Number of non-isomorphic subgroups of $p$-adic integers. What is the cardinality of the set of non-isomorphic subgroups of $p$-adic integers $\mathbb Z_p$ for a given $p$?  The obvious upper bound is $2^{2^{\aleph_0}}$. But are there $2^{2^{\aleph_0}}$ non-isomorphic subgroups?
 A: If $P$ is a set of primes, let $\mathbf{Z}[P^{-1}]$ be the subring of $\mathbf{Q}$ generated by inverses of primes in $P$. If $g$ is an element in a torsion-free abelian group $A$, define $\pi_A(g)$ as the set of primes $p$ such that $g\in p^nA$ for all $n$.
Consider a family of set of primes $(P_i)_{i\in I}$. For $J\subset I$, consider the group $G_J=\bigoplus_{j\in J}\mathbf{Z}[P_j^{-1}]$. Then for every $g\in G$, we have $\pi_{G_J}(g)=\bigcap_{j\in\text{Supp}(g)}P_j$.
Now assume that $I$ has continuum cardinal, and that no $P_i$ is equal to a finite intersection of the $P_j$ for $j\neq i$ (you can indeed suppose the intersection of the $P_i$ pairwise finite). Then you can retrieve $J$ from $G_J$ as $J=\{i\in I,\exists g\in G,\pi_{G_J}(g)=P_i\}$. Thus the $G_J$ are pairwise non-isomorphic, as $J$ ranges over subsets of $I$. You thus get $2^c$ many non-isomorphic groups.
Now you want to get them in $\mathbf{Z}_p$. Write $Q=\{\text{primes}\}\smallsetminus \{p\}$. Starting from a basis of $\mathbf{Q}_p$ over $\mathbf{Q}$ contained in $\mathbf{Z}_p$, you find a copy of the direct sum of continuum copies of $\mathbf{Z}[Q^{-1}]$ in $\mathbf{Z}_p$. Since you can require the primes in the above construction to avoid $p$, you're done.  
