# Prüfer Group as a p-adic quotient

I've been trying to get a grasp of the structure of the p-adics to begin understanding p-adic representations. I read that the Prüfer Group $$\mathbb{Z}(p^{\infty})=\underset{\underset{n}{\longrightarrow}}{\lim}\mathbb{Z}/p^n\mathbb{Z}$$ can be understood as the quotient $$\mathbb{Q}_p/\mathbb{Z}_p$$.

Could someone explain to me how this is?

My first thought is that I can understand the quotient by analogy to how $$\mathbb{Q}/\mathbb{Z}$$ gives points that are "rational divisions of the circle." When considering the $$p$$-adic quotient above as a quotient of power series base-$$p$$, we seem to get what are effectively "fractions base-$$p$$." It seems like the coefficients $$k\in\{0,1,\ldots,p-1\}$$ of the negative powers $$p^{-n}$$ of $$p$$ correspond to $$k^{th}$$ powers of $$p^{n}$$'th roots of unity. My understanding here however is quite foggy.

• What definition of the Prüfer group are you using? May 8, 2022 at 20:47
• The Prüfer group is $\bigcup_{n\ge 1} p^{-n}\Bbb{Z/Z}$ while $\Bbb{Q}_p/\Bbb{Z}_p=\bigcup_{n\ge 1} p^{-n}\Bbb{Z}_p/\Bbb{Z}_p$ May 8, 2022 at 20:57
• Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. May 8, 2022 at 21:26
• I think you mean $\varinjlim_n \mathbb{Z}/p^n\mathbb{Z}$, not $\varprojlim_n \mathbb{Z}/p^n\mathbb{Z}$ (the inverse limit is $\mathbb{Z}_p$, while the Prüfer group is the direct limit). May 8, 2022 at 22:46
• @Shaun Done, thanks. May 9, 2022 at 0:21