On infinite groups with unique minimal subgroup Let $\operatorname{Sub}(G)$  be the lattice of all subgroups of an infinite (abelian) group $G$. If $\operatorname{Sub}(G)\setminus\{\{1\}\}$ has a minimum element which is a cyclic subgroup of prime order, then according to third example in this page, it seems $G$ is isomorphic to $\Bbb Z_{p^\infty}$.
If this is a correct deduction, is there group theoretic proof for this (without using modules)?
 A: Here is a sketch of a proof. Such a group must be a torsion group, and it must be a $p$-group i.e. all elements have order a power of $p$. Since it is infinite, it has finite subgroups of arbitrarily large order, and all such subgroups are cyclic. So it has a unique subgroup of order $p^n$ for any $n$ (else two such would generate a larger non-cyclic subgroup), so it is the ascending union of its cyclic subgroups of order $p^n$ and hence isomorphic to ${\mathbb Z}_{p^\infty}$.
A: Suppose $G$ has a subgroup $G_0:=C_p$ which nontrivially intersects (hence is contained in) all other nontrivial subgroups of $G$. We define an ascending sequence of subgroups $(G_i)_{i=0}^\infty$ in addition to the $G_0$ already defined. For each $i$, if $G_i<G$ pick $g_{i+1}\in G\setminus G_i$. It must be torsion, so set $G_{i+1}$ to be the new group $\langle g_{i+1},G_i\rangle$. By induction, we can see that each $G_i$ is finite, and must be cyclic of prime power order (any other isomorphism type would imply a minimal subgroup other than $G_0$). We can also see that we can continue to pick $g_i$s indefinitely, since if $G_i=G$ for some $i$ then $G$ is $C_{p^n}$ for some $n$, which is finite, a contradiction. Finally, let $G_\infty=\bigcup_{i\ge0}G_i$. We will have $G_\infty\cong\Bbb Z(p^\infty)$. 
Suppose $g\in G\setminus G_\infty$. It must be $p$-torsion, so set $n$ is minimal so that $g^{p^n}\in G_\infty$. Let $h=g^{p^{n-1}}$ so that $h^p\in G_\infty$ but $h\not\in G_\infty$, and let $\bar{h}\in G_\infty$ be the $p$th root of $h^p$ within the subgroup $G_\infty$. Then $\langle h^{-1}\bar{h}\rangle$ is another minimal subgroup, a contradiction. Hence $G=G_\infty$.
