1

Let $\mathbb{Z}_p$ be the ring of $p$-adic integers

I understood that all closed subgroups of $\mathbb{Z}_p$ are $p^n\mathbb{Z}_p$, and multiplicative group of units of $\mathbb{Z}_p$ is $(\mathbb{Z}/(p-1)\mathbb{Z})\times\mathbb{Z}_p$

Then, why the closed subgroup of this unit group is of form $A\times p^n\mathbb{Z}_p$ where $A$ is subgroup of $\mathbb{Z}/(p-1)\mathbb{Z}$?

I don’t have any idea how to find(classify) all closed subgroup of multiplicative group of units of $\mathbb{Z}_p $ Why they must be of form $A\times p^n\mathbb{Z}_p$? Aren’t there any other possibilities?

Min
  • 11

1 Answers1

1

This does not answer why the subgroups are direct products of subgroups but only which of those (latter) can occur.To see why all subgroups must be direct products of subgroups see the comment by @Torsten Schoeneberg below and in the linked post Tobias Kildetoft´s answer. It follows all subgroups of $\mathbb{Z}/(p-1)\mathbb{Z}\times\mathbb{Z}_p$ are of the form $$A\times B$$ where $A$ is a subgroup of $\mathbb{Z}/(p-1)\mathbb{Z}$ and $B$ is a subgroup of $\mathbb{Z}_p$. Since $\mathbb{Z}/(p-1)\mathbb{Z}$ as a topological space is compact, the projection map $$\pi:\mathbb{Z}/(p-1)\mathbb{Z}\times\mathbb{Z}_p\rightarrow\mathbb{Z}_p$$ is closed and thus $B$ must be closed in $\mathbb{Z}_p$ whenever $A\times B$ is closed in $\mathbb{Z}/(p-1)\mathbb{Z}\times\mathbb{Z}_p$ and thus $B=0$ or $B=p^n\mathbb{Z}_p$ for some $n\in\mathbb{N}_0$.

Peter Melech
  • 4,353
  • My question is, why they are of form A product B? – Min Nov 15 '23 at 07:57
  • 1
    All points (x,y) where x=y form a subgroup of Euclidean plane(which is product of two real lines), but this set is definitely not a product of two subsets of real line. Subgroup of direct product need not be a direct product – Min Nov 15 '23 at 12:03
  • You are right, I see Your point. I will think about it and write later. I also edited my answer, which is not an answer to Your crucial question. – Peter Melech Nov 15 '23 at 12:33
  • 1
    To amend the missing step: Subgroups of direct products are described in the answer to https://math.stackexchange.com/q/485512/96384. In this case, it is obvious that the only chance to get an iso from a subquotient of $\mathbb Z/(p-1)$ to a subquotient of $\mathbb Z_p$ (all of whose finite subquotients have order a power of $p$) is when both subquotients are trivial, i.e. in the notation of that answer, $G_1=G_2$ and $H_1=H_2$. It follows that here, subgroups of the direct product are direct products of subgroups. – Torsten Schoeneberg Nov 15 '23 at 20:13
  • @TorstenSchoeneberg Thank You very much! The bijection between subgroups and 5-tuples yielding isomorphisms between subquotients is something I would have never come up with, but indeed it is obvious here that the subquotients must be trivial. – Peter Melech Nov 16 '23 at 12:08