If there exists a subgroup contained in all nontrivial subgroups of $G$, then $G$ is quasicyclic. 
Problem: Let $G$ be an infinite abelian group. Show that if $G$ has a nontrivial subgroup $K$ such that $K$ is contained in all nontrivial subgroups of $G$, then $G$ is a $p$- group for some prime $p$.  Moreover, $G$ is of type $p^\infty$ (quasicyclic) group.

I have the following result:

If $G$ is an infinite abelian group all of whose proper subgroup are finite, then $G$ is of type $p^\infty$ group for some prime $p$.

So I am trying to show that each of subgroup of $G$ is finite. But I can't see anymore.  Please help me to find some direction.
 A: Assume $K\le G$ is a non-trivial subgroup of $G$ and that for every non-trivial subgroup $X\le G$, we have $K\le X$.
Then $K$ itself has no nontrivial subgroup. Especially, for any $k\in K\setminus\{1\}$, we have $\langle k\rangle=K$ and thus $K\cong \mathbb Z$ or $K\cong \mathbb Z/n\mathbb Z$. But only the case with $n=p$ prime leads to $K$ having no nontrivial subgroups.
Let $g\in G\setminus\{1\}$ be arbitrary. Then $k\in\langle k\rangle =  K\le \langle g\rangle$, i.e. $k=g^n$ for some $n$. If we write $n=p^mr$ with $r$ not divisible by $p$, then $h:=g^{p^{m+1}}$ has order dividing $r$ (because $h^r=g^{pn}=k^p=1$) and thus $k\notin \langle h\rangle$, from which we conclude $h=1$, hence $g^{p^{m+1}}=1$ and the order of $g$ is a power of $p$.  This shows that $G$ is a $p$-group.
Can you see why the second claim (type $p^\infty$) follows from here if we use that $|G|=\infty$?
And where did I use that $G$ is abelian?
A: Hint: If it is as @Alexander noted, show that the group is torsion. So you can write $G$ for some prime $p$ as $$G\cong G_p$$ where $G_p=\{x\in G\mid p^nx=0\}$ for some $n\geq 0$. This can be show that the group in a $p$ group. For the rest, I think we should focus on proving $G$ is dividable.
