About pro-$p$ groups Let $p$ be a prime number.  A pro-$p$ group is the inverse limit of an inverse system of discrete finite $p$-groups.
(I've just read the definition).
I have two questions:


*

*Is it true that a the $p$-part of an abelian profinite group is a pro-$p$ group? (I think so)

*How is an abelian pro-$p$ group a $\mathbb{Z}_p$-module?

 A: EDITED: (I have inserted here at the top an answer to your first question.)
There is a $p$-part of an abelian profinite group $G$, but it’s not what you wanted to define it as. You must define it as
$$
G_p = \left\lbrace g\in G:\lim_{n\to\infty}g^{p^n}=e_G\right\rbrace\,.
$$
You easily check that this $G_p$ is a subgroup of your abelian profinite group. Verification that $G_p$ is closed in $G$, thus compact, thus a profinite group, requires a few more words.
I’ll use additive notation in $G$, and use the letter $U$ for the open subgroups that define the topology of $G$. What is the condition that $g\in G_p$?
\begin{align}
g\in G_p&\Longleftrightarrow\forall U,\exists n_0\text{ such that }\forall n\ge n_0, p^ng\in U\\
&\Longleftrightarrow\forall U,\exists n\text{ such that }p^ng\in U\\
&\Longleftrightarrow\forall U,\exists n\text{ such that }g\in p^{-n}U\\
&\Longleftrightarrow g\in\bigcap_U\bigcup_np^{-n}U\,
\end{align}
Now, for a given U, this group $\bigcup_np^{-n}U$ is the total union of an ascending chain of opens, call it $U'$. It’s open, so closed, and we’re taking an intersection of closed subgroups $U'$, and that’s closed. Therefore compact, so a profinite group, and clearly a pro-$p$-group.
ORIGINAL POST FOLLOWS:
Here’s how an abelian pro-$p$-group $G$ is a $\Bbb Z_p$-module:
Let $z\in\Bbb Z_p$, and exhibit $z$ as a $p$-adically convergent sequence of positive integers, $z=\lim_in_i$. (If you like, you can take the $n_i$’s to be the partial sums in the standard representation of $z$ as a “power series in $p$”.) Now, for $g\in G$, define $g^z=\lim_ig^{n_i}$. You need to prove a few things, but they’re easy enough.
Note that the definition of this operation had nothing to do with abelianness of $G$: it’s always defined on a pro-$p$-group. It’s just that you don’t get a module structure without the abelian condition.
