Subgroups of the $p$-adics that miss $\mathbb{Z}$

Question

I am trying to understand the complementation of $\mathbb{Z}$ in the $p$-adics. The motivation for this question is that it will help me understand the situation in $\text{Tak}(\mathbb{Z}_p)$, where $\text{Tak}$ is the Takasaki functor from abelian groups to quandles. Any complement of $\mathbb{Z}$ in $\mathbb{Z}_p$ as a group would have to be an uncountable subgroup of $\mathbb{Z}_p$ which does not contain any integer except $0$. Is there any classification or construction of subgroups of the $p$-adics for which these properties hold, if such a thing exists?

Answer 1

If we treat $\mathbb Q_p$ as a vector space over $\mathbb Q,$ we can take an uncountable basis $\{v_i\},$ with $v_0=1.$

Since we can multiply the basis elements by scalars, we can ensure the $v_i\in\mathbb Z_p.$ Indeed, we can ensure that $\|v_i\|_p=1$ for each $i,$ which is to say each $v_i$ is a unit in $\mathbb Z_p.$

Then the $\mathbb Z_{(p)}$-module generated by the $v_i, i\neq 0,$ is such a subgroup.

It is maximal in that if you add any element to it, it must generate all of $\mathbb Q_p$ as a $\mathbb Q$-vector-space, and hence we can get a non-zero integer in the corresponding group.

You can show any maximal subgroup $G$ not containing any non-zero element of $\mathbb Z$ is of this form.

This is because you can show it must be closed under multiplication by elements of $\mathbb Z_{(p)},$ or we can add in those elements.

Likewise, if $\alpha\in\mathbb Z_p$ and $p\alpha\in G,$ then $\alpha\in G,$ or again $G$ is not maximal.

And then $\mathbb Q\otimes G$ is a sub-vector space of $\mathbb Q_p$ which does not contain any non-zero rational, and thus we can make a basis for $\mathbb Q_p$ which combines a basis for $\mathbb Q\otimes G,$ one integer, and possibly more basis elements. But any other basis element is an extension we can use to make $G$ larger.