ldots).
\end{equation}
The above element is zero in the quotient $F/H$. Thus, the above element is in $H$. Thus, we must have
\begin{equation*} 
 e_{1} - 2x = k_{2}(e_{1} - 2e_{2}) + \cdots + k_{N}(e_{1} - Ne_{N})
\end{equation*}
for some integers $k_{1}, \ldots, k_{N}$. (Since the coordinates in $(1)$ after the $N$-th position are $0$, we only need to go up until $k_{N}$.)
Writing the above in tuple notation, we get
\begin{equation} \tag{$2$}
 e_{1} - 2x = (k_{2} + \cdots + k_{N}, -2k_{2}, -3k_{3}, \ldots, -Nk_{N}, 0, 0, \ldots).
\end{equation}
Equating $(1)$ and $(2)$ gives us a system of linear equations as
\begin{align*} 
 1 - 2n_{1} & = k_{2} + \cdots + k_{N}, \tag{$\ast$}\\
 2n_{2} & = 2k_{2}, \\
 2n_{3} & = 3k_{3}, \\
 & \vdots \\
 2n_{N} &= Nk_{N}.
\end{align*}
Note that the sum $k_{2} + \cdots + k_{N}$ is odd. Thus, there is some $K \in \{2, \ldots, N\}$ such that $k_{K}$ is odd. But on the other hand, we have
\begin{equation*} 
 2n_{K} = Kk_{K} \quad\text{or}\quad \frac{n_{K}}{K} = \frac{k_{K}}{2}.
\end{equation*}
Since $k_{K}$ is odd, we see that $\frac{k_{K}}{2} = \frac{n_{K}}{K}$ is not an integer. In other words, $K \nmid n_{K}$.
We now claim that there is no $y \in F$ such that $\overline{x} = K \overline{y}$.
For the sake of contradiction, assume that there is such a $y$. As before, write $y$ as
\begin{equation*} 
 y = (m_{1}, \ldots, m_{M}, 0, 0, \ldots)
\end{equation*}
with $M \ge N$.
Then, we have
\begin{equation*} 
 x - Ky = (n_{1} - Km_{1}, \ldots, n_{K} - Km_{K}, \ast, \ast, \ldots).
\end{equation*}
As before, this is equal to an element of $H$. Since $K \ge 2$, the $K$-th coordinate of $x - Ky$ must be divisible by $K$. (See $(2)$ for the appearance of an arbitrary element of $H$).
But we have already seen that
\begin{equation*} 
 n_{K} - Km_{K} \equiv n_{K} \not\equiv 0 \mod K.
\end{equation*}
This is the desired contradiction. $\QED$

Note that the same example works for (ii) as well. Indeed, the $X$ in part (i) is clearly a subset of the $X$ in part (ii) and for this $G$, we have already seen that
\begin{equation*} 
 \Div(G) \subsetneq X_{\text{(i)}} \subseteq X_{\text{(ii)}}.
\end{equation*}

Additional remark. Note that in (i), you were asking if the set of all divisible elements is precisely equal to $\Div(G)$. As I have shown, that is not the case.
This answer completely characterises what $\Div(G)$ looks like. It is the set of totally divisible elements (as defined in that answer). Essentially, just being divisible is not enough to ensure that you can find a "divisor" that is again divisible.
For example, looking at our $G$, we see that $\overline{e_{1}} = 2 \overline{e_{2}}$ but there is no $x$ such that $\overline{e_{2}} = 2 \overline{x}$.
Totally divisible ensures that you have a "compatible family of divisors" which guarantees that each divisor is itself divisible. The answer there shows that this is necessary as well.
A: To show that something is a subset of $\mathrm{Div}(G)$, you will need to use the definition: $\mathrm{Div}(G)$ is the largest divisible subgroup of $G$.
(I should point something out here: it's not obvious that such a subgroup always exists! This needs proving, but I'll assume you've done that.)
Anyway, to show that $X \subseteq \mathrm{Div}(G)$ for some set $X$, you must show that $X$ is contained in a divisible subgroup of $G$. Since $\mathrm{Div}(G)$ is the largest divisible subgroup, it will then contain $X$.
Here, in (1) we have $X = \{g \in G : n \mid g \; \forall n \in \mathbb{N}^+\}$. We need to show this is contained in a divisible subgroup of $G$. But actually, this is a divisible subgroup of $G$! Show this, and you will be done.
The same idea works for (2). Please give this a shot and let us know if you're still stuck!
