I am rather new to $p$-adics, and I am trying to show that every abelian pro-$p$ group is a $\mathbb{Z}_p$ module. I know some partial answers already exist (About pro-$p$ groups), but I am having trouble understanding how the module operation is defined and why it works.
My understanding of the problem is:
- We are given an abelian pro-$p$ group $G = \lim\limits_{\longleftarrow} G_i$, defined additively. An arbitrary element of $G$ therefore looks like $(\dots, g_2, g_1, g_0)$ where each $g_i$ has order a power of $p$ in $G_i$.
- An arbitrary element $x$ of $\mathbb{Z}_p$ can be expressed as $x_0 + x_1p+ x_2p^2 + \dots$, and identified with $(\dots, x_2, x_1, x_0)$, with each $x_i \in \mathbb{Z}/p\mathbb{Z}$
I would like to define $x \cdot g$ so that it satisfies the module axioms. The link above says from what I understand that $x \cdot g := (\dots, (x_0 + x_1p + x_2p^2) \cdot g_2, (x_0 + x_1p) \cdot g_1, x_0 \cdot g_0)$ where $n \cdot g_i$ is understood as $g_i + \dots + g_i$, added $n$ times. However, although the other axioms are fine, I have no idea why $(xy) \cdot g = x \cdot (y \cdot g)$.
Do I have the correct operation? Am I missing something obvious?
Let me offer a more conceptual but perhaps sloppier strategy for handling this.
Let $g$ be an element of your abelian pro-$p$-group $G$. What you know is that $\{p^ng\}$ is a sequence with limit $0_G$. Well, let $z\in\Bbb Z_p$; you are asking how to define $zg$. First, express $z$ as a $p$-adic limit of integers, $\{m_i\}$, so that $\{m_i-m_{i-1}\}\to0$, $p$-adically. Then it should be easy enough to show that $\{m_ig\}$ is a Cauchy sequence in $G$. The limit of this sequence should be your $zg$.