Can we explicitly describe all finite-index subgroups of $\mathbf{Z}_p^\times$?
Also, if $H \subset \mathbf{Z}_p^\times$ is a finite-index subgroup, what can we say about the isomorphism class of $H$ as a topological group?
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The structure of $\mathbb{Z}_p^\times$ is known, and this says that $H$ are actually isomorphic to subgroups of $\mathbb{Z}/(p-1)\mathbb{Z}$. – Seewoo Lee Nov 20 '22 at 23:31
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@SeewooLee: While I agree the answer to OP's question follows from your link, I think you might have misread "finite-index" as "finite". – Torsten Schoeneberg Nov 21 '22 at 00:08
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@Bun: What are your thoughts? What do you know about $\mathbb Z_p^\times$? – Torsten Schoeneberg Nov 21 '22 at 00:13
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@TorstenSchoeneberg You're right. Thank you for pointing it out. – Seewoo Lee Nov 21 '22 at 00:28
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@TorstenSchoeneberg $\mathbf{Z}_p^\times$ comes with subgroups of the form $1 + p^m \mathbf{Z}_p$, and it seems natural to guess that these are the only ones. – Bun Nov 21 '22 at 03:43
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Good start. But there is also the subgroup made up of the roots of unity, right? – Torsten Schoeneberg Nov 21 '22 at 04:28
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Right, so the next guess would be the ones I described together with a subgroup of the roots of unity. – Bun Nov 21 '22 at 09:12
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... and "together with" means we can also form direct products (of any subgroup of the unit roots with any subgroup of the form $1 + p^m \mathbb Z_p$). – Torsten Schoeneberg Nov 21 '22 at 19:13
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Can you find all the finite-index subgroups of $\mathbf Z/m\mathbf Z \times \mathbf Z_p$ where $p \nmid m$? In particular, can you show a finite-index subgroup contains some $p^r\mathbf Z_p$, which is open in $\mathbf Z_p$, so finite-index subgroups are automatically open subgroups (using the product topology on $\mathbf Z/m\mathbf Z \times \mathbf Z_p$)?
If you can describe the finite-index subgroups of $\mathbf Z/m\mathbf Z \times \mathbf Z_p$ for $p \nmid m$, then you can turn that into a description of finite-index subgroups of $\mathbf Z_p^\times$ for $p > 2$, since $\mathbf Z_p^\times = \mu_{p-1} \times (1 + p\mathbf Z_p)$ as multiplicative groups and
a) $\mu_{p-1} \cong \mathbf Z/(p-1)\mathbf Z$ (non-canonically, just as $(\mathbf Z/p\mathbf Z)^\times \cong \mathbf Z/(p-1)\mathbf Z$ non-canonically)
b) $1+p\mathbf Z_p \cong p\mathbf Z_p$ as topological groups by the $p$-adic logarithm, so $1+p\mathbf Z_p \cong \mathbf Z_p$ as topological groups by using division by $p$.
The case of $\mathbf Z_2^\times$ is a bit trickier since $\mathbf Z_2^\times = \{\pm 1\} \times (1 + 4\mathbf Z_2)$, with $\{\pm 1\} \cong \mathbf Z/2\mathbf Z$ and $1+4\mathbf Z_2 \cong 4\mathbf Z_2$ by the $2$-adic logarithm, so $\mathbf Z_2^\times \cong \mathbf Z/2\mathbf Z \times \mathbf Z_2$.
KCd
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