Assume $A$ is abelian of order $1000$ then we have :
$$A=S_2\times S_5 $$
where $S_2$ and $S_5$ are respectively the $2$ and $5$-Sylow subgroup. Now I claim that any subgroup $H$ will be written as :
$$H=H_2\times H_5 \text{ where } H_p:=H\cap S_p\text{ for } p=2,5$$
(there is something to prove here, I leave this to you).
Hence we see that there is canonical bijection between :
$$\{\text{subgroups of }G\}\text{ and }\{\text{subgroups of }S_2\}\times \{\text{subgroups of }S_5\}$$
Hence it reduces to understand the number of subgroups (for $p$ prime) of the three different abelian groups of order $p^3$ :
$$\mathbb{Z}_{p^3}\text{, }\mathbb{Z}_{p^2}\times \mathbb{Z}_{p}\text{ and } \mathbb{Z}_{p}\times \mathbb{Z}_{p}\times \mathbb{Z}_{p} $$
The first one has exactly four subgroups (hint : exactly one for each divisor). The third one has $1+\frac{p^3-1}{p-1}+\frac{p^3-1}{p-1}+1=2p^2+2p+4$ subgroups (hint : a subgroup of the third group is a $\mathbb{Z}_{p}$-subvectorial space). The second one, it is a little harder.
Set $G=\mathbb{Z}_{p^2}\times \mathbb{Z}_{p}$. Take $H$ a subgroup of $G$, the exposant of $H$ is either $1$, $p$, $p^2$. There is exactly one
subgroup of exposant $1$ (the trivial one).
Now if $H$ is a subgroup of exposant $p$ then $H$ must be included in :
$$p\mathbb{Z}_{p^2}\times \mathbb{Z}_p $$
This last group is isomorphic to $\mathbb{Z}_{p}\times \mathbb{Z}_p$ and has exactly $\frac{p^2-1}{p-1}+1=p+2$ subgroups of exposant $p$. Hence we have in $G$ $p+2$ subgroups of exposant $p$.
Now if $H$ is a subgroup of exposant $p^2$ it is either of order $p^3$ (in which case $H=G$) or is of order $p^2$ hence cyclic. It is somehow easy to show that $H$ will always contain a unique element of the form $(1,a)$ and that such an element is of order $p^2$. Hence we have as many cyclic subgroup of order $p^2$ in $G$ as the number of choice for $a$, i.e. $p$ ($a\in \mathbb{Z}_p$). Finally we have $p+1$ subgroup of exposant $p^2$. Finally we have $2p+4$ subgroups in $\mathbb{Z}_{p^2}\times \mathbb{Z}_{p}$.
To sum up
$$\mathbb{Z}_{p^3}\text{ has }4\text{ subgroups}$$
$$\mathbb{Z}_{p}\times \mathbb{Z}_{p}\times \mathbb{Z}_{p}\text{ has } 2p^2+2p+4\text{ subgroups}$$
$$\mathbb{Z}_{p^2}\times \mathbb{Z}_{p}\text{ has } 2p+4\text{ subgroups}$$
In particular if you know your abelian group of order $1000$ (e.g. its decomposition through the theorem of finitely generated abelian groups) you can easily deduce the form of the $p$-Sylow and apply the two results above. For instance take :
$$ G:=\mathbb{Z}_5\times\mathbb{Z}_{10}\times\mathbb{Z}_{20}$$
Then we have that :
$$G=(\mathbb{Z}_2\times\mathbb{Z}_4)\times(\mathbb{Z}_5\times\mathbb{Z}_5\times\mathbb{Z}_5) $$
Hence we have exactly $((2\times 2+4)+(2\times 5^2+2\times 5+4))$ subgroups, i.e. $8+50+10+4=72$ subgroups.
Proceed by induction on $n$ in the statement "if $P$ has order $p^n$ and every subgroup of $P$ is normal, then $P$ is abelian."
The base case is trivial since groups of order $p$ are cyclic.
Suppose the statement holds for groups of order $p^i$ where $i<n$, and let $P$ be a group of order $p^n$. If $P$ is cyclic, we are done. Otherwise $P$ has a normal subgroup $U\cong \Bbb Z/p\times \Bbb Z/p$. Within $U$ we have two subgroups $H_1$ and $H_2$ corresponding to $\Bbb Z/p\times\{0\}$ and $\{0\}\times \Bbb Z/p$. By assumption, $H_1$ and $H_2$ are normal, so we can consider the groups $P/H_1$ and $P/H_2$. Now any subgroup $K$ of $P/H_1$ corresponds to a subgroup $K'$ of $P$ containing $H_1$. By assumption $K'$ is normal in $P$, so the correspondence theorem says $K$ is normal in $P/H_1$. This shows all subgroups of $P/H_1$ are normal, so $P/H_1$, and similarly $P/H_2$ is abelian by the induction hypothesis.
Best Answer
The condition $G/pG = 0$ for all primes $p$ is the same as $G$ being divisible. Indeed, the condition you wrote is the same as saying that $pG = G$ for all $G$. Then for a prime $q$, we also have $qp G = q(pG) = qG = G$. Then by the fundamental theorem of arithmetic, $nG = G$ for all $n \neq 0$. In other words, the map $G \longrightarrow G$ via $g \mapsto ng$ is onto for all $n \neq 0$. This is exactly what it means to be divisible.
There are many divisible abelian groups, but no finitely generated ones by your argument. Here's one example: $\mathbb Q$. Other examples can be constructed from this one by realizing that quotients and products of divisible groups are divisible. So for instance, $\mathbb R \times \mathbb R / \mathbb Z \times \mathbb Q / \mathbb Z$ is divisible. Any $\mathbb Q$ vector space is as well. An extremely important homological fact is that every abelian group embeds into a divisible group (we say that $\mathbb Z$-Mod has "enough injectives")