Pro P Groups and the P Adic Integers - Examples

For each $n\ge 1$, reducing $x_{n+1}$ modulo $p^n$ removes only the final term $a_n{p}^n$, so $x_{n+1}<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>p^n=x_n$. Thus the tuple $(x_n{)}_{n\ge 1}$ is compatible.

By [L1], every compatible tuple defines an element of ${\mathbb{Z}}_p$. Hence the displayed digit data encodes a $p$-adic integer.

For each fixed $r\ge 1$, once $n\ge r$ the residue of $s_n$ modulo $p^r$ is already $a_0+a_{1}p+\cdots +a_{r-1}{p}^{r-1}$, which is exactly the $r$-th coordinate of $x$. So the first $r$ coordinates of $s_n$ and $x$ agree for all sufficiently large $n$.

Step 1.1 says that for every $r$, the sequence eventually lies in the basic neighbourhood of $x$ determined by the first $r$ coordinates. By [L1], that is exactly convergence to $x$ in the inverse-limit topology.

If $m>n$, then $t_m-t_n=p^n(1+p+\cdots +p^{m-n-1})$, so the first $n$ residue coordinates of $t_m$ and $t_n$ agree. By [F1], this gives $d_p(t_m,t_n)\le p^{-n}$. Hence $(t_n)$ is Cauchy.

Since ${\mathbb{Z}}_p$ is complete by [L1], the Cauchy sequence $(t_n)$ converges to some element of ${\mathbb{Z}}_p$.

The subgroup $1⊴P$ is one of the indexing subgroups in [F1], and its quotient is the initial object $P\mathrm{/}1\cong P$ of the quotient diagram: it has the natural quotient map to every $P\mathrm{/}N$.

A compatible tuple is uniquely determined by its coordinate in the initial object $P\mathrm{/}1$, and every element of $P\mathrm{/}1$ determines such a tuple by its images in the other quotients. Therefore ${\widehat{P}}^{(p)}\cong P$, and the canonical completion map is exactly this isomorphism. The conclusion is consistent with [L1], since $P$ was already pro-$p$ to begin with.

The additive group ${\mathbb{Z}}_p$ is abelian, so $[{\mathbb{Z}}_p,{\mathbb{Z}}_p]=0$. Its $p$th-power subgroup in additive notation is $p{\mathbb{Z}}_p$, which is already closed. Therefore [L1] gives $\mathrm{\Phi }({\mathbb{Z}}_p)=p{\mathbb{Z}}_p$.

Reduction modulo $p$ is a surjective homomorphism ${\mathbb{Z}}_p\to \mathbb{Z}\mathrm{/}p\mathbb{Z}$ with kernel $p{\mathbb{Z}}_p$. By step 1.1, the quotient by $\mathrm{\Phi }({\mathbb{Z}}_p)$ is therefore ${\mathbb{F}}_p$.

By coordinatewise addition in [F1], the subgroup $pG$ consists of pairs whose two coordinates are divisible by $p$. Therefore $G\mathrm{/}pG\cong ({\mathbb{Z}}_p\mathrm{/}p{\mathbb{Z}}_p)\times ({\mathbb{Z}}_p\mathrm{/}p{\mathbb{Z}}_p)\cong {\mathbb{F}}_p^2$.

The Frattini quotient of the abelian pro-$p$ group $G$ is exactly $G\mathrm{/}pG$, so [L1] says that a subset topologically generates $G$ exactly when its image spans ${\mathbb{F}}_p^2$. For a pair $(u,v)$, spanning ${\mathbb{F}}_p^2$ is the same as being a basis.

By [L1], the embedded copy of $\mathbb{Z}$ is dense in ${\mathbb{Z}}_p$.

The embedded copy of $\mathbb{Z}$ is exactly $\mathbb{Z}\cdot 1$. By [L2], its closure is all of ${\mathbb{Z}}_p$ but it is not equal to ${\mathbb{Z}}_p$ as an abstract subgroup. Thus a dense subgroup can be proper.

Therefore the stated universal claim is false, and the embedded copy of $\mathbb{Z}$ is the required counterexample.