# 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?

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I suppose the immediate thing to do would be to find a set of $2^{2^{\aleph_0}}$ sets which are of size $2^{\aleph_0}$ and their complement is $2^{\aleph_0}$; and the intersection of every two sets is "small". Now take the groups generated by such sets, and show that by the "small intersection" property these groups are distinct. Therefore there are $2^{2^{\aleph_0}}$ subgroups... No? – Asaf Karagila Nov 2 '12 at 22:39
@Asaf, I meant the set of non-isomorphic subgroups. Apologies for the misprint. – Levon Haykazyan Nov 2 '12 at 22:46
Oh non-isomorphic... with that I can't help anymore... :-) – Asaf Karagila Nov 2 '12 at 22:46
@Yves: done, thanks. – Levon Haykazyan Nov 2 '12 at 23:32
Indirectly related to this question: math.stackexchange.com/questions/119642 – Asaf Karagila Nov 2 '12 at 23:48

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.

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Thank you for this answer. I am still trying to understand the construction. A couple of questions, if I may. Please excuse my ignorance, if they are trivial or stupid. In the first line I think you should write $Z[P^{-1}]$ instead of $Z[J^{-1}]$, otherwise I don't know who $J$ is. Similarly in the second line there should be $\pi_A(g)$ instead of $\pi_J(g)$. Also if I got it correctly $A = Z[{p}^{-1}]$ is the group generated by $1/p$, so $\pi_A(g) = \{p\}$ does not hold for say $g = 1/p$. This contradicts the statement in the second paragraph. Thanks again. – Levon Haykazyan Nov 4 '12 at 12:59
thanks for the 2 typos. On the other hand, if say $P=\{p\}$, then $\mathbf{Z}[P^{-1}]$ is the ring, not the group generated by $1/p$. But it is the group generated by the family of $p^{-n}$ where $n$ ranges over natural numbers. – YCor Nov 4 '12 at 18:21